- Notice whether each question is about a mean or a proportion. - Decide whether the question concerns only sampled outages or all the provider's reported outages. - A population-mean confidence interval has a different goal from describing a sample's smallest and largest observations.

1. Choice c) estimates the population mean length of the provider's reported outages. Each outage report is one observation, not each home. 2. Choice a) asks for a proportion, not a population mean. 3. Choice b) describes the range of observed outages, not an estimate of a population mean.

c). Choice a) concerns a proportion; b) concerns the observed sample range.

- Identify the quantities needed for the center and margin of error. - The stated confidence level and sample size determine the multiplier. - What sample summary estimates the unknown population spread?

1. A one-sample \(t\)-confidence interval uses \(\bar{x}\pm t^*\frac{s}{\sqrt{n}}\). 2. The sample size determines \(df=59\), and the stated \(95\%\) confidence level determines the critical value \(t^*\). 3. The sample mean gives the center, but the sample standard deviation \(s\) is missing. 4. To construct the interval, Dr. Rojas still needs \(s\), the standard error, or equivalent variability information.

No. He also needs the sample standard deviation, standard error, or equivalent variability information. The confidence level is already given.

- Identify whether the data came from a sample or from the entire population of interest. - Confidence intervals quantify uncertainty created by sampling. - Ask whether there is any unsampled part of the stated population left to estimate.

1. A confidence interval is used to estimate an unknown population parameter from a sample. 2. Here, every member of the population of interest was measured, so the population mean is known directly. 3. The population mean commute time is \(26.7\) minutes, with no sampling uncertainty from estimating it using a subset.

No. Because the district measured the entire population of interest, the population mean is known exactly as \(26.7\) minutes; a sampling-based confidence interval is unnecessary.

- Distinguish the set of jars actually measured from all jars produced. - Which mean was calculated directly from observations? - Confidence intervals estimate a population quantity, not a value already calculated exactly.

1. The sample mean \(\bar{x}=502\,\text{g}\) is the point estimate. 2. The unknown population parameter is \(\mu\), the mean contents of all jars from the line. 3. The sample mean is known from the collected data, while the population mean is being estimated.

Point estimate: \(\bar{x}=502\,\text{g}\). Parameter: the unknown population mean \(\mu\) of jar contents. The sample mean is known.

- Decide whether the question concerns individuals or a sample average. - Use the sample size when turning a sample standard deviation into a standard error. - Interpret both spreads before comparing their numerical values.

1. The estimated standard error is \(SE=\frac{s}{\sqrt{n}}=\frac{10}{\sqrt{25}}=2\) points. 2. The sample standard deviation describes spread among individual scores. Standard error estimates how much the sample mean varies across repeated samples of the same size.

\(SE=2\) points. It measures uncertainty in the sample mean, not the spread of individual scores.

- Think about how many observations can vary freely once their sample mean is fixed. - The confidence level tells you which column to use. - Choose the table row for the degrees of freedom, not simply for the sample size.

1. A one-sample \(t\)-interval uses \(df=n-1\). 2. With \(n=16\), the degrees of freedom are \(15\). 3. The corresponding \(95\%\) critical value is \(t^*=2.131\).

Use \(df=15\) and \(t^*=2.131\).

- Look for the multiplier that determines how many standard errors the interval extends from its center. - A margin of error should be expressed in the original measurement units. - This step does not require the sample mean.

1. The margin of error equals the critical value times the standard error. 2. \(ME=2.10\cdot1.8\,\text{min}=3.78\,\text{min}\). 3. The multiplier is unitless, so the margin retains the units of the measured time.

\(ME=3.78\,\text{min}\).

- The margin of error describes an equal distance on either side of the point estimate. - Keep the endpoints in increasing order. - Check that their average equals the reported center.

1. Subtract the margin of error from the estimate for the lower endpoint: \(28.4-1.6=26.8\). 2. Add it for the upper endpoint: \(28.4+1.6=30.0\). 3. Both endpoints are the same distance from \(28.4\), so the interval is centered on the point estimate.

The interval is \((26.8,30.0)\) minutes. Its midpoint is \(28.4\,\text{min}\) because both endpoints are \(1.6\,\text{min}\) from the point estimate.

- Look at the center and half-width of the reported interval. - Identify the population parameter that the interval was designed to estimate. - Distinguish a statement about a population mean from a statement about individual observations.

1. The point estimate is the midpoint: \(\frac{14.2+17.8}{2}=16.0\,\text{min}\). 2. The margin of error is half the width: \(\frac{17.8-14.2}{2}=1.8\,\text{min}\). 3. The interval estimates a population mean, not the proportion of individual waiting times in a range. 4. A correct interpretation is that we are \(92\%\) confident the population mean weekday morning shuttle wait for students at the college is between \(14.2\) and \(17.8\) minutes.

a) Point estimate: \(16.0\,\text{min}\); margin of error: \(1.8\,\text{min}\). b) The interval concerns the population mean, not individual waits. We are \(92\%\) confident that the population mean weekday morning shuttle wait is between \(14.2\) and \(17.8\) minutes.

- Intervals built from the same sample should be centered at the same sample statistic. - Compare how far each endpoint lies from that common center. - Greater confidence requires covering a broader range of plausible parameter values.

1. Both intervals have midpoint \(35.4\), so the point estimate is \(\bar{x}=35.4\). 2. For \((34.1,36.7)\), the margin of error is \(\frac{36.7-34.1}{2}=1.3\). 3. For \((33.5,37.3)\), the margin of error is \(\frac{37.3-33.5}{2}=1.9\). 4. With the same sample information, a higher confidence level requires a larger critical value and therefore a wider interval. 5. Thus, \((34.1,36.7)\) is the \(90\%\) interval and \((33.5,37.3)\) is the \(99\%\) interval.

The point estimate is \(35.4\) for both intervals. The \(90\%\) interval is \((34.1,36.7)\) with margin of error \(1.3\). The \(99\%\) interval is \((33.5,37.3)\) with margin of error \(1.9\). Higher confidence requires a wider interval when the sample stays the same.

