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Idea of a sampling distribution

Learning goals

  • Distinguishes a sample statistic from its population parameter.
  • Explains a sampling distribution as variation across repeated samples.
  • Describes the center and spread of a sampling distribution.
  • Explains how sample size affects sampling variability.
  • Distinguishes a sampling distribution from individual observations.

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Idea of a sampling distribution 1
A school district has \(1200\) seniors. A simple random sample of \(80\) seniors is selected, and the sample proportion \(\hat p\) who plan to attend a two-year college is calculated. Identify the population, the sample, the parameter, the statistic, and the sampling distribution in this setting.

Hints

- Separate the full group of interest from the individuals actually observed. - Distinguish a fixed population quantity from a value computed from one sample. - Imagine repeating the sampling method many times to identify the distribution being requested.

Solution

1. The population is all \(1200\) seniors in the district. 2. The sample is the \(80\) selected seniors. 3. The parameter is the true proportion \(p\) of all district seniors who plan to attend a two-year college. 4. The statistic is the observed sample proportion \(\hat p\). 5. The sampling distribution is the distribution of \(\hat p\) over all possible simple random samples of \(80\) seniors from the population.

Answer

Population: all \(1200\) district seniors. Sample: the \(80\) selected seniors. Parameter: the population proportion \(p\). Statistic: the sample proportion \(\hat p\). Sampling distribution: the distribution of \(\hat p\) across all possible simple random samples of size \(80\).
Idea of a sampling distribution 2
A researcher takes one random sample of \(200\) voters and obtains \(\hat p=0.54\). The researcher calls the single value \(0.54\) “the sampling distribution of \(\hat p\).” Explain the error and describe what the sampling distribution actually represents.

Hints

- Decide whether the quoted number is one outcome or a collection of possible outcomes. - Imagine repeating the entire sampling process. - Identify what varies from sample to sample.

Solution

1. The value \(0.54\) is one realized statistic from one sample. 2. A sampling distribution contains the possible values of \(\hat p\) and their probabilities under repeated use of the same sampling method. 3. It describes sample-to-sample variability, not the result of only one sample.

Answer

The value \(0.54\) is one sample proportion, not a distribution. The sampling distribution is the probability distribution of \(\hat p\) across all possible random samples of size \(200\) drawn by the same method.
Idea of a sampling distribution 3
A finite population has mean \(\mu\). Instead of taking a partial sample, a researcher measures every member of the population and computes the mean. Describe the sampling distribution of this mean if the same census procedure is conceptually repeated.

Hints

- Identify whether any members are left unobserved. - Ask what could vary if the same complete measurement procedure were repeated. - Compare the results you would get from two repeats of the complete procedure.

Solution

1. A census includes the entire population every time. 2. The computed mean is always exactly \(\mu\). 3. Therefore, the sampling distribution places probability \(1\) at \(\mu\) and has standard deviation \(0\).

Answer

The mean is always \(\mu\). Thus \(P(\bar X=\mu)=1\), and the sampling distribution has standard deviation \(0\).
Idea of a sampling distribution 4
A sample of \(20\) school-bus trips has an average travel time of \(24\,\text{minutes}\). One of those trips lasted \(18\,\text{minutes}\). Which number is the statistic calculated from the full sample?

Hints

- Decide which number comes from a single trip. - Look for the number computed from the whole sample.

Solution

1. A statistic summarizes a sample of observations. 2. The \(18\)-minute time is one individual observation. 3. The \(24\)-minute average summarizes all \(20\) trips, so it is the sample statistic.

Answer

The sample mean, \(24\,\text{minutes}\).
Idea of a sampling distribution 5
Exactly \(40\%\) of all seniors in a school district take a foreign-language class. In one random sample, \(37\%\) take one; in another random sample, \(43\%\) do. Which percentage is the fixed population parameter, and why can the two sample percentages differ?

Hints

- Identify the statement about all seniors, not just those selected. - Think about which students are included in each random sample.

Solution

1. The population parameter is the proportion of all seniors, which is fixed at \(40\%\). 2. The \(37\%\) and \(43\%\) values are statistics from separate random samples. 3. Random samples need not contain exactly the same mix of students, so their sample proportions can differ.

Answer

The parameter is \(40\%\). The sample percentages vary because the selected students differ from sample to sample.
Idea of a sampling distribution 6
A simulation repeatedly takes random samples of students and records one average test score from each sample. The histogram summarizes \(100\) sample means. a) Does one horizontal-axis value represent an individual score or a sample mean? b) How many sample means are at least \(70\) and less than \(75\)?
Figure for problem 563495

Hints

- Recall which number is recorded after each simulated sample is taken. - Read the frequency-axis value for the specified score interval. - Pay attention to which interval contains the boundary at \(75\).

Solution

1. Each simulated sample contributes one mean, so the horizontal axis gives sample means, not individual test scores. 2. The bar for means from \(70\) up to but not including \(75\) has height \(40\).

Answer

a) A sample mean. b) \(40\) sample means.
Idea of a sampling distribution 7
A probability simulation is repeated in \(50\) independent runs of \(200\) trials each. The \(50\) estimated probabilities have mean \(0.413\) and standard deviation \(0.034\). Interpret the standard deviation in this context. What would generally happen to this run-to-run standard deviation if each run used \(2000\) trials?

Hints

- Distinguish variability among estimates from the probability being estimated. - More trials stabilize relative frequencies. - The number of independent runs and trials per run play different roles.

