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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 \(71.5\pm4.5\).
55558211
A confidence interval is reported as \(0.62\pm0.04\). Identify the point estimate and the margin of error.

Hints

- Rewrite the plus-or-minus notation in words before naming its parts. - Ask which number stays at the center and which number describes the distance from that center to an endpoint.

Solution

1. In estimate-plus-or-minus form, the center is the point estimate. 2. The amount after \(\pm\) is the margin of error. 3. Therefore, the point estimate is \(0.62\) and the margin of error is \(0.04\).

Answer

Point estimate: \(0.62\); margin of error: \(0.04\).
55558311
Panels a) and b) show two confidence intervals for the same type of population quantity. Which interval has the smaller margin of error? Explain using the displayed widths.
Figure for problem 555583

Hints

- Compare the horizontal lengths of the two displayed segments. - Think about how the distance from an interval's center to either endpoint changes when the whole interval becomes narrower.

Solution

1. Margin of error is half the width of a confidence interval. 2. Panel b) shows the shorter interval, so it has the smaller half-width. 3. Therefore, interval b) has the smaller margin of error.

Answer

Interval b), because it is narrower and therefore has the smaller half-width.
55558711
A polling organization reports a \(95\%\) confidence interval for a population proportion. In repeated-sampling terms, what does the \(95\%\) confidence level mean?

Hints

- Ask what would change if a new random sample were drawn and a new interval were computed. - Focus on the long-run performance of the interval-producing procedure rather than on individual people in the population.

Solution

1. Imagine repeating the same sampling-and-interval procedure many times. 2. About \(95\%\) of the intervals produced by that procedure would contain the true population proportion.

Answer

If the same sampling-and-interval procedure were repeated many times, about \(95\%\) of the intervals produced would contain the true population proportion.
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> What relationship between confidence level and interval width is shown, and why does that tradeoff occur when the sample is unchanged?

Hints

- Compare the endpoint differences across the three rows. - The sample itself has not changed, so focus on what the confidence level changes about the interval procedure. - Greater capture frequency and narrower intervals pull in opposite directions.

Solution

1. The interval widths are \(6\), \(8\), and \(12\) percentage points for the \(90\%\), \(95\%\), and \(99\%\) intervals. 2. As the confidence level increases, the interval becomes wider. 3. A method designed to capture the fixed population parameter more often over repeated samples must extend farther from the same sample estimate, trading precision for greater confidence.

Answer

Higher confidence produces a wider interval for the same sample because greater long-run capture requires allowing a broader range of plausible parameter values.
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.” Rewrite the final sentence so the \(95\%\) confidence level is interpreted correctly.

Hints

- Separate what is fixed after the data are collected from what would vary over repeated samples. - Interpret the percentage as a property of the interval-producing procedure. - Describe what would happen across many repetitions of the same method.

Solution

1. After the sample is observed and the interval is calculated, the population mean is fixed; the \(95\%\) refers to the long-run success rate of the interval-producing method. 2. A correct statement is that about \(95\%\) of intervals produced by this method from repeated random samples would contain the town's true mean monthly electricity use.

Answer

About \(95\%\) of intervals produced by this method from repeated random samples would contain the town's true mean monthly electricity use.
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. Does the very small reported margin of error make this result reliable for all voters? Explain what uncertainty the margin of error addresses and what problem it misses.

Hints

- Separate precision of a calculation from representativeness of the data source. - Ask what kind of randomness a margin-of-error formula is designed to quantify. - A very large sample cannot automatically repair systematic selection into the sample.

Solution

1. No. The volunteer sample is systematically unrepresentative because both visiting the site and choosing to respond are related to support for the policy. 2. The small margin of error describes random sampling variability under the assumptions behind the interval calculation. 3. It does not measure or correct voluntary-response bias or undercoverage, so high numerical precision does not make the estimate accurate for all voters.

Answer

No. The small margin of error addresses sampling variability, but it does not correct the voluntary-response and coverage bias in the sample.
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.
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.” Evaluate the claim and identify the confidence interval that would directly address equality of the two population means.

Hints

- A confidence interval gives a range of plausible values, not proof of one exact value. - Ask which single population parameter directly represents equality of the two means. - The uncertainty in a difference is not determined by visual overlap alone.

Solution

1. The claim is incorrect. Overlap of two separate confidence intervals does not prove that the population means are equal. 2. The relevant parameter for equality is the difference \(\mu_A-\mu_B\). A confidence interval for that difference directly shows whether \(0\), the equality value, is plausible.

Answer

The claim is incorrect. Overlapping separate intervals do not prove equal means; use a confidence interval for \(\mu_A-\mu_B\) and check whether it includes \(0\).
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. Explain why this report is misleading and give one way to report the interval without hiding its uncertainty.

