Confidence interval for a difference in means

Learning goals

Identifies a difference in population means as the quantity to estimate.
Checks conditions for a two-sample mean confidence interval.
Calculates the standard error of a difference in sample means.
Constructs a confidence interval for a difference in population means.
Interprets a confidence interval for a difference in population means.
Explains confidence level through repeated sampling.

Difficulty: 1 – 5

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Confidence interval for a difference in means 1

A state agency plans to estimate the difference in mean processing time between online and mailed permit applications submitted during 2025. It will use independent random samples from applications submitted by each method. Write an investigative question appropriate for a confidence interval rather than a hypothesis test.

Hints

- Identify the two population means being compared. - Use wording that asks for a range of plausible parameter values. - Avoid stating a greater-than, less-than, or not-equal claim.

Solution

1. The parameter is the difference between the two population mean processing times for applications submitted during 2025. 2. A confidence-interval question asks for a range of plausible values rather than asserting a direction or testing a claim. 3. The population consists of all online and mailed permit applications submitted to the agency during 2025.

Answer

One valid question is: “What is a plausible interval for the difference in mean processing time between all online and all mailed permit applications submitted to the agency during 2025?”

Confidence interval for a difference in means 2

Independent random samples have sample means \(\bar{x}_A=72.5\) and \(\bar{x}_B=68.0\). A confidence interval will be constructed for \(\mu_A-\mu_B\). Before calculating any standard error, what point estimate will be at the center of the interval? What value of \(\mu_A-\mu_B\) represents no population mean difference?

Hints

- Keep the population subtraction order and sample subtraction order the same. - A confidence interval for a difference is centered at the observed sample difference. - Think about what subtraction gives when two population means are equal.

Solution

1. The point estimate for \(\mu_A-\mu_B\) is the corresponding sample-mean difference, \(\bar{x}_A-\bar{x}_B\). 2. Compute \(72.5-68.0=4.5\). 3. No population mean difference means \(\mu_A=\mu_B\), so \(\mu_A-\mu_B=0\).

Answer

The confidence interval is centered at the point estimate \(4.5\). The no-difference parameter value is \(0\).

Confidence interval for a difference in means 3

A standard two-sample \(95\%\) confidence interval for \(\mu_A-\mu_B\) has point estimate \(3.2\) and margin of error \(1.1\). Write the interval and state whether \(0\) is included.

Hints

- Use the point estimate as the interval's center. - Move the margin of error the same distance to the left and right. - Check whether the no-difference value lies between the endpoints.

Solution

1. A standard confidence interval has the form point estimate plus or minus the margin of error. 2. The lower endpoint is \(3.2-1.1=2.1\), and the upper endpoint is \(3.2+1.1=4.3\). 3. Thus, the interval is \((2.1,4.3)\). 4. Since both endpoints are positive, \(0\) is not included.

Answer

The interval is \((2.1,4.3)\). It does not include \(0\).

Confidence interval for a difference in means 4

A recreation center wants a confidence interval for the population mean difference in swim-lesson times between beginners and experienced swimmers. The calculation will use \(\bar{x}_B-\bar{x}_E\). Which population parameter does the interval estimate? What does a negative difference mean?

Hints

- Replace each sample mean in the statistic with its population mean. - Use the subtraction order to interpret the sign.

Solution

1. The estimate \(\bar{x}_B-\bar{x}_E\) targets \(\mu_B-\mu_E\), the difference between the two population mean lesson times. 2. A negative value means beginners' mean lesson time is lower than experienced swimmers' mean lesson time.

Answer

The parameter is \(\mu_B-\mu_E\). A negative difference means the population mean lesson time for beginners is shorter.

Confidence interval for a difference in means 5

A survey compares mean satisfaction scores for customers who shop online and in stores. The independent sample means have estimated standard errors of \(1.2\) and \(1.6\) points. Olivia says the standard error of their difference is \(1.6-1.2=0.4\) points because the means are subtracted. Explain Olivia's error and find the correct estimated standard error of the difference.

Hints

- Imagine the two sample means changing independently across repeated samples. - Could subtracting them remove all the uncertainty in either one? - Keep track of the difference between a spread and a squared spread.

Solution

1. Subtracting two independent means does not mean subtracting their standard errors; independent variances add. 2. Square the two standard errors and combine them: \(SE_{\text{diff}}=\sqrt{1.2^2+1.6^2}\). 3. Therefore, \(SE_{\text{diff}}=\sqrt{1.44+2.56}=2\) points.

Answer

Olivia subtracted standard errors instead of adding variances. The correct standard error is \(SE_{\text{diff}}=2\) points.

Confidence interval for a difference in means 6

The estimated difference in mean backpack weights between an algebra class (A) and a biology class (B) is \(-4.5\,\text{lb}\), with margin of error \(2.3\,\text{lb}\). The two random samples contain different students. Write the confidence interval for \(\mu_A-\mu_B\). Does it suggest that algebra students carry lighter backpacks on average?

Hints

- Subtract and add the same margin to the estimated difference. - Think about what a completely negative interval says about the order of the two population means.

Solution

1. The interval is the estimate plus or minus the margin of error. 2. Its endpoints are \(-4.5-2.3=-6.8\) and \(-4.5+2.3=-2.2\), in pounds. 3. Both endpoints are negative, supporting a smaller population mean weight for the algebra class.

Answer

\((-6.8,-2.2)\,\text{lb}\). Yes; the interval favors a lower population mean for the algebra class.

Confidence interval for a difference in means 7

Each number line represents a \(95\%\) confidence interval for \(\mu_A-\mu_B\) in a different study. Which studies have intervals that rule out a population mean difference of zero? In those studies, which group has the higher estimated mean?

Hints

- Find zero on each number line and check whether the interval reaches or crosses it. - The sign of \(\mu_A-\mu_B\) tells which group's mean is larger.

Solution

1. Panel a) lies entirely below zero, so it excludes zero and favors a lower mean for A. 2. Panel b) crosses zero, so zero is a possible population mean difference under that interval. 3. Panel c) lies entirely above zero, so it excludes zero and favors a higher mean for A.

Answer

Panels a) and c) rule out zero. Panel a) favors group B; panel c) favors group A.

Confidence interval for a difference in means 8

A community college compares the time students need to complete the same computer-based registration task using two interface designs. Independent random samples are used, and a \(95\%\) confidence interval for \(\mu_A-\mu_B\) is \((-1.2,3.4)\,\text{s}\). Interpret the interval in context. Does the interval provide convincing evidence that the two population mean completion times differ? Explain.

Hints

- Keep the meaning of the subtraction \(\mu_A-\mu_B\) visible when reading both endpoints. - Ask which value would represent no population mean difference at all. - Decide whether that no-difference value is among the plausible values given by the interval.

Solution

1. The interval estimates the difference \(\mu_A-\mu_B\) in population mean completion times. 2. We are \(95\%\) confident that \(\mu_A-\mu_B\) is between \(-1.2\,\text{s}\) and \(3.4\,\text{s}\). 3. Because \(0\) is inside the interval, a population mean difference of \(0\) is plausible at this confidence level. 4. Therefore, the interval does not provide convincing evidence that the two population mean completion times differ.

Answer

We are \(95\%\) confident that Interface A's population mean completion time is from \(1.2\,\text{s}\) lower to \(3.4\,\text{s}\) higher than Interface B's. Because the interval contains \(0\), it does not provide convincing evidence of a difference in the population means.

Confidence interval for a difference in means 9

Two independent laboratories measure the same type of reference material using different procedures. For a \(98\%\) confidence interval for the population mean difference \(\mu_A-\mu_B\), software reports an estimated difference of \(2.7\,\text{units}\), standard error \(0.85\,\text{unit}\), and \(t^*=2.423\). Construct the confidence interval with endpoints to two decimal places and interpret it in terms of the two laboratory procedures.

Hints

- The software output already supplies the estimate, its uncertainty, and the multiplier needed for the interval. - Keep the interval centered at the reported difference. - Use the sign of both endpoints to describe which population mean is larger.

Solution

1. The margin of error is \(2.423(0.85)\approx2.060\,\text{units}\). 2. The interval is \(2.7\pm2.060\), or approximately \((0.64,4.76)\,\text{units}\). 3. The entire interval is positive, so Procedure A's population mean measurement is estimated to be higher than Procedure B's. 4. We are \(98\%\) confident that \(\mu_A-\mu_B\) lies between about \(0.64\) and \(4.76\) units.

Answer

The \(98\%\) confidence interval for \(\mu_A-\mu_B\) is approximately \((0.64,4.76)\,\text{units}\). We are \(98\%\) confident that Procedure A's population mean measurement is about \(0.64\) to \(4.76\) units higher than Procedure B's.

Confidence interval for a difference in means 10

A \(95\%\) confidence interval for \(\mu_U-\mu_R\), the difference in population mean commute time between urban and rural employees in a company, is \((-6.2,-1.5)\,\text{min}\). Rewrite the interval for \(\mu_R-\mu_U\), and interpret the rewritten interval in context.

Hints

- Reversing a difference does not change its magnitude, but it changes its sign. - After changing signs, put the smaller endpoint first. - Make the verbal interpretation match the new subtraction order rather than the original one.

Solution

1. Reversing the subtraction changes the sign of every possible difference. 2. Negating and reordering the endpoints gives the interval \((1.5,6.2)\,\text{min}\) for \(\mu_R-\mu_U\). 3. We are \(95\%\) confident that rural employees' population mean commute time is between \(1.5\) and \(6.2\) minutes longer than urban employees' population mean commute time.

Answer

The interval for \(\mu_R-\mu_U\) is \((1.5,6.2)\,\text{min}\). We are \(95\%\) confident that the rural population mean commute time is about \(1.5\) to \(6.2\) minutes longer than the urban population mean commute time.

Confidence interval for a difference in means 11

A study comparing two independent population means produces a \(95\%\) confidence interval of \((1,9)\) using equal sample sizes in the two groups. The researchers plan a similar study with both sample sizes multiplied by \(4\), while the population variability, confidence level, and estimated difference remain about the same. Approximately how wide should the new confidence interval be, and what interval would that suggest if it remains centered at the same estimate?

Hints

- Start by separating interval width from margin of error. - Increasing both sample sizes changes the sampling uncertainty through a square-root relationship. - Keep the center fixed only because the problem explicitly says the estimated difference remains about the same.