- Look at how many people are in the study, not just how many measurements were recorded. - Ask whether each observation under one interface has a natural partner under the other interface. - Check both the sampling basis and the distribution of the paired differences before choosing the interval.

1. Each illustrator contributes two linked measurements, so form one completion-time difference for each illustrator. 2. Let \(\mu_d\) be the population mean within-illustrator difference in completion time, with \(d=A-B\). 3. The simple random sample and the approximately symmetric, outlier-free distribution of differences support a one-sample \(t\)-interval for \(\mu_d\). 4. A two-sample interval assumes independent groups, but the two observations from the same illustrator are dependent and were intentionally paired.

Use a one-sample \(t\)-interval for \(\mu_d\), the population mean paired difference in completion time for association members, with \(d=A-B\). A two-sample interval is inappropriate because the A and B times from the same illustrator are not independent.

- Ask whether the population spread is known or estimated from the same small sample. - The relevant reference distribution changes when extra uncertainty comes from estimating that spread. - Small degrees of freedom make the tails of the reference distribution more pronounced.

1. Because the population standard deviation is unknown, the sample standard deviation must estimate it, so a \(t\)-distribution is appropriate. 2. With \(n=9\), the degrees of freedom are \(df=8\). 3. For a \(95\%\) confidence interval with \(df=8\), \(t^*\approx2.306\), which is larger than \(1.96\). 4. The larger critical value reflects additional uncertainty from estimating the population standard deviation and produces a wider interval than using \(1.96\).

Use a \(t\) critical value with \(df=8\). For \(95\%\) confidence, \(t^*\approx2.306\), which is larger than \(1.96\), so the appropriate interval is wider than Jiwoo's proposed normal-based interval.

- The software output already gives the two ingredients needed to determine the interval's half-width. - Keep the interval centered at the sample mean. - The interpretation should refer to the population mean completion time, not to individual participants.

1. The margin of error is \(1.699(1.25)=2.12375\,\text{min}\). 2. The confidence interval is \(63.7\pm2.12375\), or approximately \((61.58,65.82)\,\text{min}\). 3. We are \(90\%\) confident that the population mean module-completion time is between about \(61.58\) and \(65.82\) minutes.

The \(90\%\) confidence interval is approximately \((61.58,65.82)\,\text{min}\). We are \(90\%\) confident that the population mean completion time lies between about \(61.58\) and \(65.82\) minutes.

- Compare the desired margin of error with the current one as a ratio. - Precision changes with the square root of sample size rather than directly with sample size. - Determine what sample-size factor produces the desired change in interval half-width.

1. With confidence level and variability held about constant, margin of error is approximately proportional to \(1/\sqrt{n}\). 2. Reducing the margin of error from \(4.2\) to \(2.1\) cuts it in half. 3. To cut the margin of error in half, multiply the sample size by \(4\). 4. The required sample size is approximately \(4(36)=144\).

Approximately \(144\) observations are needed. Halving the margin of error requires about four times the sample size when the confidence level and variability stay about the same.

- Separate conditions about how the data were collected from conditions about the shape of the observed data. - Ask whether every student in the target population had a sampling mechanism that supports population inference. - Increasing sample size does not automatically remove selection bias.

1. A valid confidence interval for the university population mean requires an appropriate randomization condition, such as a random sample from that population. 2. The first \(50\) people leaving one recreation center at one time form a convenience sample, not a random sample of university students. 3. The sample size and distribution shape address the sample-data condition, but they do not remove selection bias or create random sampling. 4. Therefore, the interval procedure is not justified for inference to the full university population from this sampling method.

The interval is not justified for the full university population because the sample is a convenience sample rather than a random sample. A large, well-shaped sample does not repair a nonrandom sampling design.

- Count the independent data values that actually enter the one-sample analysis after pairing. - Each pair produces one difference, which is the observation used for inference. - Degrees of freedom should be based on the number of those differences.

1. A matched-pairs interval analyzes one sample consisting of the \(14\) within-person differences. 2. The original \(28\) measurements are not treated as \(28\) independent observations in the inference procedure. 3. A one-sample \(t\)-interval based on \(n=14\) differences uses \(df=n-1=13\).

The correct degrees of freedom are \(13\). The analysis uses the \(14\) paired differences as one sample, not the \(28\) original measurements as independent observations.

- Identify the parameter the interval was designed to estimate. - Distinguish uncertainty about a population mean from variability among individual observations. - State the conclusion about the parameter rather than about individual customers.

1. The interval was constructed to estimate a population mean, not the distribution of individual customer times. 2. Individual times can vary much more widely than the uncertainty in the estimated mean, so the confidence interval does not describe where \(95\%\) of individual observations fall. 3. The correct interpretation is that we are \(95\%\) confident that the population mean time customers spend in the store lies between \(31.4\) and \(35.8\) minutes.

Keisha's interpretation is incorrect because the interval estimates the population mean, not the range containing \(95\%\) of individual customer times. We are \(95\%\) confident that the population mean lies between \(31.4\) and \(35.8\) minutes.

- Apply the same temperature conversion to both endpoints of the interval. - Because the conversion increases as Fahrenheit temperature increases, the endpoint order stays the same. - Changing units does not change the confidence level of the interval.

1. Convert the lower endpoint: \(\frac{5}{9}(68-32)=20\) degrees Celsius. 2. Convert the upper endpoint: \(\frac{5}{9}(74-32)=\frac{70}{3}\) degrees Celsius. 3. A one-to-one linear unit conversion changes the numerical endpoints but not the confidence level. 4. The converted \(95\%\) confidence interval is exactly \(\left(20,\frac{70}{3}\right)\) degrees Celsius, or approximately \((20.00,23.33)\) degrees Celsius. 5. We are \(95\%\) confident that the population mean afternoon classroom temperature lies in this converted interval.

The exact interval is \(\left(20,\frac{70}{3}\right)\) degrees Celsius, or approximately \((20.00,23.33)\) degrees Celsius. We are \(95\%\) confident that the population mean afternoon classroom temperature lies in this interval.