Hints

- Compare the visible widths before and after rounding. - The reported precision should preserve the uncertainty communicated by the interval. - An equivalent center-and-margin form is acceptable if it retains the same endpoints.

Solution

1. The original interval has nonzero width, but rounding both endpoints to one decimal makes it appear to collapse to a single value. 2. The report should retain enough precision to show the uncertainty, for example \((2.46,2.54)\) seconds or equivalently \(2.50\pm0.04\) seconds.

Answer

The rounding hides the interval's nonzero width. A suitable report is \((2.46,2.54)\) seconds or \(2.50\pm0.04\) seconds.
54931711
A poll reports \(46\%\pm3\) percentage points. A reader interprets this as “plus or minus \(3\%\) of \(46\%\).” Give the interval intended by the poll and explain the reader's error.

Hints

- Decide whether the stated amount is added directly to the percentage or first multiplied by it. - Keep absolute percentage-point differences separate from relative percent changes. - Use the wording of the reported margin to choose the correct operation.

Solution

1. Three percentage points are added to and subtracted from \(46\%\) directly, giving \((43\%,49\%)\). 2. The reader treats \(3\) percentage points as a relative \(3\%\) change. Percentage points are absolute differences on the percentage scale, whereas a percent change is relative to the starting percentage.

Answer

The intended interval is \((43\%,49\%)\). The reader confuses an absolute change of \(3\) percentage points with a relative change of \(3\%\).
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 \((56\%,62\%)\). Identify the inconsistency and give the confidence interval that matches the stated estimate and margin of error.

Hints

- Reconstruct an interval from its center and half-width. - Then compare that result with the printed endpoints. - Equivalent forms of the same interval must have the same midpoint.

Solution

1. An interval with estimate \(54\%\) and margin of error \(3\) percentage points must be centered at \(54\%\). 2. Its endpoints are \(54-3=51\) and \(54+3=57\), so the matching interval is \((51\%,57\%)\). The printed interval is centered at \(59\%\), so it cannot represent the stated estimate.

Answer

The printed interval is inconsistent because it is centered at \(59\%\), not \(54\%\). The correct interval is \((51\%,57\%)\).
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 reader says this means every participant reduced time by between \(0.3\) and \(1.2\) hours. Explain why that interpretation is wrong and state what the interval actually estimates.

Hints

- Identify the exact population parameter named in the interval description. - Keep an interval for an average separate from a range of raw individual values. - Use the subtraction order to interpret the negative signs.

Solution

1. The interval is for the population mean of the paired changes, not for every individual's change. 2. Because the entire interval is negative for after minus before, it supports a population mean reduction between \(0.3\) and \(1.2\) hours. 3. Individual participants can have changes outside that range.

Answer

The interval estimates the population mean paired change, not the range of individual changes. It corresponds to a population mean reduction between \(0.3\) and \(1.2\) hours.
54932811
A bootstrap \(95\%\) confidence interval for the population median emergency-response time is \((24,29)\) minutes. Explain what this interval estimates and why it does not mean that \(95\%\) of individual response times lie between \(24\) and \(29\) minutes.

Hints

- Use the exact population quantity named in the interval description. - Separate uncertainty about a population parameter from the distribution of individual observations. - A confidence level is not the percentage of raw data values inside the endpoints.

Solution

1. The interval gives plausible values from \(24\) to \(29\) minutes for the population median emergency-response time under the bootstrap method. 2. A confidence interval for a population median describes uncertainty in that one population parameter; it is not an interval containing a specified percentage of individual response times.

Answer

It estimates the population median response time. It does not describe the middle \(95\%\) of individual response times or any fixed fraction of individual observations.
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.
54933711
A convenience sample gives a \(95\%\) confidence interval of \((64\%,70\%)\). Using the same sample, a researcher calculates a wider \(99\%\) interval of \((62\%,72\%)\) and claims that the higher confidence solves the sampling problem. Explain why increasing the confidence level does not repair the problem.

Hints

- Identify what changed between the two reported intervals and what did not. - Separate random variability from systematic selection into the sample. - A calculation cannot by itself repair an unrepresentative data source.

Solution

1. Both intervals are based on the same convenience sample, so both can be centered on a systematically unrepresentative estimate. 2. Raising the confidence level widens the allowance for random sampling variability; it does not correct selection bias in how the observations entered the sample. 3. A suitable inference for the target population requires a representative probability-based sampling design or another defensible way to address the selection mechanism.

Answer

Higher confidence only widens the interval for sampling uncertainty. It does not remove the selection bias of the convenience sample.
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
Daniela says, “A \(95\%\) confidence method guarantees that exactly \(95\) of the next \(100\) intervals will contain the true parameter.” Correct the statement and explain why observing \(92\) captures in one batch of \(100\) intervals would not by itself disprove a valid \(95\%\) method.