Solution

1. The original interval has width \(9-1=8\) and midpoint \(5\). 2. Confidence-interval width is approximately proportional to \(1/\sqrt{n}\) when other factors remain similar. 3. Multiplying both sample sizes by \(4\) therefore reduces the width by a factor of about \(1/2\). 4. The new width should be about \(4\), giving a margin of error of about \(2\). 5. If the estimate remains \(5\), the approximate new interval is \((3,7)\).

Answer

The new interval should be about \(4\) units wide, approximately half the original width. If it remains centered at \(5\), an approximate interval is \((3,7)\).

Confidence interval for a difference in means 12

Two independent simple random samples are used to estimate a difference in population means. The sample sizes are \(35\) and \(40\), both populations are much larger than the samples, and neither population is known to be normal. Use the large-sample guideline that each sample size of at least \(30\) supports two-sample \(t\) inference unless there is evidence of extreme skewness or influential outliers. Jordan Kim says a two-sample \(t\)-interval cannot be used because normal population distributions were not stated. Evaluate Jordan's claim.

Hints

- Apply the large-sample guideline stated in the problem to each sample separately. - Distinguish the absence of a normal-population statement from positive evidence of extreme skewness or influential outliers. - Keep the shape requirement separate from the randomization and population-size conditions.

Solution

1. The samples are independent simple random samples, and the large populations support the \(10\%\) condition. 2. The stated large-sample guideline supports two-sample \(t\) inference when both sample sizes are at least \(30\), unless there is evidence of extreme skewness or influential outliers. 3. Here, \(n_1=35\) and \(n_2=40\), and no such problematic shape information is given. 4. Therefore, lack of stated population normality does not by itself prevent use of the two-sample \(t\)-interval.

Answer

Jordan's claim is incorrect. The independent random samples have sizes \(35\) and \(40\), so the stated large-sample guideline supports a two-sample \(t\)-interval without requiring normal population distributions. The other interval conditions must also hold.

Confidence interval for a difference in means 13

Mina Park is constructing a two-sample \(t\)-confidence interval for \(\mu_1-\mu_2\). Mina calculates the standard error as \(\frac{s_1}{\sqrt{n_1}}-\frac{s_2}{\sqrt{n_2}}\) because the parameter is a difference. Explain the error and state how the estimated sampling variability should be combined for independent samples.

Hints

- The subtraction in the parameter controls the center, but variability follows a different rule. - Think about how independent variances combine when random quantities are added or subtracted. - A standard error must be nonnegative and should reflect uncertainty from both samples.

Solution

1. The parameter is a difference of means, but independent sources of variance add rather than subtract. 2. The estimated variance of \(\bar{x}_1-\bar{x}_2\) is \(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\). 3. Therefore, the standard error is \(\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\). 4. Subtracting the two separate standard errors could even create an unrealistically small or negative quantity and does not represent the variability of the difference.

Answer

Mina should add the independent variance contributions, not subtract standard errors. Use \(\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\) for the standard error of \(\bar{x}_1-\bar{x}_2\).

Confidence interval for a difference in means 14

Two independent studies estimate the same difference in population means with the same sample sizes and confidence level. Study A has sample standard deviations \(3.1\) and \(3.4\). Study B has sample standard deviations \(7.8\) and \(8.2\). Which study should have the narrower two-sample \(t\)-confidence interval, assuming the other conditions are comparable? Explain using the standard error.

Hints

- Hold the sample sizes and confidence level fixed and identify what differs between the studies. - More within-group variability creates more uncertainty in the estimated difference. - Interval width responds directly to the estimated sampling uncertainty.

Solution

1. With the same sample sizes, each estimated variance contribution is larger when the sample standard deviation is larger. 2. Study A therefore has the smaller standard error because both of its sample standard deviations are much smaller. 3. At the same confidence level, a smaller standard error generally produces a smaller margin of error. 4. Thus, Study A should have the narrower confidence interval.

Answer

Study A should have the narrower interval because its smaller sample standard deviations produce a smaller standard error and therefore a smaller margin of error.

Confidence interval for a difference in means 15

A college wants to estimate the difference in population mean weekly study time between students who commute and students who live on campus. It takes one random sample of \(28\) commuters and a separate random sample of \(28\) residential students. Lucas Meyer sorts each sample from smallest to largest, pairs the observations by rank, and proposes a paired \(t\)-confidence interval based on the \(28\) resulting differences. Explain why this procedure is inappropriate and identify the appropriate confidence-interval procedure.

Hints

- Ask whether each observation in one group is meaningfully linked to exactly one observation in the other group. - A numerical ordering does not create a study-design relationship between observations. - Identify whether the parameter describes one population of differences or two separate population means.

Solution

1. The observations are from two separate random samples, so there is no natural one-to-one link between a commuter and a residential student. 2. Sorting the samples and pairing by rank creates artificial pairs and does not turn independent observations into matched data. 3. The parameter is \(\mu_C-\mu_R\), the difference between two population means. 4. The appropriate procedure is a two-sample \(t\)-confidence interval for \(\mu_C-\mu_R\), provided its conditions are satisfied.

Answer

A paired \(t\)-interval is inappropriate because the two samples are independent and have no genuine pairing. Use a two-sample \(t\)-confidence interval for \(\mu_C-\mu_R\), assuming the required conditions hold.

Confidence interval for a difference in means 16

A valid \(95\%\) confidence interval for \(\mu_A-\mu_B\), the difference in population mean tensile strength for two materials, is \((2.4,8.0)\,\text{MPa}\). Use the interval to make decisions at \(\alpha=0.05\) for these two-sided tests: a) \(H_0:\mu_A-\mu_B=0\) versus \(H_a:\mu_A-\mu_B\ne0\) b) \(H_0:\mu_A-\mu_B=6\,\text{MPa}\) versus \(H_a:\mu_A-\mu_B\ne6\,\text{MPa}\)

Hints

- A confidence interval lists parameter values that remain plausible at the matching two-sided significance level. - Check each proposed null difference against the same interval. - A null value inside the interval and a null value outside the interval lead to different test decisions.

Solution

1. A two-sided test at \(\alpha=0.05\) rejects a null value when that value is outside the corresponding \(95\%\) confidence interval. 2. For part a, \(0\) is outside \((2.4,8.0)\,\text{MPa}\), so reject \(H_0\). 3. For part b, \(6\,\text{MPa}\) is inside \((2.4,8.0)\,\text{MPa}\), so fail to reject \(H_0\).

Answer

a) Reject \(H_0\). The interval provides convincing evidence that \(\mu_A-\mu_B\ne0\). b) Fail to reject \(H_0\). A difference of \(6\,\text{MPa}\) is compatible with the \(95\%\) confidence interval.

Confidence interval for a difference in means 17

A valid \(95\%\) confidence interval for \(\mu_A-\mu_B\) is \((-1.2,2.4)\). A later calibration review shows that every observation in Group A should have been \(5\) units larger, while Group B values were correct. What is the corrected \(95\%\) confidence interval for \(\mu_A-\mu_B\)? Explain why its width does not change.

Hints

- Separate the effect of the calibration change on the center of Group A from its effect on spread. - The same margin of error applies if the sample standard deviations and sample sizes do not change. - A constant shift in one group shifts the entire interval for the difference by that constant.

Solution

1. Adding \(5\) to every Group A observation adds \(5\) to the sample mean for Group A but does not change its sample standard deviation. 2. The estimated difference \(\bar{x}_A-\bar{x}_B\) therefore increases by \(5\), while the standard error and margin of error stay the same. 3. Add \(5\) to both endpoints of the original interval: \((-1.2+5,2.4+5)=(3.8,7.4)\).

Answer

The corrected \(95\%\) confidence interval is \((3.8,7.4)\). Its width is unchanged because adding a constant to all Group A observations shifts the estimated difference but does not change sampling variability.

Confidence interval for a difference in means 18

Dr. Samuel Ortega takes independent simple random samples of \(30\) observations from Population A and \(30\) observations from Population B. Each population contains \(500\) individuals. Keira Brown says the \(10\%\) condition fails because the combined sample size is \(60\), which is more than \(10\%\) of \(500\). Evaluate Keira's reasoning for a two-sample \(t\)-confidence interval.

Hints

- Identify which population each sample was drawn from. - The independence check concerns sampling without replacement within each population. - Do not combine sample sizes that came from different populations when checking this condition.

Solution

1. The \(10\%\) condition is checked separately within each population because each sample is drawn from its own population. 2. For Population A, \(30/500=0.06\), so the condition is satisfied. 3. For Population B, \(30/500=0.06\), so the condition is also satisfied. 4. The combined sample size of \(60\) is not compared with the size of either population for this condition.

Answer

Keira's reasoning is incorrect. Each sample is \(6\%\) of its own population, so both samples satisfy the \(10\%\) condition.

Confidence interval for a difference in means 19

A valid two-sample \(t\)-confidence interval for \(\mu_A-\mu_B\) is \((2.0,6.0)\). The sample mean for Group A is \(14.0\). What sample-mean difference was used as the point estimate, and what was the sample mean for Group B?

Hints

- The point estimate lies at the center of a standard two-sample confidence interval. - Keep the subtraction order consistent with the parameter \(\mu_A-\mu_B\). - Once the sample-mean difference is known, use the given Group A mean to recover Group B's mean.

Solution

1. A standard two-sample confidence interval is centered at \(\bar{x}_A-\bar{x}_B\). 2. The midpoint of the interval is \(\frac{2.0+6.0}{2}=4.0\), so \(\bar{x}_A-\bar{x}_B=4.0\). 3. Since \(\bar{x}_A=14.0\), solve \(14.0-\bar{x}_B=4.0\). 4. Therefore, \(\bar{x}_B=10.0\).

Answer

The point estimate is \(4.0\), and the sample mean for Group B is \(10.0\).

Confidence interval for a difference in means 20

A \(95\%\) confidence interval for \(\mu_A-\mu_B\), where the response is measured in kilograms, is \((2.4,6.8)\,\text{kg}\). Diego Morales reports, “Group A's population mean is between \(2.4\%\) and \(6.8\%\) higher than Group B's.” Explain the error and give a correct interpretation.

Hints

- Look at the parameter named in the interval and identify its units. - Subtraction produces an absolute difference, while percentages require a ratio to a baseline. - Keep the interpretation on the same measurement scale as the parameter.