- Think about the general form of a one-sample confidence interval for a mean. - Identify what value must lie exactly at the center of the interval. - Compare that required center with the midpoint of the reported endpoints.

1. A standard one-sample \(t\)-confidence interval has the form \(\bar{x}\pm t^*\frac{s}{\sqrt{n}}\), so it is symmetric about the sample mean. 2. The midpoint of the reported interval is \(\frac{45.2+54.0}{2}=49.6\). 3. Since \(49.6\ne50.0\), the reported interval is not centered at the sample mean. 4. Therefore, it cannot be a correctly calculated standard one-sample \(t\)-confidence interval from this sample.

No. The interval's midpoint is \(49.6\), not the sample mean \(50.0\), so it cannot be the standard one-sample \(t\)-confidence interval from that sample.

- Use the random-sampling and sample-shape information to check the small-sample procedure. - Use the sample size to find the standard error from the sample standard deviation. - The critical value times the standard error gives the distance from the sample mean to each endpoint.

1. With \(2.5\%\) in each tail, the central area is \(100\%-5\%=95\%\), so this is a \(95\%\) confidence interval. The random sample and stated shape support the small-sample \(t\)-procedure. 2. The standard error is \(\frac{6}{\sqrt{11}}\approx1.809\). 3. The margin of error is \(2.228\cdot1.809\approx4.031\). 4. The interval is \(32\pm4.031\), or approximately \((27.97,36.03)\).

The confidence level is \(95\%\), and the confidence interval is approximately \((27.97,36.03)\).

- Start with the relationship among margin of error, critical value, sample standard deviation, and sample size. - Treat the sample size as the unknown in the standard-error term. - Check that the resulting sample size reproduces the stated margin of error.

1. The margin of error satisfies \(4=2.00\left(\frac{10}{\sqrt{n}}\right)\). 2. This gives \(4=\frac{20}{\sqrt{n}}\), so \(\sqrt{n}=5\). 3. Therefore, \(n=25\).

The sample size was \(25\).

- The shape requirement depends on sample size. - Distinguish moderate skewness from extreme skewness or influential outliers. - Consider how a moderately large sample changes the effect of moderate skewness on the sampling distribution of the mean.

1. For a one-sample \(t\)-interval, a small sample needs stronger evidence that the data are free from problematic skewness and outliers. 2. Here, \(n=45\) is large enough for the sampling distribution of the sample mean to be approximately normal under ordinary moderate skewness. 3. The absence of extreme outliers further supports use of the procedure. 4. Therefore, Avery's claim is too restrictive; the stated shape does not prevent using a one-sample \(t\)-interval.

The claim is incorrect. With \(n=45\), moderate right skewness and no extreme outliers are compatible with using a one-sample \(t\)-confidence interval, assuming the randomization and independence conditions also hold.

- The smaller the sample, the more important strong skewness and outliers become for \(t\) inference. - Look for observations separated from the main body of the data. - Consider how one extreme observation can affect both the sample mean and sample standard deviation.

1. The sample size is small, so the shape of the sample data matters for assessing whether a one-sample \(t\)-interval is appropriate. 2. The boxplot shows a clear high outlier well beyond the upper whisker. 3. With only \(12\) observations, that outlier can strongly affect the sample mean and sample standard deviation. 4. Therefore, the visual does not support using the standard small-sample \(t\)-interval without additional justification or investigation of the outlier.

No. With only \(12\) observations, the clear high outlier makes the standard one-sample \(t\)-confidence interval poorly supported by the sample shape.

- The degrees of freedom for a one-sample \(t\) procedure come from estimating one population spread with the sample. - Find the standard error from the sample standard deviation and sample size. - Use the critical value supplied for the corrected degrees of freedom.

1. With the interval conditions assumed satisfied, a one-sample \(t\)-interval with \(n=20\) uses \(df=n-1=19\), not \(20\). 2. The standard error is \(\frac{12}{\sqrt{20}}\approx2.683\). 3. The margin of error is \(2.093\cdot2.683\approx5.616\). 4. The \(95\%\) confidence interval is \(80\pm5.616\), or approximately \((74.38,85.62)\).

Use \(19\) degrees of freedom. The \(95\%\) confidence interval is approximately \((74.38,85.62)\).

- Each patient contributes one after-minus-before difference. - Would knowing that two patients have similar before measurements tell you how similar their changes are? - Think about what you could calculate if you had every patient's two measurements.

1. The sample mean paired difference is \(\bar{d}=38-42=-4\). 2. A paired \(t\)-interval requires the sample standard deviation of the \(30\) within-patient differences. 3. Separate standard deviations of \(8\) and \(7\) do not determine the spread of the paired differences because the within-patient association matters. 4. The \(95\%\) level and \(29\) degrees of freedom determine \(t^*\), but the difference standard deviation is missing.

No. The mean change is \(-4\), but the sample standard deviation of the paired changes, or equivalent information from the raw pairs, is still needed.

- List the quantities that determine the margin of error for a one-sample \(t\)-interval. - Decide which of those quantities stay fixed when the sample changes. - Sample-to-sample variation affects more than just the sample mean.

1. For a fixed sample size and confidence level, the degrees of freedom and corresponding \(t\)-critical value are fixed. 2. However, the sample standard deviation \(s\) generally changes from one random sample to another. 3. The estimated standard error depends on \(s\), so the margin of error and interval width also vary across samples.

The intervals need not have the same width. Even with the same sample size and confidence level, different samples generally have different sample standard deviations, producing different standard errors and margins of error.

- For a small sample, first check whether the distribution is reasonably symmetric with no outliers. - Distinguish the spread of individual observations from the uncertainty of a sample mean. - Combine the estimated standard error with the supplied \(t\)-critical value.

1. The random sample and the stated sample shape support a one-sample \(t\)-interval. The sample standard deviation \(8\) describes variability among individual observations, not the uncertainty of the sample mean. 2. The standard error of the mean is \(8/\sqrt{16}=2\). 3. The margin of error is \(2.131\cdot2=4.262\). 4. The confidence interval is \(40\pm4.262\), or approximately \((35.74,44.26)\).

