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Confidence intervals

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54934011
A symmetric confidence interval has upper endpoint \(76.0\) and margin of error \(4.5\). Write the interval in endpoint form and estimate-plus-or-minus form.

Hints

- The upper endpoint is one margin above the center. - A symmetric interval extends the same distance below the center. - Check that the full width is twice the margin.

Solution

1. The point estimate is \(76.0-4.5=71.5\). 2. The lower endpoint is \(71.5-4.5=67.0\). 3. The interval is \((67.0,76.0)\), equivalently \(71.5\pm4.5\).

Answer

\((67.0,76.0)\), or \(71.5\pm4.5\).
5497775–13
The histogram shows a right-skewed bootstrap distribution for a population mean. The sample estimate is \(8.0\), the simulated standard deviation is \(1.2\), and the percentile endpoints are marked. a) Form the symmetric interval using \(8.0\pm 2\cdot 1.2\). b) Read the percentile interval. c) Which interval better reflects the displayed bootstrap shape? Explain.
Figure for problem 549777

Hints

- First apply the supplied symmetric rule around the sample estimate. - Read the marked percentile boundaries directly from the horizontal scale. - Compare the distances from \(8.0\) to each endpoint with the simulation's longer right tail.

Solution

1. The symmetric interval is \(8.0\pm 2.4\), so its endpoints are \(8.0-2.4=5.6\) and \(8.0+2.4=10.4\). Thus, the interval is \((5.6, 10.4)\). 2. The marked percentile endpoints are \(6.25\) and \(11.00\), so the percentile interval is \((6.25, 11.00)\). 3. The percentile interval better reflects the right-skewed bootstrap distribution. Its upper endpoint extends farther above the sample estimate than its lower endpoint extends below it, matching the longer right tail. The symmetric interval forces equal distances on both sides of \(8.0\).

Answer

a) \((5.6, 10.4)\) b) \((6.25, 11.00)\) c) The percentile interval, because its asymmetry reflects the longer right tail of the bootstrap distribution.
54929511
A random sample of library visitors is used to estimate the population mean visit length. The reported \(95\%\) confidence interval is written as \(14.2\pm1.1\) minutes. a) Write the interval using endpoints. b) Identify the point estimate and the margin of error. c) State the population quantity estimated by the interval.

Hints

- Move the stated distance in both directions from the center. - Separate the center of the interval from its half-width. - Identify whether the interval targets individual visits or one population summary.

Solution

1. The lower endpoint is \(14.2-1.1=13.1\) minutes and the upper endpoint is \(14.2+1.1=15.3\) minutes. 2. The point estimate is \(14.2\) minutes, and the margin of error is \(1.1\) minutes. 3. The interval estimates the population mean visit length for the population represented by the random sample.

Answer

a) \((13.1, 15.3)\) minutes. b) Point estimate: \(14.2\) minutes; margin of error: \(1.1\) minutes. c) The population mean visit length.
54929611
The number line shows a \(95\%\) confidence interval for the percentage of residents who support a proposed trail. a) Read the interval endpoints. b) Find the point estimate and margin of error. c) Is \(50\%\) a plausible value for the population percentage according to this interval? Explain.
Figure for problem 549296

Hints

- Read the two ends of the displayed segment. - The center and half-width describe the same interval in another form. - Check the proposed value's location relative to both endpoints.

Solution

1. The interval extends from \(45\%\) to \(51\%\). 2. The midpoint is \(\frac{45+51}{2}=48\), so the point estimate is \(48\%\). 3. The margin of error is \(51-48=3\) percentage points. 4. Since \(50\%\) lies inside the interval, it is plausible according to this interval.

Answer

a) \((45\%, 51\%)\). b) Point estimate: \(48\%\); margin of error: \(3\) percentage points. c) Yes. \(50\%\) is inside the interval.
54929811
The same random sample gives these intervals for a population percentage: <table><thead><tr><th>Confidence level</th><th>Interval</th></tr></thead><tbody><tr><td>90%</td><td>\((69\%, 75\%)\)</td></tr><tr><td>95%</td><td>\((68\%, 76\%)\)</td></tr><tr><td>99%</td><td>\((66\%, 78\%)\)</td></tr></tbody></table> a) Show that all three intervals have the same point estimate. b) Find the margin of error for each interval. c) Explain why greater confidence produces a wider interval when the sample is unchanged.

Hints

- Find the midpoint of each pair of endpoints. - Measure from each midpoint to either endpoint. - Think about the tradeoff between capture rate and precision.

Solution

1. Each midpoint is \(72\%\): \(\frac{69+75}{2}=72\), \(\frac{68+76}{2}=72\), and \(\frac{66+78}{2}=72\). 2. The margins of error are \(3\), \(4\), and \(6\) percentage points, respectively. 3. To make the repeated-sampling method capture the true parameter more often, the interval must extend farther from the same sample estimate.

Answer

a) Each interval is centered at \(72\%\). b) The margins of error are \(3\), \(4\), and \(6\) percentage points for \(90\%\), \(95\%\), and \(99\%\), respectively. c) Higher confidence requires a broader range of plausible parameter values from the same data.
54930811
A bootstrap confidence interval for a population proportion is \((0.32, 0.43)\). The original sample proportion was \(0.37\). a) Find the distance from \(0.37\) to each endpoint. b) Is the interval symmetric about the sample proportion? c) Explain why an asymmetric bootstrap interval is not automatically an error.

Hints

- Measure each endpoint from the reported sample statistic separately. - Symmetry requires equal distances on both sides. - Consider whether the simulated distribution itself must be perfectly symmetric.

Solution

1. The lower distance is \(0.37-0.32=0.05\), and the upper distance is \(0.43-0.37=0.06\). 2. The interval is not symmetric about \(0.37\) because the endpoint distances differ. 3. A percentile bootstrap interval follows quantiles of the simulated distribution, which may be skewed or not centered exactly at the original statistic.

Answer

a) Lower distance: \(0.05\); upper distance: \(0.06\). b) No. c) Bootstrap percentile endpoints reflect the simulated distribution's shape, so they need not be equally spaced around the sample estimate.
54930911
Each panel shows a confidence interval produced in a simulation where the true population mean is \(50\). The marker indicates the true mean. a) Which interval fails to capture the true mean? b) What fraction of these six intervals capture the true mean? c) Explain why this small set does not have to show exactly a \(95\%\) capture rate even if a \(95\%\) method was used.
Figure for problem 549309

Hints

- Check whether the marked parameter lies on or between the two endpoints of each interval. - Count captures before converting the count to a fraction. - Distinguish a long-run rate from a guaranteed short-run count.

Solution

1. Interval e), \((50.5, 54.0)\), lies entirely above \(50\) and fails to capture the true mean. 2. Five of the six intervals contain \(50\), so the observed capture fraction is \(\frac{5}{6}\approx 0.833\). 3. A \(95\%\) confidence level is a long-run proportion over many repetitions; a small batch can differ substantially because of random variation.

Answer

a) e). b) \(\frac{5}{6}\approx0.833\), or about \(83.3\%\). c) The confidence level describes long-run performance, not an exact quota in every small group of intervals.
54931311
A sensor's measurements are found to be \(1.7\) units too low because of a calibration offset. Before correction, a \(95\%\) confidence interval for the population mean is \((23.4, 25.0)\). a) Find the corrected confidence interval. b) Find the original and corrected margins of error. c) Explain why the correction changes the center but not the width.