Solution

1. The parameter \(\mu_A-\mu_B\) is an additive difference, so its units are kilograms. 2. The interval endpoints therefore describe plausible differences in population means measured in kilograms, not percentages. 3. A percent difference would require a ratio or a clearly defined baseline denominator, which this interval does not provide. 4. The correct interpretation is that we are \(95\%\) confident that Group A's population mean is between \(2.4\) and \(6.8\) kilograms greater than Group B's population mean.

Answer

The interval is about an absolute difference, not a percent difference. We are \(95\%\) confident that \(\mu_A\) exceeds \(\mu_B\) by between \(2.4\,\text{kg}\) and \(6.8\,\text{kg}\).

Confidence interval for a difference in means 21

Two independent random samples compare the mean time needed to complete the same task with Interface A and Interface B. The sample means are \(72.4\,\text{min}\) for A and \(68.9\,\text{min}\) for B. The estimated standard errors of the two sample means are \(1.2\,\text{min}\) and \(0.9\,\text{min}\), respectively. Using \(t^*=2.00\), construct a confidence interval for \(\mu_A-\mu_B\). Do not add the two standard errors directly.

Hints

- Begin with the difference between the two sample means in the stated order. - Independent estimates contribute separate sources of variability to the difference. - Use the supplied critical value after finding the uncertainty of the difference.

Solution

1. The estimated difference is \(72.4-68.9=3.5\,\text{min}\). 2. For independent sample means, the standard error of the difference is \(\sqrt{1.2^2+0.9^2}=1.5\,\text{min}\). 3. The margin of error is \(2.00\cdot1.5=3.0\,\text{min}\). 4. The confidence interval is \(3.5\pm3.0\,\text{min}\), or \((0.5,6.5)\,\text{min}\).

Answer

The confidence interval for \(\mu_A-\mu_B\) is \((0.5,6.5)\,\text{min}\).

Confidence interval for a difference in means 22

A \(95\%\) confidence interval for the difference in mean battery life between Brand A and Brand B, \(\mu_A-\mu_B\), is \((1.8,4.6)\) hours. ConsumerLab reviewer Maya Chen writes, “About \(95\%\) of individual Brand A batteries last between \(1.8\) and \(4.6\) hours longer than individual Brand B batteries.” Evaluate this statement and give an appropriate interpretation of the interval.

Hints

- Identify the parameter named by the interval before interpreting its endpoints. - A statement about population means is different from a statement about most individual observations. - Keep the subtraction order \(A-B\) visible in the verbal interpretation.

Solution

1. The interval estimates a difference between two population means, not the distribution of pairwise differences between individual batteries. 2. Therefore, Maya's statement about \(95\%\) of individual batteries is not supported by this interval. 3. An appropriate interpretation is that the data support plausible values from \(1.8\) to \(4.6\) hours for \(\mu_A-\mu_B\) at the stated confidence level.

Answer

Maya's statement is incorrect. The interval concerns the population mean difference: we are \(95\%\) confident that Brand A's mean battery life is between \(1.8\) and \(4.6\) hours greater than Brand B's mean battery life.

Confidence interval for a difference in means 23

A study compares mean test scores for two independent groups. Before analysis, the researchers subtract the same baseline value of \(50\) points from every score in both groups. Explain how this transformation affects the point estimate, standard error, and confidence interval for \(\mu_A-\mu_B\).

Hints

- Track what happens to each group mean when the same constant is subtracted. - Consider whether shifting every value changes within-group spread. - The interval is built from the estimated difference and its uncertainty.

Solution

1. Subtracting \(50\) from every score changes each group mean by \(-50\). 2. The difference of the sample means is unchanged because \((\bar{x}_A-50)-(\bar{x}_B-50)=\bar{x}_A-\bar{x}_B\). 3. Subtracting a constant does not change either group’s standard deviation, so the standard error of the difference is unchanged. 4. Because both the point estimate and its standard error are unchanged, the confidence interval for \(\mu_A-\mu_B\) is unchanged.

Answer

The point estimate, standard error, and confidence interval for \(\mu_A-\mu_B\) all remain unchanged.

Confidence interval for a difference in means 24

A \(95\%\) confidence interval for \(\mu_A-\mu_B\) is \((-7.0,-2.0)\). Ethan Miller rewrites it as \((2.0,7.0)\) because “a difference should be positive.” Explain why taking absolute values changes the meaning of the interval, and interpret the original interval in terms of \(\mu_B-\mu_A\).

Hints

- Keep track of which population is subtracted from which. - Negative endpoints carry directional information. - Reversing a difference changes the sign of every possible parameter value.

Solution

1. The sign of the interval records the subtraction order \(A-B\); it is not an error to have negative endpoints. 2. The interval \((-7.0,-2.0)\) says that \(\mu_A\) is estimated to be between \(2.0\) and \(7.0\) units lower than \(\mu_B\). 3. Reversing the parameter to \(\mu_B-\mu_A\) negates both endpoints and reverses their order, giving \((2.0,7.0)\). 4. Simply taking absolute values without changing the named parameter would misstate the direction of the comparison.

Answer

Keep \((-7.0,-2.0)\) for \(\mu_A-\mu_B\). Equivalently, the confidence interval for \(\mu_B-\mu_A\) is \((2.0,7.0)\).

Confidence interval for a difference in means 25

Independent random samples are used to estimate \(\mu_A-\mu_B\). Group A has \(n_A=10\), and Group B has \(n_B=45\). The boxplots summarize the two samples. Based on the sample sizes and displayed shapes, is a standard two-sample \(t\)-confidence interval well justified? Explain which feature matters most.

Hints

- Evaluate the two groups separately rather than using only the combined sample size. - Small samples require greater caution about skewness and outliers. - Look for any feature that could strongly influence a sample mean and standard deviation.

Solution

1. Group B is large enough that moderate nonnormality would not usually be a concern for inference about its mean. 2. Group A is small, so its shape matters strongly for a \(t\)-procedure. 3. The Group A boxplot shows a clear high outlier far beyond the rest of the data. 4. With only \(10\) observations in Group A, that outlier makes the standard two-sample \(t\)-confidence interval poorly justified without additional information or a more appropriate analysis.

Answer

No. The clear outlier in the small Group A sample is the main concern. The much larger Group B sample does not compensate for a severe shape problem in the small group.

Confidence interval for a difference in means 26

A reported \(95\%\) confidence interval for \(\mu_A-\mu_B\) is \([0,4.20]\). The lower endpoint is exactly zero, not a rounded value. Sofia Alvarez says the interval proves that \(\mu_A>\mu_B\) because none of its values is negative. Evaluate the claim.

Hints

- Which population difference represents no difference at all? - Check the square bracket at the lower endpoint. - Distinguish values that are nonnegative from values that are strictly positive.

Solution

1. The parameter value \(0\) represents equal population means. 2. The interval \([0,4.20]\) includes \(0\) exactly as its lower endpoint. 3. Because a zero population mean difference remains compatible with the reported interval, the interval does not establish a strictly positive difference at the \(95\%\) confidence level.

Answer

Sofia's claim is not justified. The interval includes \(0\), so it does not rule out equal population means.

Confidence interval for a difference in means 27

Two studies produce the same estimated population mean difference \(5\) and the same estimated standard error \(2\). For a \(95\%\) confidence interval, Study 1 uses \(10\) degrees of freedom with \(t^*=2.228\), while Study 2 uses \(100\) degrees of freedom with \(t^*=1.984\). Construct both intervals and explain why the one with fewer degrees of freedom is wider even though the point estimate and standard error are identical.

Hints

- The point estimate and standard error are fixed, so compare the critical values. - Confidence-interval width depends directly on the critical value. - Smaller degrees of freedom produce heavier \(t\) tails.

Solution

1. Study 1 has margin of error \(2.228\cdot2=4.456\), giving \((0.544,9.456)\). 2. Study 2 has margin of error \(1.984\cdot2=3.968\), giving \((1.032,8.968)\). 3. With fewer degrees of freedom, the \(t\) distribution has heavier tails, so a larger critical value is needed for the same confidence level. 4. The larger critical value makes Study 1’s interval wider.

Answer

Study 1: \((0.544,9.456)\). Study 2: \((1.032,8.968)\). Study 1 is wider because fewer degrees of freedom require a larger \(t\)-critical value.

Confidence interval for a difference in means 28

Independent random samples from populations A and B have sample means \(\bar{x}_A,\bar{x}_B\), sample standard deviations \(s_A,s_B\), and sample sizes \(n_A,n_B\). Noah Williams is choosing the standard-error expression for a two-sample \(t\)-confidence interval for \(\mu_A-\mu_B\). Which expression is appropriate? a) \(\frac{s_A}{\sqrt{n_A}}-\frac{s_B}{\sqrt{n_B}}\) b) \(\sqrt{\frac{s_A^2}{n_A}+\frac{s_B^2}{n_B}}\) c) \(\frac{s_A+s_B}{\sqrt{n_A+n_B}}\) Explain the choice.

Hints

- Combine independent sources of uncertainty at the variance level. - Subtracting point estimates does not mean subtracting their spreads. - A standard error must have the same measurement units as the mean difference.

Solution

1. Each independent sample mean contributes sampling variance to the difference. 2. Independent variances add even though the sample means are subtracted. 3. Estimating those variances with the sample standard deviations gives \(\frac{s_A^2}{n_A}+\frac{s_B^2}{n_B}\). 4. Taking the square root gives the standard error in option b).

Answer

b) \(\sqrt{\frac{s_A^2}{n_A}+\frac{s_B^2}{n_B}}\).

Confidence interval for a difference in means 29

Independent random samples give identical sample means, \(\bar{x}_A=\bar{x}_B=50\). Both groups have \(n=25\) and sample standard deviation \(10\), and both sample distributions are approximately symmetric with no outliers. Using \(t^*=2.01\), construct a confidence interval for \(\mu_A-\mu_B\), rounding endpoints to two decimal places. Explain why equal sample means do not produce a zero-width confidence interval.

Hints

- Check the sampling and shape conditions separately from the observed equality of the means. - Even when the point estimate is zero, each independent sample mean has sampling variability. - Build the margin of error from both groups' uncertainty contributions.