Leila used the sample standard deviation instead of the standard error and omitted the \(t\)-critical value. The correct interval is approximately \((35.74,44.26)\).

- Compare each claimed threshold with the entire confidence interval, not just its midpoint. - A lower-bound claim is strongest when the whole interval lies above that bound. - A value inside the interval remains plausible at the stated confidence level.

1. Every value in the interval is greater than \(10\), so the interval supports the conclusion that the population mean exceeds \(10\). 2. The value \(11\) lies inside the interval, and the interval also includes plausible mean values below \(11\), such as \(10.5\). 3. Therefore, this \(95\%\) interval does not establish that the population mean exceeds \(11\).

The interval supports \(\mu>10\), because its lower endpoint is \(10.2\). It does not establish \(\mu>11\), because plausible values in the interval lie on both sides of \(11\).

- Distinguish a data error from a genuine but unusual observation. - Ask whether the exclusion rule existed before the result was inspected. - A narrower interval is not automatically a more trustworthy interval.

1. A valid observation is part of the random sample and contains information about the population being studied. 2. Removing it because of its effect on the interval makes the analysis depend on the observed result rather than on a justified data-quality or pre-specified rule. 3. The unusual value may also indicate that the population shape is problematic for a small-sample \(t\)-procedure, which should be addressed rather than hidden by deletion. 4. Dr. Suzuki should retain the value unless there is a defensible reason for exclusion and should assess the procedure's conditions and sensitivity transparently.

Deleting a valid observation merely to narrow the interval is not justified. Precision should come from the study design and data, not from removing legitimate values after seeing how they affect the result.

- Convert the confidence level into a long-run miss rate. - Apply that rate to the number of repeated studies. - An expected count is a long-run average, not a guaranteed outcome in one batch.

1. A \(95\%\) confidence procedure captures the true population mean in about \(95\%\) of repeated random samples. 2. Therefore, it misses the true mean in about \(5\%\) of repetitions. 3. Among \(100\) independent repetitions, the expected number of misses is \(100\cdot0.05=5\). 4. The actual number varies randomly from one set of \(100\) repetitions to another, so exactly \(5\) misses is not guaranteed.

About \(5\) intervals are expected to miss the true mean, but the actual number can be higher or lower because interval coverage varies randomly across repeated samples.

- Compare the half-widths of the two intervals. - Identify which factors in the margin of error are identical for the two samples. - The remaining factor determines the ratio of their sample standard deviations.

1. Interval A has margin of error \((52-48)/2=2\). 2. Interval B has margin of error \((51-45)/2=3\). 3. With the same sample size and critical value, the margin of error is directly proportional to the sample standard deviation. 4. Therefore, \(s_B/s_A=3/2=1.5\).

Sample B had the larger sample standard deviation. Its sample standard deviation was \(1.5\) times that of Sample A.

- Hold the sample mean and standard error fixed because the data are the same. - Compare the critical values required by the two confidence levels. - Greater confidence requires covering a wider range of plausible parameter values.

1. Both intervals are based on the same sample, so they must have the same center at the sample mean. 2. Raising the confidence level from \(90\%\) to \(95\%\) requires a larger \(t\)-critical value. 3. With the same standard error, the larger critical value produces a larger margin of error. 4. Therefore, the \(95\%\) interval must be wider than the \(90\%\) interval, not narrower as reported.

The intervals cannot both be correct. For the same sample, a \(95\%\) \(t\)-confidence interval must be wider than the corresponding \(90\%\) interval.

- For a small sample, first check whether the shape supports a one-sample \(t\)-interval. - Identify the statistic that estimates the population mean; a benchmark is used only for comparison. - Build the margin of error from the sample's standard error and the supplied critical value.

1. The random sample and stated sample shape support a one-sample \(t\)-interval. A confidence interval estimates the population mean from the sample, so its point estimate is the sample mean \(\bar{x}=53\). 2. The standard error is \(10/\sqrt{25}=2\). 3. The margin of error is \(2.064\cdot2=4.128\). 4. The confidence interval is \(53\pm4.128\), or approximately \((48.87,57.13)\). 5. The benchmark \(50\) can be compared with the finished interval, but it does not determine the interval's center.

The \(95\%\) confidence interval is approximately \((48.87,57.13)\), centered at the sample mean \(53\).

- Identify every factor that enters a one-sample \(t\) margin of error. - The sample size is only one source of interval width. - A larger sample can be offset by greater sample variability.

1. The width of a one-sample \(t\)-interval depends on the product of the \(t\)-critical value and the estimated standard error. 2. The estimated standard error depends on both the sample standard deviation and the sample size. 3. Different combinations of sample size and sample variability can therefore produce the same margin of error. 4. The \(t\)-critical value can also differ when the sample sizes, and hence degrees of freedom, differ.

Equal widths do not imply equal sample sizes. Different sample standard deviations, sample sizes, and corresponding \(t\)-critical values can produce the same interval width.

- Apply the variable transformation to every plausible value of the original mean. - Multiplication by a negative constant reverses order on the number line. - Rewrite the transformed endpoints from smaller to larger value.

1. The population means satisfy \(\mu_Y=-2\mu_X\). 2. Multiplying the original endpoints by \(-2\) gives \(-6\) and \(-10\). 3. Because multiplication by a negative number reverses inequalities, the lower endpoint is \(-10\) and the upper endpoint is \(-6\). 4. Therefore, the transformed confidence interval is \((-10,-6)\).

The \(95\%\) confidence interval for \(\mu_Y\) is \((-10,-6)\).

- Compare how confidence level affects interval width. - Ask whether the reporting rule was fixed before the data were examined. - A procedure's stated coverage applies to the procedure actually chosen in advance, not to whichever result looks most favorable afterward.