Hints

- Separate a long-run proportion from a fixed count in every finite batch. - Expected behavior does not force every observed batch to match it exactly. - Ask whether random variation in capture counts is possible under a valid method.

Solution

1. A \(95\%\) confidence level is a long-run capture rate for the method, not a guarantee of exactly \(95\) successes in every batch of \(100\). 2. Even when each interval has a \(0.95\) capture probability under the method, the observed count in a finite batch varies randomly around its expected value. Therefore, \(92\) captures can occur without contradicting the long-run \(95\%\) behavior.

Answer

A \(95\%\) method captures the parameter about \(95\%\) of the time in the long run; it does not guarantee exactly \(95\) captures per \(100\). A batch with \(92\) captures can occur through ordinary sampling variation.
54934111
A \(95\%\) confidence interval for the ratio \(\mu_A/\mu_B\) is \((0.92,1.08)\). Does this interval show a clear difference between the two population means? Explain using the reference value for equality.

Hints

- For a ratio, identify the value that represents equal numerator and denominator quantities. - Check whether that reference value lies inside the entire interval. - Base the conclusion on the parameter the interval actually estimates.

Solution

1. For the ratio \(\mu_A/\mu_B\), equality of the population means corresponds to the reference value \(1\). 2. Because \(1\) lies inside \((0.92,1.08)\), equality remains plausible at this confidence level, so the interval does not show a clear direction of difference.

Answer

No. Equality corresponds to the ratio \(1\), and \(1\) lies inside \((0.92,1.08)\).
54977711
The histogram shows a 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 from the graph. c) Which interval better reflects the shape displayed by the bootstrap distribution? Explain using the histogram.
Figure for problem 549777

Hints

- First apply the supplied symmetric rule around the sample estimate. - Use the marked reference lines and their numerical positions on the horizontal scale to read the percentile endpoints. - For part c), compare how the histogram extends on the two sides of its main concentration, then compare that with the symmetry or asymmetry of each interval.

Solution

1. The symmetric interval is \(8.0\pm2.4\), 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 histogram has a longer right tail than left tail. The percentile interval is also asymmetric, extending farther above \(8.0\) than below it, while the symmetric interval forces equal distances on both sides. Therefore, the percentile interval better reflects the displayed shape.

Answer

a) \((5.6, 10.4)\) b) \((6.25, 11.00)\) c) The percentile interval, because its asymmetry reflects the longer right tail visible in the bootstrap histogram.
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.
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.
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.
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 \((55\%,65\%)\). Could unresolved nonresponse move the proportion favoring the proposal among all \(400\) selected residents outside that interval? Use the two extreme assumptions about the \(100\) nonrespondents to justify your answer.

Hints

- Keep the original selected-sample denominator \(400\) when considering the missing responses. - Test the two extreme possibilities for the \(100\) people whose responses are unknown. - Compare the resulting proportions with the reported respondent-only interval.

Solution

1. If all \(100\) nonrespondents oppose the proposal, then \(180\) of \(400\) selected residents favor it, giving \(45\%\). 2. If all \(100\) nonrespondents favor it, then \(280\) of \(400\) favor it, giving \(70\%\). 3. Both extremes lie outside \((55\%,65\%)\), so the respondent-only confidence interval does not account for uncertainty from systematic differences between respondents and nonrespondents.

Answer

Yes. Under the two extreme assumptions the full selected-sample proportion could be as low as \(45\%\) or as high as \(70\%\), so nonresponse uncertainty can extend beyond \((55\%,65\%)\).
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.
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 original 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> What does increasing the number of bootstrap resamples improve here, and what important limitation of the original sample does it not improve?

Hints

- Compare how much the endpoints move between repeated runs at each resample count. - Separate randomness created by the computer simulation from limitations of the original collected data. - Repeating a computation cannot create new information about the population.

Solution

1. The two \(10{,}000\)-resample runs have nearly identical endpoints, while the two \(100\)-resample runs differ much more. More bootstrap resamples therefore reduce Monte Carlo variation in estimating the bootstrap percentiles. 2. Repeating the bootstrap more times does not add new population observations or fix bias or lack of representativeness in the original sample.

Answer

More bootstrap resamples make the numerical endpoints more stable by reducing simulation noise. They do not repair bias or poor representativeness in the 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.

Hints

- Check zero to assess direction and the tolerance boundaries to assess practical importance. - The two questions use different reference values. - Compare the entire confidence interval with the entire practically unimportant range.

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 \(n=100\), sample mean \(20\), and confidence interval \((18,22)\). Sample 2 has \(n=300\), sample mean \(24\), and confidence interval \((22.8,25.2)\). Both samples target the same population mean. a) Find the sample-size-weighted combined sample mean. b) Thabo 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 sample mean 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 sample mean 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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