Solution

1. The independent random samples and stated sample shapes support a two-sample \(t\)-interval. The point estimate is \(50-50=0\). 2. The estimated standard error is \(\sqrt{\frac{10^2}{25}+\frac{10^2}{25}}=\sqrt{8}\approx2.828\). 3. The margin of error is \(2.01\cdot2.828\approx5.685\). 4. The confidence interval is approximately \((-5.69,5.69)\). 5. Equal observed sample means make the point estimate zero, but both sample means still have sampling uncertainty.

Answer

The confidence interval is approximately \((-5.69,5.69)\). Equal sample means do not remove the sampling variability in either estimate.

Confidence interval for a difference in means 30

From the same two independent samples, an \(80\%\) confidence interval for \(\mu_A-\mu_B\) is \((1.0,5.0)\), while a \(95\%\) confidence interval is \((-0.5,6.5)\). Explain why these intervals are not contradictory even though only the higher-confidence interval includes \(0\).

Hints

- Compare the centers of the two intervals first. - Higher confidence requires a larger margin of error for the same data. - A value can fall outside a narrow interval but inside a wider one.

Solution

1. Both intervals use the same point estimate, which is the midpoint \(3\). 2. The \(95\%\) interval uses a larger critical value than the \(80\%\) interval, so it must be wider. 3. The narrower \(80\%\) interval can exclude \(0\) while the wider \(95\%\) interval includes \(0\). 4. Greater confidence requires accepting a broader range of plausible population mean differences.

Answer

The intervals are consistent. The \(95\%\) interval is wider because it uses a larger critical value, so it can include \(0\) even though the narrower \(80\%\) interval does not.

Confidence interval for a difference in means 31

A standard two-sample \(t\)-confidence interval is reported for \(\mu_A-\mu_B\). The observed sample-mean difference is \(1\), but the reported interval is \((-2,6)\). Explain why this cannot be a standard \(t\)-confidence interval based on that point estimate.

Hints

- Recall the basic form of a standard \(t\)-confidence interval. - Find the midpoint of the reported endpoints. - Compare that midpoint with the stated sample-mean difference.

Solution

1. A standard two-sample \(t\)-confidence interval has the form point estimate plus or minus a margin of error. 2. Therefore, it must be symmetric around the observed sample-mean difference. 3. The midpoint of \((-2,6)\) is \(2\), not \(1\). 4. Thus, either the reported point estimate or at least one interval endpoint is incorrect.

Answer

The report is inconsistent. A standard \(t\)-confidence interval based on a point estimate of \(1\) must be centered at \(1\), while \((-2,6)\) is centered at \(2\).

Confidence interval for a difference in means 32

Three studies want confidence intervals for quantitative population parameters. a) A random sample of light bulbs is used to estimate one population's mean lifetime. b) The same \(25\) patients are measured before and after a therapy, and the goal is to estimate the population mean change. c) Independent random samples are taken from Factory A and Factory B, and the goal is to estimate \(\mu_A-\mu_B\). Which study calls for an independent two-sample \(t\)-confidence interval for a difference in means? Explain why the other two designs call for different mean-inference structures.

Hints

- Decide how many independent populations are represented in each study. - Repeated measurements on the same person create pairing. - Match the interval structure to the population parameter being estimated.

Solution

1. Study c) has two independent samples from two populations and targets \(\mu_A-\mu_B\), so it calls for an independent two-sample \(t\)-confidence interval. 2. Study a) has only one sampled population and one population mean, so it is a one-mean interval problem. 3. Study b) has repeated measurements on the same patients, so it should be analyzed through the within-patient differences rather than as two independent samples.

Answer

Study c) calls for the independent two-sample \(t\)-confidence interval. Study a) is a one-mean interval, and study b) is a paired mean-difference interval.

Confidence interval for a difference in means 33

A bakery compares mean checkout times from independent weekday and weekend samples. The estimated standard errors of the two sample means are \(1.5\,\text{min}\) and \(2\,\text{min}\). The chosen \(t\) critical value is \(2.12\). Find the margin of error for a confidence interval for the difference in population mean checkout times. Give the answer to two decimal places.

Hints

- Before using the multiplier, decide how the two estimated spreads combine. - The groups are independent, so think in variance units before returning to minutes. - A confidence interval extends its estimate by the same margin on both sides.

Solution

1. For independent means, combine the estimated standard errors as \(SE_{\text{diff}}=\sqrt{1.5^2+2^2}=2.5\,\text{min}\). 2. The margin of error is \(ME=t^*SE_{\text{diff}}=2.12(2.5)=5.30\,\text{min}\).

Answer

\(ME=5.30\,\text{min}\).

Confidence interval for a difference in means 34

Lakeside Hospital compares two shift schedules in two ways. a) Twelve nurses each work both schedules, with their own hours recorded under each. b) Sixteen nurses work the old schedule and sixteen different nurses work the new one. The nurses are randomly selected from separate large groups. Both groups' shift lengths are roughly symmetric with no outliers. The sample standard deviations are \(1.2\,\text{h}\) for the old schedule and \(1.6\,\text{h}\) for the new schedule. Which study can use an independent two-mean confidence interval? For that study, find the estimated standard error of the difference in sample means.

Hints

- Ask whether the same nurse appears in both sets of measurements. - For the study with separate groups, think about how each sample's spread affects the difference of their means. - The result must be in hours, not squared hours.

Solution

1. In a), each nurse supplies two matched times. A paired-mean interval, not an independent two-mean interval, fits those measurements. 2. In b), the nurses in the two groups are different, so an independent two-mean interval is appropriate. 3. For b), the estimated standard error is \(\sqrt{\frac{1.2^2}{16}+\frac{1.6^2}{16}}=\sqrt{0.25}=0.5\,\text{h}\).

Answer

Study b) can use an independent two-mean interval; its estimated standard error is \(0.5\,\text{h}\). Study a) needs a paired-mean interval.

Confidence interval for a difference in means 35

Harbor Market (A) and Mill Street Market (B) each take independent random samples of customer purchases. The totals are roughly symmetric with no extreme outliers. Harbor has \(n_A=16\), \(\bar{x}_A=35\) dollars, and \(s_A=8\) dollars. Mill Street has \(n_B=25\), \(\bar{x}_B=30\) dollars, and \(s_B=10\) dollars. Construct a \(95\%\) confidence interval for \(\mu_A-\mu_B\) using the supplied critical value \(t^*=2.03\). Round the endpoints to two decimals.

Hints

- Start by subtracting the two sample means in the requested order. - Both sample means vary. How would leaving either one out affect your estimate of uncertainty? - Use the supplied multiplier for the margin and extend equally from the estimate.

Solution

1. The point estimate is \(35-30=5\) dollars. 2. The independent two-mean estimated standard error is \(\sqrt{\frac{8^2}{16}+\frac{10^2}{25}}=\sqrt{8}\approx2.8284\) dollars. 3. The margin of error is \(2.03\sqrt{8}\approx5.7417\) dollars. 4. The interval is \(5\pm5.7417\), or approximately \((-0.74,10.74)\) dollars.

Answer

The \(95\%\) interval for \(\mu_A-\mu_B\) is approximately \((-0.74,10.74)\) dollars.

Confidence interval for a difference in means 36

These two intervals estimate the same population mean difference, \(\mu_A-\mu_B\), using the same independent-group data. One has \(90\%\) confidence and the other has \(99\%\). Which panel shows the \(99\%\) interval? Which interval rules out zero? Explain from the graph.

Hints

- Compare how far each interval extends from its center. - Find zero on the number line and check whether it lies between an interval's endpoints. - Think about the tradeoff between confidence and precision when data stay fixed.

Solution

1. The \(99\%\) interval in panel b) is wider because greater confidence requires a larger margin of error for the same data. 2. The \(90\%\) interval in a) lies above zero, whereas the \(99\%\) interval in b) crosses zero. 3. Only panel a) rules out a mean difference of zero. Greater confidence can come at the cost of a less definite statement about the sign.

Answer

Panel b) shows the \(99\%\) interval because it is wider. Only panel a) rules out zero.

Confidence interval for a difference in means 37

A city compares average water pressure at two pumping stations using independent random samples. The number line gives a \(95\%\) two-mean \(t\)-confidence interval for \(\mu_A-\mu_B\), in pounds per square inch (psi). Using the same data and method, the city could run two separate two-sided tests at \(\alpha=0.05\): one with \(H_0:\mu_A-\mu_B=0\) and one with \(H_0:\mu_A-\mu_B=4\,\text{psi}\). Which null hypotheses would be rejected? Explain using the interval.

Hints

- Think about which proposed population differences remain plausible under the interval. - The interval and both tests use the same observations and the same \(t\) method. - Consider each proposed difference separately rather than comparing them with each other.

Solution

1. A two-sided \(t\) test at \(\alpha=0.05\) rejects a null value exactly when that value is outside the matching \(95\%\) \(t\)-confidence interval. 2. The displayed interval runs from \(1\) to \(5\,\text{psi}\). 3. Zero is outside the interval, so reject \(H_0:\mu_A-\mu_B=0\). 4. Four is inside the interval, so do not reject \(H_0:\mu_A-\mu_B=4\,\text{psi}\).

Answer

Reject the null difference of \(0\), because it is outside the interval. Do not reject the null difference of \(4\,\text{psi}\), because it is inside. The \(95\%\) interval gives the same decisions as the matching two-sided \(5\%\) tests.

Confidence interval for a difference in means 38

A city library system compares checkout wait times at two large branches. Independent random samples are taken from more than \(1000\) customers at each branch. At Branch A, \(n_A=28\), \(\bar{x}_A=11.4\,\text{min}\), and \(s_A=2.5\,\text{min}\). At Branch B, \(n_B=32\), \(\bar{x}_B=10.1\,\text{min}\), and \(s_B=2.1\,\text{min}\). Both sample distributions are roughly symmetric with no outliers. Construct and interpret a \(90\%\) confidence interval for \(\mu_A-\mu_B\), the difference in population mean checkout wait times. Give the interval endpoints to two decimal places.

Hints

- First decide whether the two samples are linked or independent. - Check whether the sampling plan and the sample distributions support inference about two population means. - Keep the subtraction order consistent from the estimate through the interpretation.