1. The confidence level determines the procedure's long-run coverage and should be chosen for a substantive reason rather than selected to produce a desired conclusion. 2. A lower confidence level produces a narrower interval and is therefore more likely to exclude a benchmark. 3. Choosing the reported level only after inspecting which interval gives the preferred result makes the reporting rule data-dependent and overstates the strength of the evidence. 4. The pre-specified \(95\%\) interval should remain the primary interval, with any additional intervals clearly identified as secondary analyses.

Selecting the \(80\%\) interval only because it excludes the benchmark is a post hoc choice that makes the evidence look stronger than the pre-specified \(95\%\) analysis. The planned \(95\%\) interval should remain the primary result.

- Identify the experimental or observational unit that appears twice. - Ask whether each “before” value has a specific matching “after” value. - Express the target as the mean of one set of within-unit differences.

1. Each commuter provides a matched pair of measurements, so the analysis should use one difference per commuter, such as \(d=\text{before}-\text{after}\). 2. The population parameter is \(\mu_d\), the population mean of those paired travel-time differences. 3. The two measurements from the same commuter are related, so they are not independent observations from two separate groups. 4. A confidence interval should therefore be built from the sample of \(18\) paired differences, provided the conditions for inference about a mean difference are satisfied.

Analyze the \(18\) within-commuter differences. The parameter is \(\mu_d\), the population mean paired difference. The before and after measurements are linked within each commuter, so an independent two-sample analysis would ignore the pairing.

- Keep each student's before and after values together. - Apply the subtraction in the order given in the question. - A paired study analyzes one change per person.

1. Subtract before from after for each student: \(3\), \(2\), and \(4\). 2. The mean difference is \(\bar{d}=\frac{3+2+4}{3}=3\) problems. 3. A paired interval would estimate \(\mu_d\), the population mean after-minus-before change per student.

The differences are \(3,2,4\); their mean is \(\bar{d}=3\) problems. The parameter is \(\mu_d\), the population mean after-minus-before change.

- Begin by converting the sample standard deviation into uncertainty in the sample mean. - Use the given multiplier to find how far the interval extends from its center. - The interpretation should concern all boxes in the population, not individual box weights.

1. The estimated standard error is \(\frac{8}{\sqrt{16}}=2\,\text{lb}\). 2. The margin of error is \(2.131\cdot2=4.262\,\text{lb}\). 3. The interval is \(50\pm4.262\), or approximately \((45.74,54.26)\,\text{lb}\). 4. We are \(95\%\) confident that the population mean box weight lies between those values.

\(SE=2\,\text{lb}\), \(ME=4.262\,\text{lb}\), and the \(95\%\) interval is approximately \((45.74,54.26)\,\text{lb}\). We are \(95\%\) confident that the population mean box weight lies within this interval.

- The reported standard deviation is already for the within-person differences. - Keep the new-minus-old sign when finding the interval center. - Interpret a negative difference in terms of travel time.

1. Analyze the nine within-person changes as one sample, not as independent groups. 2. The standard error is \(\frac{3}{\sqrt{9}}=1\,\text{min}\). 3. The margin of error is \(2.306\,\text{min}\), so the interval is \(-2.5\pm2.306\), or approximately \((-4.81,-0.19)\,\text{min}\). 4. Negative values mean the new route takes less time on average.

The \(95\%\) paired-mean confidence interval is approximately \((-4.81,-0.19)\,\text{min}\). It suggests the new route reduces population mean travel time.

- Read the interval bands relative to the marked reference value. - An interval rules out a value only if that value is outside its endpoints. - A whole interval on one side of the target supports a directional comparison.

1. Panel a) runs from \(7\) to \(9\), entirely below T at \(10\). 2. Panel b) runs from \(8\) to \(12\) and contains T. 3. Therefore, a) both rules out the target and suggests the population mean wait is below it; b) does not rule the target out.

Panel a) rules out the target and lies entirely below it. Panel b) includes the target.

- The two travel times from the same route are linked, so keep each route's observations together. - Use the distribution of the route-by-route differences, not two independent groups. - Separate what the interval estimates from what the before/after design can establish causally.

1. Because the observations are paired by route, analyze the \(16\) route differences as one sample. The random sampling and shape conditions are satisfied. 2. Use a one-sample \(t\)-interval for the population mean difference with \(df=15\). For \(95\%\) confidence, \(t^*\approx2.131\). 3. The standard error is \(\frac{2.4}{\sqrt{16}}=0.6\,\text{min}\), so the margin of error is \(2.131(0.6)\approx1.279\,\text{min}\). 4. The interval is \(1.8\pm1.279\), or approximately \((0.52,3.08)\,\text{min}\). 5. We are \(95\%\) confident that, for routes like those sampled, the population mean old-minus-new travel-time difference is between about \(0.52\) and \(3.08\) minutes. The interval describes the before/after difference but does not by itself establish that the signal plan caused it.

The \(95\%\) confidence interval is approximately \((0.52,3.08)\,\text{min}\) for \(\mu_d\), where \(d=\text{old time}-\text{new time}\). We are \(95\%\) confident that the population mean before/after difference lies in this range for routes like those sampled; the uncontrolled before/after design does not establish causation.

- The population standard deviation is not given, so use the sample's estimate of variability appropriately. - A small sample makes the shape information relevant before constructing the interval. - Build the interval around the sample mean using a confidence-level multiplier and the estimated sampling uncertainty.

1. The random sample and sample shape support a one-sample \(t\)-interval with \(df=11\). 2. For \(98\%\) confidence, \(t^*\approx2.718\). 3. The standard error is \(\frac{1.5}{\sqrt{12}}\approx0.433\,\text{min}\), so the margin of error is \(2.718(0.433)\approx1.177\,\text{min}\). 4. The interval is \(8.6\pm1.177\), or approximately \((7.42,9.78)\,\text{min}\). 5. We are \(98\%\) confident that the population mean restoration time for jobs like those sampled is between about \(7.42\) and \(9.78\) minutes.

The \(98\%\) confidence interval is approximately \((7.42,9.78)\,\text{min}\). We are \(98\%\) confident that the population mean restoration time lies in this interval.