Hints

- Apply the calibration adjustment to every plausible mean, not only to the midpoint. - Compare endpoint differences before and after the shift. - An additive change affects location differently from spread.

Solution

1. Add \(1.7\) to both endpoints: \(23.4+1.7=25.1\) and \(25.0+1.7=26.7\). 2. The original margin of error is \(\frac{25.0-23.4}{2}=0.8\). The corrected interval has the same width, so its margin is also \(0.8\). 3. Adding a constant shifts every possible mean by the same amount but does not scale distances between values.

Answer

a) \((25.1, 26.7)\). b) Both margins of error are \(0.8\). c) An additive calibration shifts the interval without stretching it.
54931411
A \(95\%\) confidence interval for the proportion of a county's \(12{,}000\) registered bicycles that need a safety repair is \((0.28, 0.34)\). a) Convert the interval to percentages. b) Convert it to an interval for the number of registered bicycles needing repair. c) Explain why the count interval should not be interpreted as an exact inventory.

Hints

- Translate a proportion into a percentage by changing its scale. - Apply each plausible proportion to the known population size. - Preserve the distinction between an estimate and a complete count.

Solution

1. The proportion interval is \((28\%, 34\%)\). 2. The count endpoints are \(12{,}000\cdot 0.28=3360\) and \(12{,}000\cdot 0.34=4080\). 3. The resulting interval \((3360, 4080)\) is an estimate derived from sample uncertainty, not a direct count of every bicycle.

Answer

a) \((28\%, 34\%)\). b) Approximately \((3360, 4080)\) bicycles. c) It is an inferential range based on a sample, not a census total.
54931611
A calculated confidence interval is \((2.46, 2.54)\) seconds. A report rounds both endpoints to one decimal place and prints \((2.5, 2.5)\) seconds. a) Explain why the rounded interval is misleading. b) Find the original midpoint and margin of error. c) Give one more appropriate way to report the interval. More than one format is acceptable.

Hints

- Compare the displayed width before and after rounding. - Use the average and half-difference of the original endpoints. - Keep enough digits to preserve the uncertainty being communicated.

Solution

1. Rounding both endpoints to one decimal makes the interval appear to have zero width, hiding the uncertainty in the original calculation. 2. The midpoint is \(\frac{2.46+2.54}{2}=2.50\) seconds, and the margin of error is \(2.54-2.50=0.04\) seconds. 3. The interval should retain enough decimal places to show its width, such as \((2.46, 2.54)\) seconds or \(2.50\pm0.04\) seconds.

Answer

a) The rounding collapses a nonzero interval to a single displayed value. b) Midpoint: \(2.50\) seconds; margin of error: \(0.04\) seconds. c) One appropriate report is \((2.46, 2.54)\) seconds. The equivalent form \(2.50\pm 0.04\) seconds is also acceptable.
54931711
A poll reports \(46\%\pm3\) percentage points. A reader interprets this as “plus or minus \(3\%\) of \(46\%\).” a) Write the interval intended by the poll. b) Write the interval produced by the reader's relative-percent interpretation. c) Explain the difference between percent and percentage points in this context.

Hints

- Decide whether the stated amount is added directly or first multiplied by the estimate. - Calculate the reader's proposed relative change separately. - Compare absolute and relative ways of describing a change in a percentage.

Solution

1. Three percentage points means subtracting and adding \(3\) directly: the intended interval is \((43\%, 49\%)\). 2. Three percent of \(46\%\) is \(0.03\cdot 46=1.38\) percentage points, so the reader's interval is \((44.62\%, 47.38\%)\). 3. A percentage-point change is an absolute difference between percentages, while a percent change is relative to the starting percentage.

Answer

a) \((43\%, 49\%)\). b) \((44.62\%, 47.38\%)\). c) Percentage points are absolute differences on the percentage scale; percent changes are relative differences.
54932011
To check a confidence-interval method, a computer repeatedly samples from a population whose parameter is known. Of \(1000\) intervals, \(948\) contain the parameter. a) Estimate the method's capture rate and miss rate. b) Is this result reasonably consistent with a method intended to have \(95\%\) confidence? Explain. c) At the observed miss rate, how many misses would be expected among \(2500\) intervals?

Hints

- Use intervals containing the parameter divided by all simulated intervals. - The capture and miss rates must add to one. - Apply the observed long-run rate to the new number of repetitions.

Solution

1. The estimated capture rate is \(\frac{948}{1000}=0.948=94.8\%\). 2. The miss rate is \(1-0.948=0.052=5.2\%\). 3. A capture rate of \(94.8\%\) is close to the intended \(95\%\), so the result is reasonably consistent with the method. 4. At a \(5.2\%\) miss rate, the expected number among \(2500\) intervals is \(2500\cdot 0.052=130\).

Answer

a) Capture rate: \(94.8\%\); miss rate: \(5.2\%\). b) Yes. The simulated capture rate is close to \(95\%\). c) About \(130\) misses.
54932111
A report gives a sample estimate of \(54\%\), a margin of error of \(3\) percentage points, and a confidence interval of \((56\%, 62\%)\). a) Show why the three reported pieces of information are inconsistent. b) Write the interval that matches the estimate and margin of error. c) Find the midpoint and margin of error of the printed interval.

Hints

- Reconstruct the interval from the stated center and half-width. - Find the center of the printed endpoints independently. - All equivalent forms of one interval must agree on the same midpoint.

Solution

1. An estimate of \(54\%\) with a margin of \(3\) percentage points should have endpoints \(54-3=51\) and \(54+3=57\), not \(56\) and \(62\). 2. The matching interval is \((51\%, 57\%)\). 3. The printed interval has midpoint \(\frac{56+62}{2}=59\%\) and margin of error \(62-59=3\) percentage points.

Answer

a) The printed interval is centered at \(59\%\), not \(54\%\). b) \((51\%, 57\%)\). c) Midpoint: \(59\%\); margin of error: \(3\) percentage points.
54932311
A paired study reports a \(95\%\) confidence interval of \((-1.2, -0.3)\) hours for the population mean change, defined as after minus before. a) Interpret the sign of the interval. b) Restate the result as a positive range for the estimated mean reduction. c) Explain why the interval does not imply that every participant reduced time by between \(0.3\) and \(1.2\) hours.

Hints

- Use the stated subtraction order to translate a negative sign. - Reverse the wording when expressing a decrease as a positive amount. - Keep an interval for an average distinct from a range of raw observations.

Solution

1. Every plausible mean change is negative, so the population mean after value is lower than the population mean before value. 2. A change from \(-1.2\) to \(-0.3\) hours corresponds to a mean reduction from \(0.3\) to \(1.2\) hours. 3. The interval estimates the population mean of paired changes, not the range of individual changes.