Solution

1. The independent random samples satisfy the randomization and \(10\%\) conditions, and the sample shapes support two-sample \(t\) inference. 2. The estimated difference is \(11.4-10.1=1.3\,\text{min}\). 3. The standard error is \(\sqrt{\frac{2.5^2}{28}+\frac{2.1^2}{32}}\approx0.601\,\text{min}\). Technology gives about \(53\) degrees of freedom and \(t^*\approx1.674\) for \(90\%\) confidence. 4. The margin of error is \(1.674(0.601)\approx1.006\,\text{min}\), giving \(1.3\pm1.006\), or approximately \((0.29,2.31)\,\text{min}\). 5. We are \(90\%\) confident that the population mean wait at Branch A is between about \(0.29\) and \(2.31\) minutes longer than at Branch B.

Answer

The \(90\%\) confidence interval for \(\mu_A-\mu_B\) is approximately \((0.29,2.31)\,\text{min}\). We are \(90\%\) confident that Branch A's population mean checkout wait is about \(0.29\) to \(2.31\) minutes longer than Branch B's.

Confidence interval for a difference in means 39

A museum wants a confidence interval for the difference in population mean visit length between members and nonmembers. It takes independent random samples of \(22\) members from a population of \(180\) members and \(24\) nonmembers from a population of \(500\) nonmembers. Both sample distributions are roughly symmetric with no outliers. Is a two-sample \(t\)-interval justified from this sampling plan? Check the relevant conditions and identify any condition that fails.

Hints

- Check each population separately when deciding whether sampling without replacement is close enough to independent sampling. - A condition can hold for one sample and fail for the other. - The shape information addresses a different requirement from the population-size requirement.

Solution

1. The randomization condition is met because both groups were sampled randomly and independently. 2. For members, \(10\%\) of the population is \(18\), but the sample size is \(22\), so the \(10\%\) condition fails for that group. 3. For nonmembers, \(10\%\) of \(500\) is \(50\), so the sample size \(24\) satisfies the \(10\%\) condition. 4. The sample-data condition is reasonable because both samples are described as roughly symmetric with no outliers. 5. Since the \(10\%\) condition fails for the member sample, the usual two-sample \(t\)-interval is not justified under the stated sampling-without-replacement plan.

Answer

No. Random sampling and sample-shape conditions are satisfied, and the nonmember sample meets the \(10\%\) condition. The member sample does not: \(22>0.10(180)=18\). Therefore, the usual two-sample \(t\)-interval is not justified from the stated plan.

Confidence interval for a difference in means 40

A regional training program compares assessment scores for two independently sampled groups. Group A has \(n_A=35\), \(\bar{x}_A=72.4\), and \(s_A=8.1\). Group B has \(n_B=30\), \(\bar{x}_B=68.9\), and \(s_B=7.5\). The sampling conditions for two-sample \(t\) inference are satisfied. Construct a \(95\%\) confidence interval for \(\mu_A-\mu_B\), with endpoints to two decimal places, and state whether the interval provides convincing evidence that the population means differ.

Hints

- Start with the observed difference in sample means using the requested subtraction order. - Combine the two estimated sampling-variance contributions because the samples are independent. - After constructing the interval, compare it with the value representing equal population means.

Solution

1. The observed mean difference is \(72.4-68.9=3.5\). 2. The standard error is \(\sqrt{\frac{8.1^2}{35}+\frac{7.5^2}{30}}\approx1.936\). 3. Technology gives approximately \(63\) degrees of freedom and \(t^*\approx1.999\) for \(95\%\) confidence. 4. The margin of error is \(1.999(1.936)\approx3.870\), so the interval is approximately \((-0.37,7.37)\). 5. Because \(0\) is in the interval, the interval does not provide convincing evidence that the two population mean assessment scores differ.

Answer

The \(95\%\) confidence interval for \(\mu_A-\mu_B\) is approximately \((-0.37,7.37)\). Because it contains \(0\), it does not provide convincing evidence of a difference in the population mean assessment scores.

Confidence interval for a difference in means 41

Dr. Hannah Lee has two independent random samples and wants a confidence interval for the difference in population means. The sample standard deviations are \(4.2\) and \(9.8\). Marcus Boyd claims that a Welch two-sample \(t\)-interval cannot be used because the sample standard deviations are too different. Evaluate Marcus's claim. State whether equal population standard deviations are required for a Welch two-sample \(t\)-interval, and identify what conditions should be checked instead.

Hints

- Distinguish the assumptions of a Welch interval from those of a pooled two-sample interval. - Ask whether each sample's variability can contribute separately to the estimated uncertainty. - Keep the sampling-design conditions separate from any equal-variance assumption.

Solution

1. Marcus's claim is incorrect. A Welch two-sample \(t\)-interval for independent means does not require the two population standard deviations to be equal. 2. Its standard error uses each sample's variability separately rather than pooling the two standard deviations under an equal-variance assumption. 3. For the stated sampling design, the relevant checks are independent random samples, the \(10\%\) condition when sampling without replacement, and sample-distribution shapes or sample sizes that make \(t\) inference reasonable. 4. Different sample standard deviations alone do not invalidate the procedure.

Answer

Marcus's claim is incorrect. Equal population standard deviations are not required for a Welch two-sample \(t\)-interval. For these independent random samples, check the \(10\%\) condition when applicable and whether each sample's shape or size makes \(t\) inference reasonable.

Confidence interval for a difference in means 42

Using the same two independent samples, Elena Rossi calculates two confidence intervals for \(\mu_1-\mu_2\): \((0.5,3.9)\) and \((-0.2,4.6)\). One is a \(90\%\) interval and the other is a \(95\%\) interval. Match each interval to its confidence level. Then explain why a claim that \(\mu_1>\mu_2\) is supported by one interval but not by the other.

Hints

- Intervals from the same data share the same center but need not share the same width. - Higher confidence requires a larger range of plausible parameter values. - Compare each interval with the population-difference value representing equality.

Solution

1. Both intervals have midpoint \(2.2\), so they are centered at the same estimated mean difference. 2. The wider interval, \((-0.2,4.6)\), must have the higher confidence level, so it is the \(95\%\) interval. 3. The narrower interval, \((0.5,3.9)\), is the \(90\%\) interval. 4. The \(90\%\) interval is entirely positive, so it supports \(\mu_1>\mu_2\) at that confidence level. 5. The \(95\%\) interval includes \(0\), so equal population means remain plausible at the higher confidence level.

Answer

\((0.5,3.9)\) is the \(90\%\) interval, and \((-0.2,4.6)\) is the \(95\%\) interval. The \(90\%\) interval is entirely positive, but the \(95\%\) interval includes \(0\), so the stronger confidence requirement no longer gives convincing evidence that \(\mu_1>\mu_2\).

Confidence interval for a difference in means 43

A two-sample \(t\)-confidence interval for \(\mu_1-\mu_2\) is reported as \((1.1,7.3)\). The critical value used was \(t^*=2.01\). Recover the point estimate, margin of error, and standard error that were used to construct the interval. Give the standard error to three decimal places.

Hints

- The interval's center recovers the estimated difference in sample means. - The distance from the center to either endpoint is the interval's half-width. - Use the reported critical multiplier to work backward from half-width to estimated sampling uncertainty.

Solution

1. The point estimate is the midpoint: \(\frac{1.1+7.3}{2}=4.2\). 2. The margin of error is half the width: \(\frac{7.3-1.1}{2}=3.1\). 3. Since margin of error equals the critical value times the standard error, the standard error is \(\frac{3.1}{2.01}\approx1.542\).

Answer

Point estimate: \(4.2\). Margin of error: \(3.1\). Standard error: approximately \(1.542\).

Confidence interval for a difference in means 44

Two study designs will estimate \(\mu_1-\mu_2\) with the same confidence level. For planning purposes, use population standard deviation \(10\) for each group, and suppose the same total of \(100\) observations is available. Design A uses \(50\) observations in each group. Design B uses \(20\) in Group 1 and \(80\) in Group 2. Compare the standard errors under the two designs and explain which design should produce the narrower confidence interval when the other factors are similar.

Hints

- With equal planning values for population variability, each group's contribution depends on how many observations that group receives. - Compare the variance contributions before thinking about interval width. - A fixed total sample size is not enough by itself to determine precision; allocation matters.

Solution

1. For Design A, the planning standard error is \(\sqrt{\frac{10^2}{50}+\frac{10^2}{50}}=2\). 2. For Design B, the planning standard error is \(\sqrt{\frac{10^2}{20}+\frac{10^2}{80}}=2.5\). 3. With the same confidence level and similar degrees of freedom, the smaller standard error leads to the smaller margin of error. 4. Therefore, the balanced \(50/50\) design should produce the narrower confidence interval.

Answer

Design A has planning standard error \(2\), while Design B has planning standard error \(2.5\). The balanced design should produce the narrower confidence interval because its sampling variability is smaller under the stated planning assumptions.

Confidence interval for a difference in means 45

In a randomized experiment, volunteers are assigned to one of two exercise routines. A \(95\%\) confidence interval for \(\mu_A-\mu_B\), the difference in mean improvement between the routines for participants like those in the experiment, is \((-12,-4)\) points. The volunteers were not randomly sampled from all adults. Interpret the interval and explain what the random assignment does and does not allow the researchers to conclude.

Hints

- Use the sign of \(A-B\) to translate the interval into words. - Separate the role of random assignment from the role of random sampling. - An interval can estimate a treatment contrast even when the target population for generalization is limited.

Solution

1. The interval indicates that Routine A's mean improvement for participants like those in the experiment is estimated to be between \(4\) and \(12\) points lower than Routine B's. 2. Because the interval excludes \(0\), the data support a difference between the treatment means. 3. Random assignment supports a causal interpretation of the treatment difference for participants like those in the experiment. 4. Because the volunteers were not randomly sampled from all adults, the interval does not automatically justify generalizing the effect to the entire adult population.

Answer

We are \(95\%\) confident that Routine A's mean improvement is about \(4\) to \(12\) points lower than Routine B's for participants like those in the experiment. Random assignment supports a causal interpretation, but the lack of random sampling limits broad generalization.

Confidence interval for a difference in means 46

A furniture manufacturer compares the flexural strength of boards from two suppliers. Independent random samples are taken from large shipments. Supplier A has \(n_A=42\), \(\bar{x}_A=15.8\,\text{kN}\), and \(s_A=4.5\,\text{kN}\). Supplier B has \(n_B=38\), \(\bar{x}_B=12.1\,\text{kN}\), and \(s_B=3.9\,\text{kN}\). The conditions for two-sample \(t\) inference are satisfied, and the appropriate \(95\%\) critical value is \(t^*=1.991\). Construct a \(95\%\) confidence interval for \(\mu_A-\mu_B\), with endpoints to two decimal places. Does this interval establish that Supplier A's population mean flexural strength exceeds Supplier B's by at least \(5\,\text{kN}\)?