- Focus on what changes from sample to sample and what stays fixed. - Confidence level is a property of a procedure carried out repeatedly. - Distinguish the randomness present before sampling from the fixed interval obtained afterward.

1. The confidence level describes the long-run success rate of the interval-building procedure under repeated random sampling. 2. If the procedure is repeated many times under the same conditions, approximately \(95\%\) of the resulting intervals will capture the fixed population mean. 3. After one sample has been observed and one interval has been calculated, both endpoints are fixed and the population mean is also fixed. 4. The particular interval either contains the population mean or it does not; the \(95\%\) refers to the procedure's repeated-sampling performance, not a probability assigned to the fixed parameter after calculation.

In repeated random sampling, about \(95\%\) of intervals produced by the procedure would contain the population mean. For one already-computed interval, the population mean is fixed and the interval is fixed, so the frequentist \(95\%\) confidence level is not a \(95\%\) probability that the parameter lies in that particular interval.

- Keep the definition of the paired difference visible when interpreting negative values. - Use the sample of within-person changes as the single sample for inference. - Distinguish evidence of a before/after population mean difference from evidence that the change itself caused that difference.

1. Analyze the \(30\) paired differences using a one-sample \(t\)-interval with \(df=29\). 2. For \(95\%\) confidence, \(t^*\approx2.045\), and the standard error is \(\frac{6.0}{\sqrt{30}}\approx1.095\,\text{min}\). 3. The margin of error is \(2.045(1.095)\approx2.240\,\text{min}\). 4. The confidence interval is \(-2.4\pm2.240\), or approximately \((-4.64,-0.16)\,\text{min}\). 5. Because the entire interval is negative for \(d=\text{after}-\text{before}\), the data support a lower population mean task time after the workflow change than before it. This before/after comparison alone does not establish that the workflow change caused the decrease.

The \(95\%\) confidence interval is approximately \((-4.64,-0.16)\,\text{min}\). Since the interval is entirely negative for \(d=\text{after}-\text{before}\), it supports lower population mean task time after the workflow change than before it, but it does not by itself establish causation.

- Interval half-width depends on more than one factor. - Larger samples affect both estimated sampling variability and degrees of freedom. - Consider how each of those changes influences the margin of error.

1. With similar sample standard deviations, Lucas's larger sample gives a smaller standard error because the estimated sampling spread decreases as \(n\) increases. 2. Lucas also has more degrees of freedom, so the \(95\%\) \(t\) critical value is smaller and closer to the standard normal critical value. 3. Both the smaller standard error and the smaller critical value reduce the margin of error. 4. Therefore, Lucas's confidence interval should usually be narrower.

Lucas's interval should usually be narrower because \(n=40\) gives a smaller standard error and a smaller \(t\) critical value than \(n=10\), assuming the sample standard deviations are similar.

- The data and estimated standard error do not change when only the confidence level changes. - Compare the two critical values as a multiplicative factor. - Keep the new interval centered at the same sample mean.

1. The sample and standard error are unchanged, so the margin of error changes in proportion to the critical value. 2. The \(99\%\) margin of error is \(2.4\left(\frac{2.797}{2.064}\right)\approx3.25\). 3. The \(99\%\) interval is \(50.6\pm3.25\), or approximately \((47.35,53.85)\). 4. The \(99\%\) interval is wider because higher confidence requires a larger critical value.

The \(99\%\) margin of error is approximately \(3.25\), giving a confidence interval of approximately \((47.35,53.85)\).

- Decide which distribution should provide the critical value when the population standard deviation is not known. - The number of observations determines the degrees of freedom for this interval. - Keep the sample mean as the center and recompute only the uncertainty term with the appropriate critical value.

1. The population standard deviation is unknown, so the interval should use a \(t\)-critical value rather than the normal critical value \(1.645\). 2. The random-sample and sample-shape conditions support a one-sample \(t\)-interval with \(df=17\). 3. For \(90\%\) confidence and \(17\) degrees of freedom, \(t^*\approx1.740\). 4. The standard error is \(\frac{5.4}{\sqrt{18}}\approx1.273\,\text{kWh}\), so the margin of error is \(1.740(1.273)\approx2.214\,\text{kWh}\). 5. The confidence interval is \(21.6\pm2.214\), or approximately \((19.39,23.81)\,\text{kWh}\).

The proposed interval incorrectly uses a normal critical value. The appropriate \(90\%\) one-sample \(t\)-confidence interval is approximately \((19.39,23.81)\,\text{kWh}\).

- The center of a confidence interval identifies the point estimate used to build it. - The distance from the center to either endpoint is the margin of error. - Relate that margin of error to the sample size, critical value, and unknown sample standard deviation.

1. The sample mean is the midpoint of the confidence interval: \(\bar{x}=\frac{102.4+107.6}{2}=105.0\) seconds. 2. The margin of error is half the interval width: \(ME=\frac{107.6-102.4}{2}=2.6\) seconds. 3. For a one-sample \(t\)-interval, \(ME=t^*\frac{s}{\sqrt{n}}\). 4. Solving for \(s\) gives \(s=\frac{2.6\sqrt{25}}{2.064}\approx6.30\) seconds.

The sample mean is \(105.0\) seconds, and the sample standard deviation is approximately \(6.30\) seconds.

- Confidence level affects the critical value, while sample size affects the standard error. - Compute the two uncertainty terms rather than assuming higher confidence always means a wider interval across different studies. - Compare the resulting margins of error directly.

1. For Plan A, the margin of error is \(1.711\left(\frac{8}{\sqrt{25}}\right)=1.711(1.6)\approx2.738\). 2. For Plan B, the margin of error is \(2.626\left(\frac{8}{\sqrt{100}}\right)=2.626(0.8)\approx2.101\). 3. Plan B uses a larger critical value because it has higher confidence, but its much larger sample produces a smaller standard error. 4. Therefore, Plan B has the smaller margin of error and the narrower expected interval.

Plan A has expected margin of error approximately \(2.738\), and Plan B has expected margin of error approximately \(2.101\). Plan B is narrower despite using \(99\%\) confidence.