Answer

a) The interval supports a negative population mean change. b) The estimated population mean reduction is between \(0.3\) and \(1.2\) hours. c) Individual changes may fall outside this range because the interval concerns their population mean.
54932811
A bootstrap \(95\%\) confidence interval for the population median emergency-response time is \((24, 29)\) minutes. a) State the parameter being estimated. b) Give a correct contextual interpretation of the interval. c) Explain why neither “the population mean is between \(24\) and \(29\)” nor “\(95\%\) of response times are between \(24\) and \(29\)” is justified.

Hints

- Use the exact statistic named in the interval description. - Separate a population center from a percentage of individual observations. - Do not substitute one measure of center for another.

Solution

1. The parameter is the median response time for the population of emergency responses represented by the sample. 2. The interval gives plausible values from \(24\) to \(29\) minutes for that population median under the bootstrap method. 3. A median and a mean are different parameters, so the interval does not estimate the mean. 4. An interval for a population median does not describe the central \(95\%\) of individual response times.

Answer

a) The population median emergency-response time. b) Plausible values for that median range from \(24\) to \(29\) minutes under the stated method. c) The interval estimates neither the mean nor the spread of individual response times.
54933111
A town has complete, accurate records for all \(8000\) households, and \(2960\) have solar panels. Find the town proportion and explain whether a sampling confidence interval is needed. Also identify two nonsampling errors that could matter if the records were imperfect.

Hints

- Compare the records with the entire target population, not with a sample. - Sampling intervals address uncertainty created by random sampling. - A census can still contain measurement or coverage errors.

Solution

1. The town proportion is \(\frac{2960}{8000}=0.37=37\%\). 2. Because the records cover the entire target population and are assumed accurate, there is no sampling variability and no sampling confidence interval is needed. 3. Possible nonsampling errors include misclassification, outdated records, duplicate records, and missing households.

Answer

The town proportion is \(37\%\). No sampling confidence interval is needed for a complete, accurate census. Imperfect records could still contain errors such as misclassification and missing or outdated households.
54933611
A confidence interval for a population proportion is \((0.412, 0.468)\). Convert it to percentages and give its point estimate and margin of error on both the proportion and percentage scales.

Hints

- Multiply each endpoint by \(100\) to convert to percentages. - The point estimate is the midpoint. - The margin of error is the distance from the midpoint to either endpoint.

Solution

1. Multiplying both endpoints by \(100\) gives \((41.2\%, 46.8\%)\). 2. The midpoint is \(\frac{0.412+0.468}{2}=0.440\), and the margin is \(0.468-0.440=0.028\). 3. On the percentage scale, the point estimate is \(44.0\%\) and the margin is \(2.8\) percentage points.

Answer

The interval is \((41.2\%, 46.8\%)\). Its center and margin are \(0.440 \pm 0.028\), equivalently \(44.0\% \pm 2.8\) percentage points.
54933811
Interval A is \((180, 220)\), and Interval B is \((18, 22)\). Each estimates a different positive population mean. Compare their absolute and relative precision.

Hints

- Find each interval's midpoint and half-width. - Divide each half-width by its own midpoint. - Absolute width and relative width answer different comparison questions.

Solution

1. Interval A has center \(200\) and margin \(20\). Interval B has center \(20\) and margin \(2\). 2. A's relative margin is \(\frac{20}{200}=10\%\), and B's relative margin is \(\frac{2}{20}=10\%\). 3. A has a larger absolute margin, but the intervals have equal relative precision.

Answer

A is \(200\pm20\), and B is \(20\pm2\). Both margins are \(10\%\) of their point estimates, so the intervals have equal relative precision even though A is wider in absolute units.
54933911
A student says, “A \(95\%\) confidence method guarantees that exactly \(95\) of the next \(100\) intervals will contain the true parameter.” a) Correct the statement. b) If \(100\) intervals are independently produced by a method with a true \(95\%\) capture rate, find the expected numbers of captures and misses. c) Explain why observing \(92\) captures would not by itself disprove the method.

Hints

- Separate an expected long-run rate from a guaranteed count. - Apply the two complementary probabilities to \(100\) repetitions. - Expected values do not force every observed batch to equal them.

Solution

1. The \(95\%\) level describes a long-run capture proportion, not an exact requirement for every set of \(100\) intervals. 2. The expected number of captures is \(100\cdot0.95=95\), and the expected number of misses is \(100\cdot0.05=5\). 3. Actual counts vary randomly around their expected values; \(92\) captures can occur even when the long-run capture probability is \(0.95\).

Answer

a) Over many repetitions, about \(95\%\) of intervals from the method capture the parameter; a particular set need not contain exactly \(95\). b) Expected captures: \(95\); expected misses: \(5\). c) Short-run random variation can produce \(92\) captures under a valid \(95\%\) method.
54929711
A random sample of \(400\) customers gives a confidence interval for a population percentage with a margin of error of \(5\) percentage points. Assume the margin of error varies inversely with the square root of the sample size. a) What sample size would reduce the margin of error to \(2.5\) percentage points? b) What margin of error would be expected with a sample size of \(900\)? c) Explain why doubling the sample size would not cut the margin of error in half.

Hints

- Compare the old and new half-widths as a ratio. - Changes in sample size act through a square root. - Use the original sample and margin as a reference pair.

Solution

1. Cutting the margin of error from \(5\) to \(2.5\) is a factor of \(\frac{1}{2}\), so the sample size must be multiplied by \(2^2=4\). Thus \(n=4\cdot 400=1600\). 2. For \(n=900\), the margin is \(5\sqrt{\frac{400}{900}}=5\cdot\frac{2}{3}=\frac{10}{3}\approx 3.33\) percentage points. 3. Doubling \(n\) multiplies the margin by \(\frac{1}{\sqrt{2}}\approx 0.707\), not by \(\frac{1}{2}\).

Answer

a) \(1600\). b) About \(3.33\) percentage points. c) Margin of error follows an inverse square-root relationship, so doubling \(n\) reduces it by a factor of \(1/\sqrt{2}\), not \(1/2\).
54929911
The histogram shows \(500\) bootstrap means for daily water use, in gallons, from one sample. The two reference lines mark the \(2.5\)th and \(97.5\)th percentiles of the bootstrap distribution. a) Use the percentile method to report a \(95\%\) confidence interval. b) Find the interval's midpoint and margin of error. c) Explain what the bootstrap process is approximating.
Figure for problem 549299

Hints

- Read the two marked percentile locations as the interval boundaries. - Use the center and half-width of those boundaries. - Ask what repeated resampling is meant to imitate.

Solution

1. The percentile endpoints are \(18.2\) and \(21.8\), so the interval is \((18.2, 21.8)\) gallons. 2. The midpoint is \(\frac{18.2+21.8}{2}=20.0\) gallons. 3. The margin of error is \(21.8-20.0=1.8\) gallons. 4. Resampling with replacement from the observed sample approximates how the sample mean would vary across repeated samples from the population.

Answer

a) \((18.2, 21.8)\) gallons. b) Midpoint: \(20.0\) gallons; margin of error: \(1.8\) gallons. c) The bootstrap distribution approximates repeated-sample variation of the sample mean.
54930011
A report states: “The \(95\%\) confidence interval for the town's mean monthly electricity use is \((710, 760)\) kilowatt-hours. Therefore, there is a \(95\%\) probability that the true mean lies in this particular interval.” a) Identify the problem with the statement. b) Write one correct repeated-sampling interpretation. Equivalent wording is acceptable. c) State one condition about the data-collection process that is needed before the interval can represent the town.