Hints

- Keep the order of subtraction in the estimate consistent with the parameter named in the question. - Combine the two groups' contributions to the uncertainty of the estimated difference. - To assess the “at least \(5\)” claim, compare every value allowed by the interval with \(5\,\text{kN}\), not just the point estimate.

Solution

1. The estimated difference is \(15.8-12.1=3.7\,\text{kN}\). 2. The standard error is \(\sqrt{\frac{4.5^2}{42}+\frac{3.9^2}{38}}\approx0.939\,\text{kN}\). 3. The margin of error is \(1.991\cdot0.939\approx1.870\,\text{kN}\). 4. The \(95\%\) confidence interval is \(3.7\pm1.870\), or approximately \((1.83,5.57)\,\text{kN}\). 5. The interval does not establish that the population mean difference is at least \(5\,\text{kN}\), because values below \(5\,\text{kN}\) are included in the interval.

Answer

The \(95\%\) confidence interval for \(\mu_A-\mu_B\) is approximately \((1.83,5.57)\,\text{kN}\). No, this interval does not establish that Supplier A's population mean exceeds Supplier B's by at least \(5\,\text{kN}\).

Confidence interval for a difference in means 47

Marine biologist Dr. Amina Yusuf wants a \(95\%\) confidence interval for the difference in population mean shell length between two independent coastal populations. Independent random samples contain \(45\) shells from Population A and \(12\) shells from Population B. Population A's sample distribution is moderately right-skewed. Population B's sample distribution is roughly symmetric with no outliers. Each population is much larger than its sample. Is a two-sample \(t\)-confidence interval appropriate based on the information given? Explain.

Hints

- Check the two groups separately when considering whether their sample means have suitable sampling distributions. - A sample of size \(45\) and a sample of size \(12\) do not need the same shape evidence. - Also verify that the sampling design supports independence within and between groups.

Solution

1. The two samples are independent random samples, and each sample is less than \(10\%\) of its population. 2. Population A has \(n=45\), so moderate skewness is not a serious obstacle to an approximately normal sampling distribution for its sample mean. 3. Population B has only \(n=12\), but its sample distribution is roughly symmetric with no outliers, which supports small-sample \(t\) inference. 4. Therefore, the stated conditions support a two-sample \(t\)-confidence interval for the difference in population means.

Answer

Yes. The randomization and independence conditions are satisfied, the larger sample can tolerate moderate skewness, and the smaller sample is roughly symmetric with no outliers. A two-sample \(t\)-confidence interval is appropriate.

Confidence interval for a difference in means 48

Riverton Materials Lab compares the mean width of polymer test strips from two suppliers. Supplier A is summarized in centimeters: \(n_A=30\), \(\bar{x}_A=12.4\,\text{cm}\), and \(s_A=1.8\,\text{cm}\). Supplier B is summarized in millimeters: \(n_B=30\), \(\bar{x}_B=118\,\text{mm}\), and \(s_B=16\,\text{mm}\). The conditions for two-sample \(t\) inference are satisfied. Use \(t^*=2.002\) for a \(95\%\) confidence interval. Construct the \(95\%\) confidence interval for \(\mu_A-\mu_B\) in centimeters, with endpoints to two decimal places.

Hints

- Put both groups on the same measurement scale before combining their summaries. - Convert both the center and the spread for the group reported in millimeters. - After the units match, use the usual two-sample interval structure.

Solution

1. Convert Supplier B's statistics to centimeters: \(118\,\text{mm}=11.8\,\text{cm}\) and \(16\,\text{mm}=1.6\,\text{cm}\). 2. The estimated difference is \(12.4-11.8=0.6\,\text{cm}\). 3. The standard error is \(\sqrt{\frac{1.8^2}{30}+\frac{1.6^2}{30}}\approx0.440\,\text{cm}\). 4. The margin of error is \(2.002\cdot0.440\approx0.880\,\text{cm}\). 5. The interval is \(0.6\pm0.880\), or approximately \((-0.28,1.48)\,\text{cm}\).

Answer

The \(95\%\) confidence interval for \(\mu_A-\mu_B\) is approximately \((-0.28,1.48)\,\text{cm}\).

Confidence interval for a difference in means 49

A transit authority compares mean interior noise levels for two train-car designs. Independent random samples satisfy the conditions for two-sample \(t\) inference. Design A has \(n_A=40\), \(\bar{x}_A=52\,\text{dB}\), and sample variance \(s_A^2=16\,\text{dB}^2\). Design B has \(n_B=35\), \(\bar{x}_B=49\,\text{dB}\), and sample variance \(s_B^2=25\,\text{dB}^2\). Use \(t^*=2.00\). Construct the confidence interval for \(\mu_A-\mu_B\) associated with this critical value. Give the interval endpoints to two decimal places.

Hints

- Notice whether each variability summary is a standard deviation or a variance. - The standard-error formula uses variance contributions from the two independent samples. - Center the interval at the sample-mean difference before adding and subtracting the margin of error.

Solution

1. The estimated difference is \(52-49=3\,\text{dB}\). 2. Because the given variability values are sample variances, the standard error is \(\sqrt{\frac{16}{40}+\frac{25}{35}}\approx1.056\,\text{dB}\). 3. The margin of error is \(2.00\cdot1.056\approx2.111\,\text{dB}\). 4. The confidence interval is \(3\pm2.111\), or approximately \((0.89,5.11)\,\text{dB}\).

Answer

The confidence interval for \(\mu_A-\mu_B\) is approximately \((0.89,5.11)\,\text{dB}\).

Confidence interval for a difference in means 50

Two independent random samples satisfy the conditions for two-sample \(t\) inference. Group 1 has \(n_1=20\), \(\bar{x}_1=102.3\), and \(s_1=1.8\). Group 2 has \(n_2=20\), \(\bar{x}_2=101.7\), and \(s_2=2.1\). Use \(t^*=2.024\) for a \(95\%\) confidence interval for \(\mu_1-\mu_2\). Construct the interval with endpoints to two decimal places. Then evaluate the statement, “Because the interval contains \(0\), the two population means have been shown to be equal.”

Hints

- Build the interval around the observed difference in sample means. - Combine the two estimated variance contributions to find the standard error. - Distinguish “not enough evidence of a difference” from “proof of equality.”

Solution

1. The estimated difference is \(102.3-101.7=0.6\). 2. The standard error is \(\sqrt{\frac{1.8^2}{20}+\frac{2.1^2}{20}}\approx0.618\). 3. The margin of error is \(2.024\cdot0.618\approx1.252\). 4. The interval is \(0.6\pm1.252\), or approximately \((-0.65,1.85)\). 5. Since \(0\) is in the interval, the data do not provide convincing evidence of a nonzero difference at the corresponding level. This does not prove the population means are exactly equal.

Answer

The \(95\%\) confidence interval is approximately \((-0.65,1.85)\). Including \(0\) means equality is compatible with the data; it does not prove that \(\mu_1=\mu_2\).

Confidence interval for a difference in means 51

Priya Nair is planning a two-sample confidence interval for a difference in population means. The two groups will have equal sample size \(n\), both population standard deviations are expected to be about \(10\), and the planning calculation uses critical value \(t^*\approx2.00\). What minimum equal sample size is needed in each group so that the planned margin of error is at most \(2\) units?

Hints

- Write the standard error using the same unknown sample size for both groups. - Combine the two equal variance contributions before applying the critical value. - Translate “margin of error at most \(2\)” into an inequality and solve for \(n\).

Solution

1. With equal sample sizes and expected standard deviations of \(10\), the planned standard error is \(\sqrt{\frac{10^2}{n}+\frac{10^2}{n}}=\sqrt{\frac{200}{n}}\). 2. The planned margin of error is \(2.00\sqrt{\frac{200}{n}}\). 3. Require \(2.00\sqrt{\frac{200}{n}}\le2\), so \(\sqrt{\frac{200}{n}}\le1\). 4. This gives \(n\ge200\).

Answer

The plan needs at least \(200\) observations in each group.

Confidence interval for a difference in means 52

Independent random samples are taken from two large populations. Group A has \(n_A=12\), \(\bar{x}_A=30.4\), and \(s_A=3.0\). Group B has \(n_B=11\), \(\bar{x}_B=27.1\), and \(s_B=2.6\). The boxplots show the sample shapes. Use \(t^*=2.086\) for a \(95\%\) confidence interval for \(\mu_A-\mu_B\). Do the boxplots support a small-sample two-sample \(t\)-interval? If so, construct the interval. Round the interval endpoints to two decimal places.

Hints

- Use the visual to assess each small sample separately for strong skewness and outliers. - Keep the point estimate in the same subtraction order as the requested parameter. - Combine the two groups' estimated variance contributions when finding the standard error.

Solution

1. Both boxplots are reasonably symmetric and show no outliers, so the sample-shape condition is reasonable for the two small samples. 2. The estimated difference is \(30.4-27.1=3.3\). 3. The standard error is \(\sqrt{\frac{3.0^2}{12}+\frac{2.6^2}{11}}\approx1.168\). 4. The margin of error is \(2.086\cdot1.168\approx2.437\). 5. The \(95\%\) confidence interval is \(3.3\pm2.437\), or approximately \((0.86,5.74)\).

Answer

Yes. The boxplots show no strong skewness or outliers. The \(95\%\) confidence interval for \(\mu_A-\mu_B\) is approximately \((0.86,5.74)\).

Confidence interval for a difference in means 53

A two-sample \(t\)-confidence interval for \(\mu_1-\mu_2\) has total width \(8\) units. The critical value is \(t^*=2.00\), both sample sizes are \(25\), and Group 1 has sample standard deviation \(s_1=6\). What sample standard deviation \(s_2\) for Group 2 is consistent with this interval width?

Hints

- Convert total interval width into a margin of error first. - Use the critical value to recover the standard error. - The squared standard error is the sum of the two group variance contributions.