- Summarize the raw sample with its mean and standard deviation first. - The sample is small, so use the information about the sample's shape when deciding whether \(t\) inference is appropriate. - Build the interval by combining the point estimate with its margin of error.

1. The sample mean is \(\bar{x}\approx5.044\,\text{kWh}\), and the sample standard deviation is \(s\approx0.230\,\text{kWh}\). 2. With \(n=9\), use a one-sample \(t\)-interval with \(df=8\). For \(90\%\) confidence, \(t^*\approx1.860\). 3. The standard error is \(\frac{0.230}{\sqrt{9}}\approx0.0766\,\text{kWh}\). 4. The margin of error is \(1.860\cdot0.0766\approx0.142\,\text{kWh}\). 5. The interval is approximately \((4.90,5.19)\,\text{kWh}\).

The \(90\%\) confidence interval for the population mean daily energy use is approximately \((4.90,5.19)\,\text{kWh}\).

- Use the visual to check for strong skewness or outliers before relying on a small-sample \(t\) procedure. - Compute the estimated sampling variability from the sample standard deviation and sample size. - Combine the sample mean with the critical value and standard error.

1. The boxplot is roughly symmetric and shows no outliers, so the sample-shape condition is reasonable for a small-sample one-sample \(t\)-interval. 2. The standard error is \(\frac{3.1}{\sqrt{14}}\approx0.829\,\text{s}\). 3. The margin of error is \(2.160\cdot0.829\approx1.790\,\text{s}\). 4. The \(95\%\) confidence interval is \(50.2\pm1.790\), or approximately \((48.41,51.99)\,\text{s}\).

Yes. The boxplot is roughly symmetric with no outliers, so a one-sample \(t\)-interval is reasonable. The \(95\%\) confidence interval is approximately \((48.41,51.99)\,\text{s}\).

- With two observations, calculate the sample mean and sample standard deviation carefully. - The population-normality statement matters because the sample is extremely small. - Notice how the very small degrees of freedom affect the critical value and margin of error.

1. The sample mean is \(\bar{x}=12\), and the sample standard deviation is \(s=\sqrt{8}\approx2.828\). 2. The standard error is \(\frac{2.828}{\sqrt{2}}=2\). 3. The margin of error is \(12.706\cdot2=25.412\). 4. The confidence interval is \(12\pm25.412\), or approximately \((-13.41,37.41)\). 5. The normal-population condition makes the \(t\) procedure valid, but with only \(2\) observations the critical value and resulting uncertainty are extremely large.

The \(95\%\) confidence interval is approximately \((-13.41,37.41)\). The interval is extremely wide because the sample provides very little information about the population mean.

- Identify whether observations in the two conditions come from the same individuals. - Use both the sampling design and the stated shape of the within-student differences when checking the small-sample conditions. - Base the interval on the variability among the changes, not on separate uncertainty calculations for the two conditions.

1. Each after measurement is naturally linked to the before measurement from the same student, so the analysis should use the within-student differences. The random sample and stated shape support a small-sample one-sample \(t\)-interval for those differences. 2. The standard error is \(\frac{2.1}{\sqrt{12}}\approx0.606\,\text{ms}\). 3. The margin of error is \(2.201\cdot0.606\approx1.334\,\text{ms}\). 4. The confidence interval is \(-3.4\pm1.334\), or approximately \((-4.73,-2.07)\,\text{ms}\). 5. Separate intervals would ignore the within-student association and would not directly estimate the population mean change.

The \(95\%\) confidence interval for the population mean paired difference is approximately \((-4.73,-2.07)\,\text{ms}\). The paired analysis is appropriate because each student supplies both measurements, so the relevant observations are the within-student differences.

- Use the sign of the entire interval to interpret the direction of the mean paired effect. - Separate the role of random assignment from the role of random sampling. - Causal conclusions and population generalizations require different features of a study design.

1. The entire interval is negative, so the data support a lower mean reaction time under caffeine than under placebo for the population represented by the experiment. 2. Random assignment of treatment order supports a causal interpretation of the caffeine-placebo difference under the experimental conditions. 3. Because the participants were volunteers rather than a random sample from all adults, the interval does not by itself justify generalizing the estimated effect to all adults.

The interval supports a causal decrease in mean reaction time of about \(12\) to \(20\) milliseconds under caffeine for the experimental setting. The volunteer sample limits how broadly that effect can be generalized to all adults.

- The matching determines the observational unit used for inference, so work with one difference from each pair. - For a small number of pairs, check the shape of the paired differences before using a \(t\)-interval. - Use the variability among the pairwise differences to measure uncertainty.

1. The matched design is analyzed through the \(15\) pairwise differences, not as \(30\) independent plot yields. The random selection of pairs and stated shape of the differences support the small-sample paired \(t\)-procedure. 2. The standard error is \(\frac{3.5}{\sqrt{15}}\approx0.904\,\text{kg}\). 3. The margin of error is \(2.145\cdot0.904\approx1.938\,\text{kg}\). 4. The confidence interval is \(2.8\pm1.938\), or approximately \((0.86,4.74)\,\text{kg}\). 5. We are \(95\%\) confident that the population mean matched-pair yield difference, A minus B, for comparable plot pairs on this farm lies between about \(0.86\,\text{kg}\) and \(4.74\,\text{kg}\).

The \(95\%\) confidence interval is approximately \((0.86,4.74)\,\text{kg}\) for \(\mu_d\). We are \(95\%\) confident that the population mean matched-pair yield difference, A minus B, lies in this interval.

- Recover the margin of error from the interval endpoints and center. - Compare the margin of error with the standard error to infer the critical value. - Match the inferred critical value to the supplied choices.

1. The interval has margin of error approximately \(33.48-30=3.48\). 2. The standard error is \(6/\sqrt{10}\approx1.897\). 3. The implied critical value is \(3.48/1.897\approx1.834\). 4. This matches \(t^*=1.833\), so the interval is the \(90\%\) confidence interval.

The reported interval is a \(90\%\) confidence interval.