Hints

- Separate what is fixed from what would vary over repeated samples. - Describe the behavior of many intervals made by the same procedure. - Precision from a formula cannot repair a poor data source.

Solution

1. After the sample is observed and the interval is calculated, the population mean is fixed; the \(95\%\) describes the long-run success rate of the method, not a changing probability for this fixed interval. 2. A correct interpretation is that about \(95\%\) of intervals produced by this method from repeated random samples would contain the town's true mean monthly electricity use. 3. The sample must represent the town through an appropriate random sampling process and should not have substantial selection or measurement bias.

Answer

a) The \(95\%\) refers to the method's long-run capture rate, not a probability assigned to the fixed parameter after this interval is calculated. b) One correct interpretation is: About \(95\%\) of intervals from repeated random samples using this method would contain the true mean. c) For example, the data should come from a representative random sample without major bias.
54930111
A confidence interval for a population mean temperature is \((18.5, 20.0)\) degrees Celsius. Convert the entire interval to degrees Fahrenheit using \(F=1.8C+32\). a) Find the Fahrenheit endpoints. b) Find the Fahrenheit midpoint and margin of error. c) Explain why converting only the Celsius midpoint and leaving the margin unchanged would be incorrect.

Hints

- Apply the same increasing linear conversion to both endpoints. - Distances are affected by the multiplicative part of a unit conversion. - Check the transformed center against the average of the transformed endpoints.

Solution

1. The lower endpoint is \(1.8\cdot 18.5+32=65.3\) degrees Fahrenheit. 2. The upper endpoint is \(1.8\cdot 20.0+32=68.0\) degrees Fahrenheit. 3. The midpoint is \(\frac{65.3+68.0}{2}=66.65\) degrees Fahrenheit, and the margin of error is \(\frac{68.0-65.3}{2}=1.35\) degrees Fahrenheit. 4. Adding \(32\) shifts the center, while multiplying by \(1.8\) also scales every distance, including the margin of error.

Answer

a) \((65.3, 68.0)\) degrees Fahrenheit. b) Midpoint: \(66.65\) degrees Fahrenheit; margin of error: \(1.35\) degrees Fahrenheit. c) The factor \(1.8\) scales the interval's half-width, so the numerical margin must change.
54930211
Two independent random samples from the same population produce the confidence intervals shown. Both use the same confidence level and method. <table><thead><tr><th>Interval</th><th>Endpoints</th></tr></thead><tbody><tr><td>A</td><td>\((42.1, 49.9)\)</td></tr><tr><td>B</td><td>\((44.5, 48.5)\)</td></tr></tbody></table> a) Find the midpoint and margin of error of each interval. b) Which interval most likely came from the larger sample? Explain. c) Why can the two point estimates differ even though the samples came from the same population?

Hints

- Use the average of the endpoints for each center. - Compare the half-widths while holding the confidence method fixed. - Remember that random samples need not contain the same observations.

Solution

1. Interval A has midpoint \(\frac{42.1+49.9}{2}=46.0\) and margin of error \(49.9-46.0=3.9\). 2. Interval B has midpoint \(\frac{44.5+48.5}{2}=46.5\) and margin of error \(48.5-46.5=2.0\). 3. Interval B most likely came from the larger sample because larger samples generally have less sampling variability and narrower intervals when the confidence level and method are fixed. 4. Random samples contain different observations, so their sample estimates can differ through sampling variability even when they come from the same population.

Answer

a) A: midpoint \(46.0\), margin \(3.9\). B: midpoint \(46.5\), margin \(2.0\). b) Interval B, because it is narrower under the same method and confidence level. c) Different random samples can produce different estimates through sampling variability.
54930311
A sample estimate of a population mean is \(32.0\). A simulation of repeated samples estimates the standard deviation of the sample statistic as \(1.4\). The simulated distribution is approximately symmetric and bell-shaped. Use the rule “estimate \(\pm\) two simulated standard deviations” to form an approximate \(95\%\) confidence interval. Then state the assumptions behind this shortcut.

Hints

- Translate the stated rule into a distance from the sample estimate. - Use the same distance on both sides because the simulated distribution is symmetric. - Check whether the shape and simulation design justify the approximation.

Solution

1. Two simulated standard deviations equal \(2\cdot 1.4=2.8\). 2. The lower endpoint is \(32.0-2.8=29.2\), and the upper endpoint is \(32.0+2.8=34.8\). 3. The approximation relies on the simulated sampling distribution being centered appropriately, approximately symmetric and bell-shaped, and generated from a model that represents the sampling process.

Answer

The approximate \(95\%\) confidence interval is \((29.2, 34.8)\). The shortcut requires an appropriate simulation model and an approximately symmetric, bell-shaped sampling distribution.
54930511
An online poll of \(50{,}000\) website visitors reports that \(81\%\) support a policy, with a margin of error of \(0.4\) percentage points. Participation was voluntary, and the website mainly attracts supporters of the policy. a) Write the reported interval. b) Explain why the very small margin of error does not make the interval reliable for all voters. c) Identify the type of error that the reported margin addresses and one type it does not address.

Hints

- Move the stated half-width in both directions from the reported percentage. - Separate interval precision from the representativeness of the data source. - Ask which uncertainty the mathematical calculation was designed to measure.

Solution

1. The reported interval is \((80.6\%, 81.4\%)\). 2. The volunteer sample is systematically unrepresentative because site visitors and respondents are more likely to support the policy. A large sample makes random sampling variation small but does not remove selection bias. 3. The margin of error addresses random sampling variability under the assumed sampling method. It does not account for voluntary-response bias or undercoverage.

Answer

a) \((80.6\%, 81.4\%)\). b) The sample is biased toward supporters, so high precision around the sample result does not imply accuracy for all voters. c) The margin addresses sampling variability, not selection bias such as voluntary response or undercoverage.
54930611
The number line shows a \(95\%\) confidence interval for the difference \(\mu_A-\mu_B\) in mean processing time, measured in minutes. a) Read the interval and its point estimate. b) Does the interval support the claim that Method A has a smaller population mean processing time than Method B? Explain. c) Translate the interval into a statement about how much faster A may be.
Figure for problem 549306

Hints

- Interpret the sign using the stated subtraction order. - Locate zero relative to the entire interval. - Convert negative A-minus-B values into positive statements about how much smaller A is.

Solution

1. The interval is \((-4.0, -0.8)\) minutes, with midpoint \(\frac{-4.0+(-0.8)}{2}=-2.4\) minutes. 2. Every value in the interval is negative, so the interval supports \(\mu_A<\mu_B\). 3. Since \(\mu_A-\mu_B\) is between \(-4.0\) and \(-0.8\), Method A's mean processing time is estimated to be between \(0.8\) and \(4.0\) minutes shorter.

Answer

a) \((-4.0, -0.8)\) minutes; point estimate \(-2.4\) minutes. b) Yes. The entire interval is below \(0\). c) Method A is estimated to be \(0.8\) to \(4.0\) minutes faster on average.
54930711
A filling machine is considered acceptable only if its population mean fill is at least \(499.0\) milliliters. A random sample produces a \(95\%\) confidence interval of \((499.2, 500.6)\) milliliters. a) Find the point estimate and margin of error. b) Does the interval support the conclusion that the machine meets the mean-fill requirement? Explain. c) Does the interval prove that every container has at least \(499.0\) milliliters? Explain.