Solution

1. A total interval width of \(8\) gives margin of error \(4\). 2. Since \(ME=t^*SE\), the standard error is \(SE=4/2.00=2\). 3. Therefore, \(4=SE^2=\frac{6^2}{25}+\frac{s_2^2}{25}\). 4. This gives \(4=1.44+\frac{s_2^2}{25}\), so \(\frac{s_2^2}{25}=2.56\). 5. Thus, \(s_2^2=64\) and \(s_2=8\).

Answer

The Group 2 sample standard deviation is \(8\).

Confidence interval for a difference in means 54

Two independent random samples satisfy the conditions for two-sample \(t\) inference. Group 1 has \(n_1=60\), \(\bar{x}_1=28.4\), and \(s_1=5.2\). Group 2 has \(n_2=150\), \(\bar{x}_2=26.9\), and \(s_2=4.1\). Use \(t^*=1.98\) for a \(95\%\) confidence interval for \(\mu_1-\mu_2\). Construct the interval, rounding endpoints to three decimal places, and state whether it excludes \(0\).

Hints

- Compute the point estimate in the requested subtraction order. - Unequal sample sizes enter separately in the two variance contributions. - Keep enough precision in the margin of error to decide whether the lower endpoint is above or below \(0\).

Solution

1. The estimated difference is \(28.4-26.9=1.5\). 2. The standard error is \(\sqrt{\frac{5.2^2}{60}+\frac{4.1^2}{150}}\approx0.750\). 3. The margin of error is \(1.98\cdot0.750\approx1.485\). 4. The interval is \(1.5\pm1.485\), or approximately \((0.015,2.985)\). 5. The lower endpoint is slightly positive, so the interval excludes \(0\), though only narrowly.

Answer

The \(95\%\) confidence interval is approximately \((0.015,2.985)\). It excludes \(0\) by a small amount.

Confidence interval for a difference in means 55

A two-sample confidence interval for \(\mu_A-\mu_B\) is being planned. Current sample sizes are \(n_A=n_B=40\), with planning standard deviations \(s_A=12\) and \(s_B=4\). The study can collect \(20\) additional observations from only one group. Ignoring the small change in the \(t\)-critical value, which choice will reduce the interval's margin of error more: increasing A to \(60\), or increasing B to \(60\)? Support your choice by comparing the estimated standard errors. Round noninteger standard errors to three decimal places.

Hints

- Compare how much each group contributes to the uncertainty of the difference. - Recompute the uncertainty under each proposed sample-size change. - Increasing the sample size of the group contributing more variability can have a larger effect.

Solution

1. With \(40\) observations in each group, the estimated standard error is \(\sqrt{\frac{12^2}{40}+\frac{4^2}{40}}=2.000\). 2. If A increases to \(60\), the standard error becomes \(\sqrt{\frac{12^2}{60}+\frac{4^2}{40}}\approx1.673\). 3. If B increases to \(60\), the standard error becomes \(\sqrt{\frac{12^2}{40}+\frac{4^2}{60}}\approx1.966\). 4. Increasing the higher-variability group A produces the larger reduction in standard error and therefore the larger reduction in margin of error.

Answer

Add the \(20\) observations to group A. The estimated standard error falls from \(2.000\) to about \(1.673\), compared with about \(1.966\) if the observations are added to group B.

Confidence interval for a difference in means 56

Separate \(95\%\) confidence intervals for two population means are \((48,52)\) for \(\mu_A\) and \((44,46)\) for \(\mu_B\). Aisha Rahman subtracts endpoints and reports \((48-46,52-44)=(2,8)\) as a \(95\%\) confidence interval for \(\mu_A-\mu_B\). Explain why this endpoint-subtraction method does not generally produce the desired \(95\%\) two-sample confidence interval.

Hints

- A confidence interval for a difference is a new inference problem, not just endpoint arithmetic on two other intervals. - Think about how the uncertainty from two independent estimates is combined. - The confidence level attached to each separate interval does not transfer directly to their endpoint difference.

Solution

1. Each separate interval has its own uncertainty statement about one population mean. 2. The uncertainty of \(\bar{x}_A-\bar{x}_B\) must be based on the standard error of the difference, which combines the two groups' variability and sample sizes. 3. Subtracting the endpoints of two separate \(95\%\) intervals does not preserve a \(95\%\) confidence level for the difference. 4. The original sample sizes and variability information, or an equivalent standard error for the difference, are needed to construct the proper two-sample interval.

Answer

\((2,8)\) is not automatically a \(95\%\) confidence interval for \(\mu_A-\mu_B\). A valid two-sample interval must be built from the estimated difference and the standard error of that difference.

Confidence interval for a difference in means 57

Independent random samples give \(n_A=500\), \(\bar{x}_A=50\), \(s_A=10\) and \(n_B=12\), \(\bar{x}_B=44\), \(s_B=5\). The Group B sample distribution is approximately symmetric with no outliers. Software that computes Welch degrees of freedom is unavailable, so use the conservative choice \(df=\min(n_A-1,n_B-1)=11\), with \(t^*=2.201\), to construct a \(95\%\) confidence interval for \(\mu_A-\mu_B\). Round the interval endpoints to two decimal places.

Hints

- Check the shape condition separately for the small sample; the large sample is less sensitive to moderate nonnormality. - Let each independent group contribute separately to the standard error. - The conservative degrees-of-freedom rule affects the critical value, not the point estimate.

Solution

1. The independent random samples, the large Group A sample, and the stated shape of the small Group B sample support a two-sample \(t\)-interval. The estimated difference is \(50-44=6\). 2. The estimated standard error is \(\sqrt{\frac{10^2}{500}+\frac{5^2}{12}}\approx1.511\). 3. Using the conservative \(11\) degrees of freedom, the margin of error is \(2.201\cdot1.511\approx3.326\). 4. The confidence interval is \(6\pm3.326\), or approximately \((2.67,9.33)\).

Answer

Using the conservative degrees of freedom, the \(95\%\) confidence interval is approximately \((2.67,9.33)\).

Confidence interval for a difference in means 58

Two independent random samples were planned with \(50\) observations per group from populations of at least \(500\) members each, but some responses are missing. The usable samples have \(n_A=40\), \(\bar{x}_A=30\), \(s_A=6\) and \(n_B=45\), \(\bar{x}_B=27\), \(s_B=5\). Assume the missing responses do not introduce bias. Using \(t^*=2.00\), construct a \(95\%\) confidence interval for \(\mu_A-\mu_B\). Round the interval endpoints to two decimal places.

Hints

- Base the standard error on the observations actually contributing data to the analysis. - Start with the difference of the two usable-sample means. - Combine the two groups' estimated sampling variability before applying the critical value.

Solution

1. The interval must use the actual usable sample sizes, \(40\) and \(45\), rather than the planned sizes. 2. The estimated difference is \(30-27=3\). 3. The estimated standard error is \(\sqrt{\frac{6^2}{40}+\frac{5^2}{45}}\approx1.206\). 4. The margin of error is \(2.00\cdot1.206\approx2.413\). 5. The confidence interval is \(3\pm2.413\), or approximately \((0.59,5.41)\).

Answer

Using the \(40\) and \(45\) usable observations, the \(95\%\) confidence interval is approximately \((0.59,5.41)\).

Confidence interval for a difference in means 59

Engineers consider two manufacturing processes practically equivalent if their population mean outputs differ by less than \(1\) unit in either direction. A \(95\%\) confidence interval for \(\mu_A-\mu_B\) is \((-0.30,0.40)\). What does the interval say about statistical equality and about the engineers’ practical-equivalence criterion?

Hints

- Compare the interval with both \(0\) and the practical bounds supplied by the engineers. - Exact equality and practical equivalence are different claims. - Use the entire interval rather than only its midpoint.

Solution

1. The interval contains \(0\), so it does not indicate a statistically detectable nonzero difference at the matching two-sided \(5\%\) level. 2. Every value in the interval lies between \(-1\) and \(1\). 3. Thus, the entire set of plausible mean differences under this interval falls inside the engineers’ stated practical-equivalence range. 4. The interval does not prove the population means are exactly equal; it supports that any difference is small enough to meet the stated practical criterion.

Answer

The interval does not establish exact equality, but it lies entirely inside the practical-equivalence range \((-1,1)\). The data therefore support the conclusion that the population mean difference is practically small under the stated criterion.

Confidence interval for a difference in means 60

A two-sample study estimates \(\mu_A-\mu_B\) as \(4\). Both groups have size \(25\). With the original measurement system, \(s_A=10\) and \(s_B=6\). For planning purposes, suppose a redesigned measurement system would reduce those standard deviations to \(5\) and \(3\) without changing the population mean difference. Use \(t^*=2.01\) for both intervals. Compare the original interval with the planned redesigned-system interval. Round interval endpoints to two decimal places, and explain the effect of reducing measurement variability.

Hints

- Keep the point estimate fixed and compare only the uncertainty terms. - See how the standard error changes when both sample standard deviations are reduced by the same factor. - The margin of error scales with the standard error when the critical value is unchanged.

Solution

1. With the original system, the standard error is \(\sqrt{\frac{10^2}{25}+\frac{6^2}{25}}\approx2.332\), so the margin of error is \(2.01\cdot2.332\approx4.688\). 2. The original interval is approximately \((-0.69,8.69)\). 3. Under the planning assumption for the redesigned system, the standard error is \(\sqrt{\frac{5^2}{25}+\frac{3^2}{25}}\approx1.166\), so the margin of error is approximately \(2.344\). 4. The planned redesigned-system interval is approximately \((1.66,6.34)\). 5. Halving both within-group standard deviations halves the standard error and margin of error, producing a much more precise interval around the same point estimate.

Answer

Original interval: approximately \((-0.69,8.69)\). Planned redesigned-system interval: approximately \((1.66,6.34)\). Under the stated planning assumption, reducing within-group variability makes the interval substantially narrower.

Confidence interval for a difference in means 61

Population A contains exactly \(200\) units, and every unit has been measured, so its population mean is known to be \(\mu_A=50\). Population B is much larger. A random sample of \(25\) units from B has \(\bar{x}_B=45\) and \(s_B=10\), with an approximately symmetric distribution and no outliers. Using \(t^*=2.064\), construct a \(95\%\) confidence interval for \(\mu_A-\mu_B\), rounding endpoints to two decimal places. Explain why only Population B contributes sampling uncertainty.

Hints

- Separate the mean known from the census from the mean estimated using the random sample. - For the sample of size \(25\), check that the stated shape supports a \(t\)-procedure. - Build the interval using only the sampling uncertainty from Population B.

