Free math worksheets

- 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.

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.

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.

- 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.

1. The margin of error is \(2.423(0.85)\approx2.060\) units. 2. The interval is \(2.7\pm2.060\), or approximately \((0.64,4.76)\) 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 confident at the level associated with the reported critical value that \(\mu_A-\mu_B\) lies between about \(0.64\) and \(4.76\) units.

The confidence interval for \(\mu_A-\mu_B\) is approximately \((0.64,4.76)\) units. At the stated confidence level, the data indicate that Procedure A's population mean measurement is about \(0.64\) to \(4.76\) units higher than Procedure B's.

- 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.

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.

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.

- 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.

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 \((3,7)\).

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

- 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.

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.

The student'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.

- 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.

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.

The student 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 \(\bar{x}_1-\bar{x}_2\).

- 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.

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.

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.

- 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.

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.

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.

- 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.

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 \((2.4,8.0)\), so reject \(H_0\). 3. For part b, \(6\) is inside \((2.4,8.0)\), so fail to reject \(H_0\).

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

- 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.

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: \((-1.2+5,2.4+5)=(3.8,7.4)\).

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.

- 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.

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.

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

- 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.

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, \(\bar{x}_B=10.0\).

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

- 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.

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.

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 \(6.8\,\text{kg}\).

- 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.

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

The confidence interval for \(\mu_A-\mu_B\) is \((0.5,6.5)\) minutes.

- 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.

1. The interval estimates a difference between two population means, not the distribution of pairwise differences between individual batteries. 2. Therefore, the reviewer’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.

The reviewer’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.

- 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.

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.

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

- 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.

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.

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

- 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.

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.

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.

- Determine whether an interval includes its endpoints. - Identify what the value \(0\) means for a difference in population means. - A claim of a strictly positive difference requires the entire interval to lie above \(0\).

1. The parameter value \(0\) corresponds to no difference between the population means. 2. Because \(0\) is an endpoint of the reported interval, it is included in the set of plausible values at the stated confidence level. 3. Therefore, the interval does not support a strictly positive difference at the \(95\%\) confidence level. 4. The fact that all other displayed values are positive does not remove the included null value.

The claim is not justified. Because \(0\) is included as an endpoint of the \(95\%\) confidence interval, the interval does not establish that \(\mu_A-\mu_B>0\).

- 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.

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.

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.

- 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.

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).

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

- 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.

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.

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

- 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.

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.

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.

- 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.

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.

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 \(2\).

- 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.

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.

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.

- 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.

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.

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.

- 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.

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.

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.

- 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. - Large differences in sample spreads may deserve context-specific attention, but they do not by themselves violate a validity condition for the Welch procedure.

1. The student'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. The relevant checks are independent random samples or random assignment, the \(10\%\) condition when sampling without replacement, and appropriate sample-distribution shapes or sufficiently large sample sizes. 4. Different sample standard deviations alone do not invalidate the procedure.

The student's claim is incorrect. Equal population standard deviations are not required for a Welch two-sample \(t\)-interval. Check randomization, the \(10\%\) condition when applicable, and the sample-data shape or sample-size condition instead.

- 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.

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.

\((0.5,3.9)\) 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 \(\mu_1>\mu_2\).

- 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.

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 \(\frac{3.1}{2.01}\approx1.542\).

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

- With equal 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.

1. For Design A, the standard error is \(\sqrt{\frac{10^2}{50}+\frac{10^2}{50}}=2\). 2. For Design B, the 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.

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

- 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.

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.

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.

- 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.

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.

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 \(5\,\text{kN}\).

- 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.

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.

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.

- 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.

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 \((-0.28,1.48)\,\text{cm}\).

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

- 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.

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 \((0.89,5.11)\,\text{dB}\).

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

- 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.”

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.

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

- 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 \(n\).

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 \(n\ge200\).

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

- 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.

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 \((0.86,5.74)\).

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

- 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.

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 \(s_2=8\).

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

- 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 \(0\).

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.

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

- 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.

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.

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.

- 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.

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.

\((2,8)\) 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.

- 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.

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 \((2.67,9.33)\).

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

- 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.

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 \((0.59,5.41)\).

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

- 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.

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.

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.

- 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.

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. With 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 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.

Original interval: approximately \((-0.69,8.69)\). Redesigned-system interval: approximately \((1.66,6.34)\). Reducing within-group variability makes the interval substantially narrower.

- 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.

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 confidence interval is \(5\pm4.128\), or approximately \((0.87,9.13)\).

The 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.

- 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.

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.

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.

- 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.

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 \((0.98,1.78)\).

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