Hints

- Compare the requirement with the least plausible mean in the interval. - Use the midpoint and half-width to describe the interval. - Keep a claim about an average separate from a claim about every observation.

Solution

1. The midpoint is \(\frac{499.2+500.6}{2}=499.9\) milliliters, and the margin of error is \(500.6-499.9=0.7\) milliliters. 2. The entire interval is above \(499.0\), so the interval supports the conclusion that the population mean meets the requirement. 3. The interval estimates the population mean and gives no guarantee about every individual container.

Answer

a) Point estimate: \(499.9\) milliliters; margin of error: \(0.7\) milliliters. b) Yes. The lower endpoint, \(499.2\), exceeds the requirement. c) No. The interval concerns the population mean, not each individual fill.
54931011
A study reports a \(95\%\) confidence interval of \((54, 62)\) for Group A's population mean and \((60, 68)\) for Group B's population mean. A reader says, “Because the intervals overlap, the population means must be equal.” a) Identify the logical error. b) State what can and cannot be concluded from the overlap alone. c) What additional interval would directly address whether the population means differ?

Hints

- Separate a range of plausible values from a proof of exact equality. - The two interval estimates are not the same as an interval for their subtraction. - Think about which parameter directly represents equality of two means.

Solution

1. Overlap does not prove equality; confidence intervals contain ranges of plausible values, not exact population means. 2. The overlap shows that some values are plausible for both means separately, but it does not determine the uncertainty in their difference. 3. A confidence interval for \(\mu_A-\mu_B\) would directly address whether zero is a plausible population difference.

Answer

a) Overlap of separate intervals does not imply equal parameters. b) The intervals share plausible values, but overlap alone does not settle whether the means differ. c) Use a confidence interval for \(\mu_A-\mu_B\) and examine whether it includes \(0\).
54931211
Two simple random samples each contain \(200\) people. Sample T comes from a town of \(1000\) people, and Sample C comes from a city of \(1{,}000{,}000\) people. Simulations using sampling without replacement give margins of error of \(3.6\) percentage points for T and \(4.0\) percentage points for C. a) Which interval is expected to be narrower? b) Explain why the town's smaller population can reduce sampling variability here. c) Why would changing the city population from \(1{,}000{,}000\) to \(2{,}000{,}000\) have almost no effect with the same sample size?

Hints

- Compare the supplied simulated half-widths directly. - Calculate what fraction of each population is sampled. - Consider when removing one sampled person meaningfully changes what remains.

Solution

1. The town interval is expected to be narrower because its simulated margin of error is \(3.6\) rather than \(4.0\) percentage points. 2. Sampling \(200\) of \(1000\) people observes \(20\%\) of the town. Without replacement, each selected person removes a meaningful part of the remaining uncertainty. 3. A sample of \(200\) is a tiny fraction of either city population, so sampling without replacement behaves almost like sampling from an effectively unlimited population.

Answer

a) The town interval. b) The sample covers \(20\%\) of the town, so sampling without replacement noticeably reduces variability. c) In either city, \(200\) is a negligible population fraction, so the margin changes very little.
54931511
A bootstrap distribution contains \(1000\) estimates arranged from least to greatest. Selected percentile values are shown. <table><thead><tr><th>Percentile</th><th>5th</th><th>10th</th><th>50th</th><th>90th</th><th>95th</th></tr></thead><tbody><tr><td>Estimate</td><td>\(0.41\)</td><td>\(0.43\)</td><td>\(0.47\)</td><td>\(0.51\)</td><td>\(0.52\)</td></tr></tbody></table> a) Use the percentile method to form a central \(90\%\) confidence interval. b) Form a central \(80\%\) confidence interval. c) Compare their widths and explain the difference.

Hints

- Split the excluded percentage equally between the two tails. - Match each tail percentage to the corresponding percentile row. - Compare endpoint differences after forming both intervals.

Solution

1. A central \(90\%\) interval leaves \(5\%\) in each tail, so it is \((0.41, 0.52)\). 2. A central \(80\%\) interval leaves \(10\%\) in each tail, so it is \((0.43, 0.51)\). 3. The widths are \(0.52-0.41=0.11\) and \(0.51-0.43=0.08\). 4. The \(90\%\) interval is wider because greater confidence requires retaining more of the simulated distribution.

Answer

a) \((0.41, 0.52)\). b) \((0.43, 0.51)\). c) Their widths are \(0.11\) and \(0.08\); the higher-confidence interval is wider.
54931811
A random sample selects \(400\) residents, but only \(300\) respond. Among respondents, \(180\) favor a proposal, and a confidence interval based only on respondents is \((55\%, 65\%)\). a) Find the respondent sample proportion. b) If all \(100\) nonrespondents oppose the proposal, what proportion of the full selected sample favors it? If all favor it, what proportion favors it? c) Explain why the reported confidence interval does not account for this uncertainty.

Hints

- Keep the respondent denominator separate from the original selected-sample denominator. - Use extreme assumptions to see the full range that missing responses could allow. - Ask whether the interval calculation modeled who failed to respond.

Solution

1. The respondent proportion is \(\frac{180}{300}=0.60=60\%\). 2. If all nonrespondents oppose, the full-sample proportion is \(\frac{180}{400}=45\%\). If all favor, it is \(\frac{180+100}{400}=70\%\). 3. The interval's margin reflects sampling variation under assumptions about the responding data; it does not include the possible systematic difference between respondents and nonrespondents.

Answer

a) \(60\%\). b) The extreme full-sample proportions are \(45\%\) and \(70\%\). c) The interval measures sampling uncertainty, not unresolved nonresponse bias.
54931911
A manufacturer needs evidence that a population mean exceeds \(100\). Panels a), b), and c) show the 90%, 95%, and 99% confidence intervals, respectively, from the same sample. a) Which intervals support the claim that the mean exceeds \(100\)? b) What is the highest confidence level among those that support the claim? c) Explain why the \(99\%\) interval leads to a less decisive conclusion than the \(95\%\) interval.
Figure for problem 549319

Hints

- For each interval, compare its lower endpoint with the required threshold. - Choose the greatest confidence level that still keeps the full interval on one side. - Relate higher confidence to a wider set of plausible values.

Solution

1. The \(90\%\) interval \((102, 108)\) and the \(95\%\) interval \((101, 109)\) lie entirely above \(100\), so both support the claim. 2. The highest supporting confidence level is \(95\%\). 3. The \(99\%\) interval \((99, 111)\) includes values below \(100\); its greater width reflects the higher long-run capture requirement.

Answer

a) The \(90\%\) and \(95\%\) intervals. b) \(95\%\). c) The wider \(99\%\) interval includes \(100\) and values below it, so the threshold conclusion is no longer supported at that level.
54932511
In a survey, \(52\%\) of all \(800\) respondents support a proposal, and \(52\%\) of the \(80\) respondents ages \(18\)–\(24\) support it. The \(95\%\) intervals are \((48.5\%, 55.5\%)\) overall and \((41\%, 63\%)\) for ages \(18\)–\(24\). a) Compare the point estimates and margins of error. b) Explain why the subgroup interval is much wider even though its point estimate is the same. c) What sample-design change would most directly improve the subgroup's precision?