- Compare the calculated interval with the range the parameter is known to occupy. - Could a possible true mean lie in one interval but not the other? - Decide whether an impossible endpoint alone tells you the shape or sample size was inappropriate.

1. The standard \(t\)-calculation produces \((-5,20)\). The parameter is known to lie in \([0,100]\). 2. Since the true parameter is inside \([0,100]\), it is contained in \((-5,20)\) exactly when it is contained in \((-5,20)\cap[0,100]=[0,20)\). 3. Intersecting with a known parameter range therefore preserves the original procedure's coverage for all possible true parameter values. The restricted interval is not the unmodified standard \(t\)-interval. 4. A negative endpoint is physically impossible as a mean, but it does not by itself prove that the \(t\)-method is invalid. The random-sampling and distribution-shape conditions still need to be checked.

Restricting the interval to \([0,20)\) does not change whether it contains any possible true population mean. The negative endpoint alone does not prove that the \(t\)-approximation is unsuitable; its conditions must be checked separately.

- In a matched design, each pair contributes one difference; check the sampling and shape conditions for those differences. - Compare the spread of those differences with the spread of the raw measurements. - Shared variation within a pair can cancel when a difference is formed.

1. The random selection and stated shape of the paired differences support a paired \(t\)-interval. The paired standard error is \(4/\sqrt{30}\approx0.730\). 2. The paired margin of error is \(2.045\cdot0.730\approx1.493\). 3. The paired confidence interval is \(2.5\pm1.493\), or approximately \((1.01,3.99)\). 4. If the two sets of measurements were incorrectly treated as independent with standard deviations \(10\) and \(10\), the estimated standard error would be \(\sqrt{10^2/30+10^2/30}\approx2.582\), much larger than the paired standard error. 5. Matching can remove shared within-pair variation, so inference based on the differences can be substantially more precise.

The \(95\%\) paired confidence interval is approximately \((1.01,3.99)\). The paired standard error is approximately \(0.730\), compared with approximately \(2.582\) if the matching were incorrectly ignored.

- For a paired procedure, inspect the distribution of the differences rather than the two original measurements separately. - Consider both the sample size and whether the plot shows an isolated extreme value. - Build the standard error from \(s_d\) and the number of pairs.

1. The histogram is somewhat right-skewed, but the sample size is \(40\) and there is no isolated extreme outlier, so the displayed shape does not by itself make a paired \(t\)-interval unreasonable. 2. The standard error is \(\frac{3.6}{\sqrt{40}}\approx0.569\,\text{s}\). 3. The margin of error is \(2.023\cdot0.569\approx1.151\,\text{s}\). 4. The interval is \(1.6\pm1.151\), or approximately \((0.45,2.75)\,\text{s}\).

No. The displayed shape does not make a paired \(t\)-interval unreasonable. The \(95\%\) confidence interval for \(\mu_d\) is approximately \((0.45,2.75)\,\text{s}\).

- Decide whether these are unrelated groups or repeated measurements on the same people. - Ask what individual quantity could describe how much time each employee saved. - Check the changes for outliers before using their mean and standard deviation.

1. The same employees were measured twice, so use a paired \(t\)-interval on within-employee differences. 2. The five differences are \(2,3,1,3,1\), with \(\bar{d}=2\) and sample standard deviation \(s_d=1\) minute. 3. The differences \(1,1,2,3,3\) are reasonably symmetric with no outliers. The standard error is \(\frac{1}{\sqrt{5}}\approx0.4472\) minute. 4. The margin of error is \(2.132\cdot0.4472\approx0.9535\) minute. 5. The interval is approximately \((1.05,2.95)\) minutes for the population mean time saved.

Use a paired \(t\)-interval on before-minus-after differences. The changes are roughly symmetric without outliers; the \(90\%\) interval is approximately \((1.05,2.95)\,\text{min}\).

- Work with the within-person differences rather than treating the \(16\) times as unrelated observations. - For a small number of differences, check the shape information before using the interval procedure. - Preserve the definition of \(d\) when turning the numerical interval into words.

1. The paired differences have mean \(\bar{d}=2.1125\,\text{s}\) and standard deviation \(s_d\approx0.9583\,\text{s}\). 2. The random sample and shape of the differences support a one-sample \(t\)-interval with \(df=7\). 3. For \(90\%\) confidence, \(t^*\approx1.895\), and the standard error is \(0.9583/\sqrt{8}\approx0.3388\,\text{s}\). 4. The margin of error is approximately \(1.895(0.3388)=0.642\,\text{s}\). 5. The interval is approximately \((1.47,2.75)\,\text{s}\). We are \(90\%\) confident that Interface A's population mean adjustment time is about \(1.47\) to \(2.75\) seconds longer than Interface B's for musicians like those sampled.

The \(90\%\) confidence interval for \(\mu_d\) is approximately \((1.47,2.75)\,\text{s}\). We are \(90\%\) confident that Interface A has a population mean adjustment time about \(1.47\) to \(2.75\) seconds longer than Interface B.

- Check the random-sampling and sample-shape information before constructing the \(t\)-interval. - Apply the correction factor to the usual standard error, not to the sample mean. - Because the correction factor is less than \(1\), anticipate how it will affect the margin of error and interval width.

1. The random sample and stated sample shape support a one-sample \(t\)-interval. The sample is \(24/200=12\%\) of the population, so the usual \(10\%\) guideline is not satisfied. 2. The unadjusted standard error is \(\frac{9.6}{\sqrt{24}}\approx1.960\). 3. The finite-population-adjusted standard error is \(\frac{9.6}{\sqrt{24}}\sqrt{\frac{200-24}{200-1}}\approx1.843\). 4. The margin of error is \(2.069(1.843)\approx3.813\). 5. The confidence interval is \(52.4\pm3.813\), or approximately \((48.59,56.21)\). The adjustment narrows the interval because sampling without replacement from a substantial fraction of the population reduces sampling variability.

The adjusted standard error is approximately \(1.843\), compared with the unadjusted value \(1.960\). The \(95\%\) confidence interval is approximately \((48.59,56.21)\).