Solution

1. Because Population A was fully measured, \(\mu_A=50\) is known without sampling error. The random sample and stated shape for Population B support a one-sample \(t\)-interval for its mean. 2. The point estimate for \(\mu_A-\mu_B\) is \(50-45=5\). 3. The only estimated standard error comes from \(\bar{x}_B\): \(10/\sqrt{25}=2\). 4. The margin of error is \(2.064\cdot2=4.128\). 5. The \(95\%\) confidence interval is \(5\pm4.128\), or approximately \((0.87,9.13)\).

Answer

The \(95\%\) confidence interval for \(\mu_A-\mu_B\) is approximately \((0.87,9.13)\). Population A contributes no sampling uncertainty because its mean is known from a census.

Confidence interval for a difference in means 62

A \(95\%\) confidence interval for \(\mu_A-\mu_B\) is \((-1.4,0.6)\). A design requirement says Process A is acceptable as long as its population mean is not more than \(2\) units below Process B’s mean. Does the confidence interval rule out a mean disadvantage for A of \(2\) units or more? Explain without claiming that the two means are equal.

Hints

- Translate the practical requirement into a value of \(\mu_A-\mu_B\). - Compare that threshold with the full confidence interval. - Meeting a practical bound is different from proving exact equality.

Solution

1. A disadvantage of \(2\) units for A corresponds to \(\mu_A-\mu_B=-2\); larger disadvantages correspond to values below \(-2\). 2. The entire confidence interval \((-1.4,0.6)\) lies above \(-2\). 3. Thus, differences of \(-2\) or less are outside the interval’s plausible range at the stated confidence level. 4. The interval still includes \(0\) and both small negative and positive differences, so it does not establish that the population means are equal or that A is better.

Answer

Yes. The interval lies entirely above \(-2\), so it rules out a mean disadvantage of \(2\) units or more for A at the stated confidence level. It does not prove the two population means are equal.

Confidence interval for a difference in means 63

Independent random samples of completion times, in seconds, are shown below. Each source population has more than \(100\) comparable tasks, and both population distributions are approximately normal. <table><tr><th>Method A</th><td>8</td><td>10</td><td>9</td><td>11</td><td>12</td><td>10</td></tr><tr><th>Method B</th><td>5</td><td>7</td><td>6</td><td>8</td><td>5</td><td>7</td><td>6</td></tr></table> Use \(t^*=2.18\) to construct a \(95\%\) confidence interval for \(\mu_A-\mu_B\). Round the interval endpoints to two decimal places.

Hints

- Compute a mean and sample standard deviation for each group before constructing the interval. - Keep the subtraction order \(A-B\) throughout the calculation. - Combine the two independent variance contributions when finding the standard error.

Solution

1. The sample summaries are \(\bar{x}_A=10\), \(s_A\approx1.414\), \(\bar{x}_B\approx6.286\), and \(s_B\approx1.113\). 2. The estimated difference is \(10-6.286\approx3.714\,\text{s}\). 3. The standard error is \(\sqrt{\frac{1.414^2}{6}+\frac{1.113^2}{7}}\approx0.714\,\text{s}\). 4. The margin of error is \(2.18\cdot0.714\approx1.557\,\text{s}\). 5. The \(95\%\) confidence interval is approximately \((2.16,5.27)\,\text{s}\).

Answer

The \(95\%\) confidence interval for \(\mu_A-\mu_B\) is approximately \((2.16,5.27)\,\text{s}\).

Confidence interval for a difference in means 64

Riverton High (group A) and Cedar High (group B) take independent random samples of students in the same math course. The number of practice quizzes each student completed is approximately normal in both large school populations: <table><tr><th>Riverton (A)</th><th>Cedar (B)</th></tr><tr><td>9</td><td>7</td></tr><tr><td>11</td><td>8</td></tr><tr><td>10</td><td>9</td></tr><tr><td>12</td><td>8</td></tr></table> Construct a \(95\%\) interval for \(\mu_A-\mu_B\). Use the supplied approximate Welch critical value \(t^*=2.57\). Round the endpoints to two decimals.

Hints

- Calculate one sample mean and sample variance for each column. - The difference changes when either sample mean changes. How should both sources of variation affect your uncertainty? - Do not round the standard error before forming the interval endpoints.

Solution

1. The sample means are \(\bar{x}_A=10.5\) and \(\bar{x}_B=8\), so the point estimate is \(2.5\). 2. The sample variances are \(s_A^2=\frac{5}{3}\) and \(s_B^2=\frac{2}{3}\). 3. The standard error is \(\sqrt{\frac{5}{12}+\frac{2}{12}}=\sqrt{\frac{7}{12}}\approx0.7638\). 4. The margin is \(2.57\sqrt{\frac{7}{12}}\approx1.9629\). Thus the interval is approximately \((0.54,4.46)\).

Answer

The \(95\%\) interval is approximately \((0.54,4.46)\) quizzes.

Confidence interval for a difference in means 65

An after-school reading program compares the number of books read by students at Eastview High (A) and Northside High (B). Each school randomly selects students from its own large student body; no student attends both schools. Eastview has \(n_A=16\), \(\bar{x}_A=24\), and \(s_A=8\). Northside has \(n_B=16\), \(\bar{x}_B=29\), and \(s_B=12\). The samples show no marked skew or outliers. Choose the correct confidence-interval procedure and find a \(95\%\) interval for the population mean difference, A minus B. Use the supplied critical value \(t^*=2.06\). Round the endpoints to two decimals.

Hints

- Decide first whether observations are matched across schools. - Calculate the difference of sample means in the stated order. - Use both groups' sizes and spreads when you work out the uncertainty in their difference.

Solution

1. Different students are sampled from the two schools, so use an independent two-mean t interval rather than a paired interval. 2. The estimate is \(24-29=-5\) books. 3. The estimated standard error is \(\sqrt{\frac{8^2}{16}+\frac{12^2}{16}}=\sqrt{13}\approx3.6056\) books. 4. The margin is \(2.06\sqrt{13}\approx7.4274\), giving \(-5\pm7.4274\), or \((-12.43,2.43)\). 5. Since zero lies inside the interval, these data do not settle which school's population mean is higher.

Answer

Independent two-mean t interval: \((-12.43,2.43)\) books for \(\mu_A-\mu_B\). Zero is included.

Confidence interval for a difference in means 66

Three researchers want to study population means: I. A call center takes a random sample of calls. It wants to know whether calls last more than \(5\,\text{min}\) on average. II. A swim club measures the same swimmers before and after lessons and wants to estimate the mean change in lap time. III. A library wants to estimate the difference in mean weekly study hours for students at Bayview High (A) and Pinecrest High (B). Each school randomly selects students from its own large student body. The populations are approximately normal, and no student attends both schools. The observations for III are: <table><tr><th>Bayview (A)</th><th>Pinecrest (B)</th></tr><tr><td>8</td><td>6</td></tr><tr><td>10</td><td>9</td></tr><tr><td>9</td><td>8</td></tr><tr><td>13</td><td>7</td></tr><tr><td>10</td><td>10</td></tr></table> a) Choose the appropriate inference method for I, II and III. Give one reason for each choice. b) For III, construct a \(95\%\) confidence interval for \(\mu_A-\mu_B\), using the supplied critical value \(t^*=2.31\). Round the endpoints to two decimals. c) Explain what the interval suggests about which school has the higher population mean study time.

Hints

- For each study, distinguish one set of observations, repeated measurements on the same people, and measurements from different people. - Ask whether each research question needs an estimate with uncertainty or a decision about a benchmark. - In III, summarize each raw-data column separately before finding the difference and its standard error.

Solution

1. Study I concerns one population mean compared with a benchmark, so use a one-mean test. Study II measures each swimmer twice and seeks a mean change, so use a paired-mean interval. Study III compares independent group means and seeks an estimate, so use an independent two-mean interval. 2. In III, the sample means are \(\bar{x}_A=10\) and \(\bar{x}_B=8\), giving estimate \(10-8=2\,\text{h}\). 3. The sample variances are \(s_A^2=3.5\) and \(s_B^2=2.5\). The estimated standard error is \(\sqrt{\frac{3.5}{5}+\frac{2.5}{5}}=\sqrt{1.2}\approx1.0954\,\text{h}\). 4. The margin of error is \(2.31\sqrt{1.2}\approx2.5304\,\text{h}\), yielding the interval \(2\pm2.5304\), or approximately \((-0.53,4.53)\,\text{h}\). 5. Because the interval includes zero, the data do not give a clear direction for the population mean difference.

Answer

a) I: one-mean test (one mean compared with a target); II: paired-mean interval (repeat times for the same swimmers); III: independent two-mean interval (different students at two schools). b) \((-0.53,4.53)\,\text{h}\) for \(\mu_A-\mu_B\). c) The interval includes zero, so either school's population mean could be higher.

Confidence interval for a difference in means 67

Two independent random samples of measurements are shown below. Each source population contains more than \(100\) measurements. <table><tr><th>Group A</th><td>12.1</td><td>11.8</td><td>12.5</td><td>13.0</td><td>12.7</td><td>11.9</td><td>12.4</td><td>12.2</td></tr><tr><th>Group B</th><td>10.8</td><td>11.1</td><td>10.5</td><td>11.4</td><td>10.9</td><td>11.2</td><td>10.7</td></tr></table> Both sample distributions are roughly symmetric with no outliers. Use \(t^*=2.164\) to construct a \(95\%\) confidence interval for \(\mu_A-\mu_B\). Round the interval endpoints to two decimal places.

Hints

- Summarize each raw sample before building the interval. - Keep the group order consistent when finding the point estimate. - For independent groups, combine the two estimated variance contributions in the standard error.

Solution

1. The sample summaries are \(\bar{x}_A=12.325\), \(s_A\approx0.406\), \(\bar{x}_B\approx10.943\), and \(s_B\approx0.310\). 2. The estimated difference is \(12.325-10.943\approx1.382\). 3. The standard error is \(\sqrt{\frac{0.406^2}{8}+\frac{0.310^2}{7}}\approx0.185\). 4. The margin of error is \(2.164\cdot0.185\approx0.401\). 5. The interval is approximately \((0.98,1.78)\).

Answer

The \(95\%\) confidence interval for \(\mu_A-\mu_B\) is approximately \((0.98,1.78)\).

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Confidence interval for a difference in means

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