Hints

- Measure each interval's half-width from its common center. - Focus on the number of observations supporting each estimate. - Improve precision by increasing information for the specific group of interest.

Solution

1. Both point estimates are \(52\%\). The overall margin of error is \(55.5-52=3.5\) percentage points, and the subgroup margin is \(63-52=11\) percentage points. 2. The subgroup estimate is based on only \(80\) respondents, so it has greater sampling variability than the estimate based on \(800\). 3. Selecting more people ages \(18\)–\(24\), possibly by intentionally oversampling that age group and weighting appropriately for overall estimates, would improve subgroup precision.

Answer

a) Both estimates are \(52\%\); the margins are \(3.5\) and \(11\) percentage points. b) The subgroup has a much smaller sample and therefore greater sampling variability. c) Increase the subgroup sample size, such as through planned oversampling.
54932611
Two proposed \(95\%\) confidence-interval methods are tested on \(2000\) simulated samples from populations with known parameters. <table><thead><tr><th>Method</th><th>Capture rate</th><th>Average width</th></tr></thead><tbody><tr><td>A</td><td>\(94.7\%\)</td><td>\(10.0\,\text{units}\)</td></tr><tr><td>B</td><td>\(78.4\%\)</td><td>\(6.0\,\text{units}\)</td></tr></tbody></table> a) Which method better matches its claimed confidence level? b) Why is Method B's smaller average width not enough to make it preferable? c) State the tradeoff that a useful interval method must manage.

Hints

- Compare each simulated capture rate with the advertised level. - A narrow interval is useful only if it captures often enough. - Think of precision and reliability as two separate performance measures.

Solution

1. Method A better matches the claimed \(95\%\) level because its simulated capture rate is \(94.7\%\), close to \(95\%\). 2. Method B misses the true parameter in \(21.6\%\) of simulations, so its narrow intervals obtain apparent precision by failing to capture often enough. 3. A useful method must balance narrow intervals with a capture rate close to the stated confidence level.

Answer

a) Method A. b) Method B is too narrow to achieve its claimed coverage; its capture rate is only \(78.4\%\). c) The method must balance precision, measured by width, with reliable long-run coverage.
54933011
An initial random sample of \(100\) observations gives a \(95\%\) confidence interval of \((48, 60)\). After collecting a new random sample of \(400\) observations, the interval is \((57, 63)\). a) Find each point estimate and margin of error. b) Compare the margins with the inverse square-root relationship. c) Explain why the larger-sample interval is narrower but need not be contained inside the first interval.

Hints

- Calculate centers and half-widths before comparing the intervals. - Relate the sample-size factor to the square root of that factor. - Precision controls width, not the exact location of a new random estimate.

Solution

1. The first interval has point estimate \(54\) and margin \(6\). The second has point estimate \(60\) and margin \(3\). 2. The sample size is multiplied by \(4\), and the margin is divided by \(2=\sqrt{4}\), matching inverse square-root scaling. 3. A larger sample reduces sampling variability around its own estimate, but a new random sample can have a different center; narrower does not mean nested.

Answer

a) First: estimate \(54\), margin \(6\). Second: estimate \(60\), margin \(3\). b) Quadrupling the sample size halves the margin, as expected. c) The new sample estimate can shift, so its narrower interval is not required to lie within the earlier interval.
54933211
A study will estimate the difference between two population means using a total of \(200\) observations. The two populations have similar variability. Simulations predict these margins of error: <table><thead><tr><th>Group 1 size</th><th>Group 2 size</th><th>Margin for difference</th></tr></thead><tbody><tr><td>\(100\)</td><td>\(100\)</td><td>\(4.0\)</td></tr><tr><td>\(120\)</td><td>\(80\)</td><td>\(4.2\)</td></tr><tr><td>\(150\)</td><td>\(50\)</td><td>\(5.2\)</td></tr></tbody></table> a) Which allocation gives the most precise interval? b) Explain why an increasingly unbalanced allocation widens the interval when group variability is similar. c) Give one circumstance in which an unequal allocation might still be reasonable.

Hints

- Use the smallest simulated half-width as the precision criterion. - A difference estimate relies on both component estimates. - Think about practical reasons the two groups may not have equal sampling costs or variability.

Solution

1. The \(100\)-and-\(100\) allocation gives the smallest margin of error, \(4.0\). 2. The uncertainty in a difference depends on information from both groups; a small group becomes the weak part of the comparison and raises overall sampling variability. 3. An unequal allocation may be reasonable if one population is much more variable, much cheaper to sample, rarer, or of special interest.

Answer

a) \(100\) observations in each group. b) A very small group contributes substantial uncertainty to the estimated difference. c) For example, unequal costs, unequal variability, or a rare subgroup may justify unequal allocation.
54933511
For an individual-level simple random sample, a survey estimate is \(61\%\) with a margin of error of \(2.0\) percentage points. A planned cluster sample is expected from simulations to have \(1.6\) times as much margin of error at the same nominal sample size. a) Find the expected cluster-sample margin of error. b) Write the expected confidence interval centered at \(61\%\). c) Explain why the number of sampled individuals alone does not determine interval precision.

Hints

- Apply the simulated design factor to the original half-width. - Keep the sample estimate fixed while changing the margin. - Consider whether observations selected together provide fully separate information.

Solution

1. The cluster-sample margin is \(1.6\cdot2.0=3.2\) percentage points. 2. The interval is \((61-3.2, 61+3.2)=(57.8\%, 64.2\%)\). 3. People within the same cluster may be similar, so clustered observations provide less independent information than the same number of individuals selected broadly.

Answer

a) \(3.2\) percentage points. b) \((57.8\%, 64.2\%)\). c) Dependence within clusters reduces effective information and can widen the interval.
54933711
A convenience sample gives an interval \((64\%, 70\%)\) at \(95\%\) confidence. Using the same biased sample, a researcher calculates a wider \(99\%\) interval of \((62\%, 72\%)\) and claims that the higher confidence solves the sampling problem. a) Explain why the claim is incorrect. b) What uncertainty is increased confidence addressing? c) What change is needed to make an interval suitable for the target population?

Hints

- Identify what remained unchanged between the two intervals. - Separate a wider allowance for random variation from correction of a shifted sample. - Repair the data-collection process rather than only the interval calculation.

Solution

1. Both intervals are based on the same convenience sample, so both can be centered on a systematically unrepresentative estimate. 2. Higher confidence addresses random sampling variability by widening the range generated by the interval method. 3. The study needs an appropriate probability sample or another design that represents the target population; changing only the confidence level cannot remove selection bias.

Answer

a) A wider interval from the same convenience sample does not remove selection bias. b) It addresses random sampling uncertainty, not systematic sample-selection error. c) Use a representative probability-based sampling design.
54934111
A \(95\%\) confidence interval for the ratio \(\mu_A/\mu_B\) is \((0.92, 1.08)\). a) Explain the meaning of the reference value \(1\). b) Does the interval show a clear difference between the population means? c) Restate the interval as an approximate percentage comparison of A with B.

Hints

- For a ratio, equality is represented by a value different from zero. - Check whether the equality value lies inside the full interval. - Compare each endpoint with one to translate it into a relative percentage.

Solution

1. A ratio of \(1\) means \(\mu_A=\mu_B\). 2. The interval contains \(1\), so it does not show a clear direction of difference at this confidence level. 3. A ratio of \(0.92\) means A is \(8\%\) lower than B, and a ratio of \(1.08\) means A is \(8\%\) higher than B.

Answer

a) \(1\) represents equal population means. b) No; the interval includes \(1\). c) A's population mean is plausibly from about \(8\%\) lower to \(8\%\) higher than B's.
54934211
A bootstrap \(95\%\) confidence interval for a population interquartile range is \((12, 18)\) points. a) State what population quantity is being estimated. b) Give a correct interpretation of the interval. c) Explain why the interval does not say that the middle \(50\%\) of individual scores lie between \(12\) and \(18\).

Hints

- Identify whether the parameter is a location, a proportion, or a measure of spread. - Distinguish the size of an interval from the raw values that form it. - Keep a confidence interval for a statistic separate from a descriptive interval of observations.

Solution

1. The parameter is the population interquartile range, which measures the width of the middle half of the population distribution. 2. The interval gives plausible values from \(12\) to \(18\) points for that population spread measure under the bootstrap method. 3. The endpoints estimate the size of the interquartile range, not the score values that mark the first and third quartiles.

Answer

a) The population interquartile range. b) Plausible values for that spread measure are \(12\) to \(18\) points. c) The interval estimates a width, not the locations of the middle \(50\%\) of individual scores.
54934311
The same sample is bootstrapped twice with each of two resample counts. <table><thead><tr><th>Bootstrap resamples</th><th>Run 1 interval</th><th>Run 2 interval</th></tr></thead><tbody><tr><td>\(100\)</td><td>\((42, 55)\)</td><td>\((39, 57)\)</td></tr><tr><td>\(10{,}000\)</td><td>\((41.1, 55.8)\)</td><td>\((41.0, 55.9)\)</td></tr></tbody></table> a) Which resample count gives more stable numerical endpoints? b) Explain why increasing bootstrap resamples improves this stability. c) Name one source of error that more bootstrap resamples do not fix.

Hints

- Compare how much the endpoints change between repeated runs at each setting. - Separate randomness from the computer simulation from randomness in the original sample. - Repeating calculations cannot improve the data that were originally collected.

Solution

1. The \(10{,}000\)-resample runs give much more similar endpoints, so they are more numerically stable. 2. More resamples reduce Monte Carlo variation caused by using a finite random set of bootstrap samples to estimate percentiles. 3. More resamples do not correct bias or lack of representativeness in the original sample, and they do not add new population information.

Answer

a) \(10{,}000\) resamples. b) More resamples estimate the bootstrap distribution's percentiles with less simulation noise. c) They do not fix a biased or unrepresentative original sample.
54934411
A simple random sample of \(625\) households from a county of \(10{,}000\) households finds that \(231\) support a recycling fee. A simulation-based \(95\%\) margin of error is \(3.8\) percentage points. a) Find the sample proportion. b) Construct the confidence interval for the population proportion. c) Convert the interval to an estimated range of county households. d) State the interpretation in context.

Hints

- Begin with supporters divided by sampled households. - Apply the margin on the percentage scale to both sides of the estimate. - Use the known county total only after forming the population-proportion interval.

Solution

1. The sample proportion is \(\frac{231}{625}=0.3696=36.96\%\). 2. The endpoints are \(36.96\%-3.8\%=33.16\%\) and \(36.96\%+3.8\%=40.76\%\). 3. Applying the endpoints to \(10{,}000\) households gives \(3316\) to \(4076\) households. 4. The method gives a \(95\%\) confidence interval of \((33.16\%, 40.76\%)\) for the county proportion, corresponding to about \(3316\)–\(4076\) households.

Answer

a) \(36.96\%\). b) \((33.16\%, 40.76\%)\). c) About \(3316\) to \(4076\) households. d) The interval gives plausible values for the countywide support proportion under the random-sampling method.
54930411
A random-sampling calculation gives a \(95\%\) confidence interval of \((48, 52)\) units for a population mean. A calibration audit shows that the measuring instrument may shift every reading by an unknown amount between \(-1.5\) and \(1.5\) units. a) Find the confidence interval's point estimate and sampling margin of error. b) Form a conservative range that allows for both the reported interval and any calibration shift in the stated range. c) Explain why the expanded range is not simply an ordinary \(95\%\) confidence interval.

Hints

- First describe the original interval by its center and half-width. - Consider the most negative shift at the lower end and the most positive shift at the upper end. - Keep random sampling uncertainty separate from systematic measurement uncertainty.

Solution

1. The point estimate is \(\frac{48+52}{2}=50\) units, and the sampling margin of error is \(2\) units. 2. The smallest adjusted endpoint is \(48-1.5=46.5\), and the largest is \(52+1.5=53.5\), giving the conservative range \((46.5, 53.5)\). 3. The original confidence level describes repeated-sampling variability under its sampling method. The calibration allowance is a separate systematic uncertainty range, so combining the two does not automatically preserve a \(95\%\) repeated-sampling interpretation.

Answer

a) Point estimate: \(50\) units; sampling margin: \(2\) units. b) \((46.5, 53.5)\) units. c) The added calibration allowance represents systematic uncertainty, not the random sampling process used to define the \(95\%\) confidence level.
54931111
A survey has a fixed setup cost of \(\$800\) and costs \(\$12\) for each completed response. The total budget is \(\$5600\). For the planned method, the margin of error in percentage points is approximately \(\frac{80}{\sqrt{n}}\), where \(n\) is the number of completed responses. a) Find the largest affordable sample size. b) Find the corresponding approximate margin of error. c) If the desired margin is \(3\) percentage points, determine whether the budget is sufficient.

Hints

- Remove the fixed cost before finding how many responses can be purchased. - Substitute the affordable count into the precision relationship. - For the target precision, solve the inequality before checking cost.

Solution

1. The response budget is \(5600-800=4800\) dollars, so the largest sample size is \(n=\frac{4800}{12}=400\). 2. The margin of error is \(\frac{80}{\sqrt{400}}=\frac{80}{20}=4\) percentage points. 3. A margin of \(3\) requires \(\frac{80}{\sqrt{n}}\le 3\), so \(n\ge\left(\frac{80}{3}\right)^2\approx 711.1\), requiring at least \(712\) responses. 4. The cost would be \(800+12\cdot 712=\$9344\), so the budget is not sufficient.

Answer

a) \(400\) responses. b) About \(4\) percentage points. c) No. At least \(712\) responses would be needed, costing \(\$9344\).
54932211
A \(95\%\) confidence interval for mean fuel economy is \((28.4, 30.0)\) miles per gallon. Convert fuel economy to liters per \(100\) kilometers using \(L=\frac{235.2}{M}\), where \(M\) is miles per gallon. a) Convert the interval endpoints and write the liters-per-\(100\)-kilometers interval in increasing order. b) Explain why the endpoint order reverses. c) Is the converted interval symmetric about the conversion of the original midpoint?

Hints

- Apply the conversion to both endpoints separately. - Check whether the conversion increases or decreases as its input increases. - A nonlinear transformation need not preserve equal endpoint distances.

Solution

1. At \(30.0\) miles per gallon, \(L=\frac{235.2}{30.0}=7.84\). At \(28.4\) miles per gallon, \(L=\frac{235.2}{28.4}\approx8.28\). 2. The converted interval is approximately \((7.84, 8.28)\) liters per \(100\) kilometers. 3. The function \(L=235.2/M\) decreases as \(M\) increases, so the larger miles-per-gallon endpoint becomes the smaller liters-per-\(100\)-kilometers endpoint. 4. The original midpoint is \(29.2\), whose conversion is \(\frac{235.2}{29.2}\approx8.05\). Its distances to \(7.84\) and \(8.28\) differ, so the converted interval is not symmetric about that value.

Answer

a) Approximately \((7.84, 8.28)\) liters per \(100\) kilometers. b) The conversion is decreasing, so larger miles-per-gallon values produce smaller liters-per-\(100\)-kilometers values. c) No. A reciprocal conversion does not preserve symmetry.
54932411
From the same sample, an analyst calculates these intervals for a population difference: <table><thead><tr><th>Confidence level</th><th>Interval</th></tr></thead><tbody><tr><td>90%</td><td>\((0.4, 4.6)\)</td></tr><tr><td>95%</td><td>\((-0.2, 5.2)\)</td></tr><tr><td>99%</td><td>\((-1.4, 6.4)\)</td></tr></tbody></table> After seeing the intervals, the analyst reports only the \(90\%\) interval because it excludes \(0\). a) Which intervals support a clear positive difference? b) Explain why selecting the confidence level after seeing the endpoints weakens the inference. c) State one defensible reporting plan. Equivalent plans are acceptable if they avoid data-dependent level selection.

Hints

- Check the location of zero relative to each complete interval. - Ask whether the reporting rule was fixed before random data determined the endpoints. - A defensible procedure should not change merely because one result is more favorable.

Solution

1. Only the \(90\%\) interval lies entirely above \(0\); the \(95\%\) and \(99\%\) intervals include \(0\). 2. Choosing the confidence level because it gives the desired conclusion makes the selection depend on random sample results and hides the less decisive intervals. 3. The confidence level and primary decision rule should be chosen before examining results, and the report should disclose the planned interval and any additional levels shown.

Answer

a) Only the \(90\%\) interval. b) Post hoc level selection favors the most decisive-looking result and no longer follows a prespecified inference rule. c) One defensible plan is to preselect the confidence level and report it transparently, with any other intervals clearly labeled as additional analyses.
54932711
A symmetric calculation for a population proportion gives \(0.98\pm 0.05\), or \((0.93, 1.03)\). a) Identify the impossible part of the interval. b) Explain why simply replacing \(1.03\) by \(1.00\) changes the stated procedure. c) Suggest one simulation-based approach that respects the parameter's range. Equivalent valid approaches are acceptable.

Hints

- Recall the allowable range of the population parameter. - Changing an endpoint after calculation can change the method's long-run behavior. - Consider constructing endpoints from simulated values that obey the parameter bounds.

Solution

1. A population proportion cannot exceed \(1\), so the upper endpoint \(1.03\) is impossible. 2. Truncating the endpoint changes the interval after it has been calculated and can alter the method's long-run capture behavior; it is not the same symmetric procedure. 3. A bootstrap or repeated-sampling simulation can generate plausible proportion estimates and use appropriate percentiles, all of which remain within \([0, 1]\).

Answer

a) The endpoint \(1.03\) is impossible for a proportion. b) Truncation creates a different, asymmetric procedure whose coverage must be checked. c) One valid approach is a simulation-based percentile interval from plausible resampled proportions.
54932911
A confidence interval for \(\log_{10}(C)\), where \(C\) is a positive concentration, is \((1.2, 1.5)\). a) Convert the interval to the original concentration scale. b) Explain why the converted interval is asymmetric on the original scale. c) Find the multiplicative factor from the lower endpoint to the upper endpoint.

Hints

- Undo the logarithm at each endpoint separately. - An exponential transformation preserves order but changes spacing. - Use exponent rules when comparing the two transformed endpoints.

Solution

1. Exponentiating both endpoints gives \(10^{1.2}\approx15.85\) and \(10^{1.5}\approx31.62\), so the interval is approximately \((15.85, 31.62)\). 2. Exponentiation converts equal additive distances on the logarithmic scale into unequal additive distances on the original scale. 3. The endpoint ratio is \(\frac{10^{1.5}}{10^{1.2}}=10^{0.3}\approx2.00\).

Answer

a) Approximately \((15.85, 31.62)\). b) The exponential transformation does not preserve equal additive distances. c) The upper endpoint is about \(2.00\) times the lower endpoint.
54933311
A \(95\%\) confidence interval for the difference \(\mu_A-\mu_B\) is \((-0.4, 0.6)\). Differences between \(-1\) and \(1\) unit are considered practically unimportant. a) Does the interval show a clear direction for the difference? b) Is every value in the interval practically unimportant under the stated rule? c) Explain how “no clear direction” can coexist with evidence of practical equivalence.
Figure for problem 549333

Hints

- Check zero to assess direction and the tolerance boundaries to assess importance. - Two different decision questions can use two different reference values. - Look at the entire confidence interval, not only its midpoint.

Solution

1. The interval contains \(0\), so it does not show a clear positive or negative direction. 2. Every value from \(-0.4\) to \(0.6\) lies inside the practically unimportant range \([-1, 1]\). 3. The data leave uncertainty about which mean is slightly larger, but they restrict the plausible difference to values too small to matter practically.

Answer

a) No; the interval includes \(0\). b) Yes; the whole interval is within \([-1, 1]\). c) The sign is uncertain, but all plausible differences are smaller than the practical threshold.
54933411
Sample 1 has \(n=100\), point estimate \(20\), and confidence interval \((18, 22)\). Sample 2 has \(n=300\), point estimate \(24\), and confidence interval \((22.8, 25.2)\). Both samples target the same population mean. a) Find the sample-size-weighted combined point estimate. b) A student averages the two lower endpoints and the two upper endpoints and reports \((20.4, 23.6)\) as a combined interval. Explain why this is not a valid construction. c) What information or procedure is needed to form a valid combined interval?

Hints

- Weight each point estimate by how many observations produced it. - An interval is more than two numbers; its endpoints come from a sampling procedure. - Combine the information first, then recalculate uncertainty using a justified method.

Solution

1. The combined point estimate is \(\frac{100\cdot20+300\cdot24}{400}=23\). 2. Averaging endpoints does not correctly combine sampling variances, dependence, confidence methods, or raw observations; it has no established capture rate. 3. A valid interval requires pooling the underlying data or combining estimates with a justified variance-based method and then recalculating uncertainty.

Answer

a) \(23\). b) Endpoint averaging has no justified confidence level and ignores how uncertainty should be combined. c) Pool the data and recompute the interval, or use a valid method based on the estimates' sampling variances.

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