Solution

1. The standard deviation \(0.034\) describes typical run-to-run variation of the estimated probability around its mean. 2. Increasing the number of trials per run reduces sampling variability. 3. The run-to-run standard deviation would generally become smaller with \(2000\) trials.

Answer

The value \(0.034\) measures typical variation among run estimates. With \(2000\) trials per run, that variation should generally decrease.
Idea of a sampling distribution 8
A simulation repeatedly takes a random sample of the same size from a population and records the sample mean. The results of \(1000\) repetitions are summarized below. <table> <tr><th>Sample mean</th><th>\(8\)</th><th>\(9\)</th><th>\(10\)</th><th>\(11\)</th><th>\(12\)</th></tr> <tr><th>Frequency</th><td>\(50\)</td><td>\(220\)</td><td>\(430\)</td><td>\(240\)</td><td>\(60\)</td></tr> </table> a) Use the simulation to estimate \(P(\bar X\ge11)\). b) Estimate the mean of the sampling distribution. c) Explain what would change if the simulation were repeated with many more samples.

Hints

- Convert relevant frequencies to relative frequencies. - Treat the table as an empirical probability distribution. - Distinguish increasing the number of simulated samples from increasing the size of each original sample.

Solution

1. The simulated probability of \(\bar X\ge11\) is \(\frac{240+60}{1000}=0.30\). 2. The simulated sampling-distribution mean is \(\frac{8(50)+9(220)+10(430)+11(240)+12(60)}{1000}=10.04\). 3. With many more repetitions, random simulation noise would generally decrease and the empirical distribution would more closely approximate the true sampling distribution.

Answer

a) \(0.30\). b) \(10.04\). c) More repetitions would usually make the simulated distribution more stable and closer to the true sampling distribution.
Idea of a sampling distribution 9
A simulation uses \(500\) random samples of size \(25\) to approximate the sampling distribution of a sample mean. Consider two changes: a) Increase the number of simulated samples from \(500\) to \(10{,}000\), keeping sample size \(25\). b) Increase each sample size from \(25\) to \(100\), keeping \(500\) simulated samples. Explain the main effect of each change.

Hints

- Decide which change alters the statistic being repeatedly computed. - Separate simulation precision from the mathematical variability of the statistic. - Track what remains fixed in each scenario.

Solution

1. In a), the target sampling distribution is unchanged because the sample size remains \(25\). More repetitions make the simulation's picture of that distribution smoother and more stable. 2. In b), the target sampling distribution changes because the statistic is now based on samples of size \(100\). Its spread is smaller, and its shape is generally closer to normal. 3. The number of repetitions controls simulation accuracy; the sample size controls the actual sampling distribution.

Answer

a) More repetitions improve the empirical approximation to the same sampling distribution. b) Larger samples create a different sampling distribution with less variability and generally a more normal shape.
Idea of a sampling distribution 10
For samples of size \(16\), the sampling distribution of \(\bar X\) has mean \(12\) and standard deviation \(1.5\). Let \(S\) be the sample sum from the same samples. a) Find the mean and standard deviation of the sampling distribution of \(S\). b) If a particular sample has \(\bar x=14\), what is its sample sum?

Hints

- Write the exact relationship between the sum and mean for a fixed sample size. - Apply that transformation to both the center and spread. - Use the same relationship for the particular observed sample.

Solution

1. For every sample, \(S=16\bar X\). 2. Therefore, \(E(S)=16(12)=192\). 3. The standard deviation is \(16(1.5)=24\). 4. If \(\bar x=14\), then \(S=16(14)=224\).

Answer

a) Mean \(192\); standard deviation \(24\). b) The sample sum is \(224\).
Idea of a sampling distribution 11
A population has standard deviation \(15\). A simulation repeatedly takes independent random samples of size \(25\) from this population and records their means. The simulated sample means have standard deviation \(3.2\). Compare the simulated spread with the theoretical standard deviation of the sampling distribution. Explain why the values need not be exactly equal.

Hints

- Compute the theoretical sampling spread from the population spread and sample size. - Compare the two standard deviations numerically. - Recall that a finite simulation has random simulation variability.

Solution

1. For independent observations, the theoretical standard deviation of the sample mean is \(\frac{15}{\sqrt{25}}=3\). 2. The simulated standard deviation \(3.2\) is \(0.2\) larger than the theoretical value. 3. A finite simulation has random simulation variability, so its empirical standard deviation need not equal the theoretical standard deviation exactly.

Answer

The theoretical standard deviation is \(3\). The simulated value \(3.2\) is slightly larger, which can occur because a finite simulation does not reproduce the theoretical sampling distribution exactly.
Idea of a sampling distribution 12
A population has mean \(70\) and standard deviation \(12\). Compare the distribution of one randomly selected observation \(X\) with the sampling distribution of \(\bar X\) for independent random samples of size \(36\). a) Compare their means. b) Compare their standard deviations. c) Explain why the sampling distribution is not the distribution of the \(36\) raw observations in one sample.

Hints

- Identify what one value represents in each distribution. - Apply the sample-size effect only to the average. - Separate within-sample variation from sample-to-sample variation.

Solution

1. Both \(X\) and \(\bar X\) have mean \(70\). 2. The standard deviation of \(X\) is \(12\), while the standard deviation of \(\bar X\) is \(12/\sqrt{36}=2\). 3. The sampling distribution contains one mean from each repeated sample, whereas the raw-data distribution contains individual observations within one sample.

Answer

a) Both means are \(70\). b) The standard deviations are \(12\) for an individual observation and \(2\) for the sample mean. c) The sampling distribution describes statistics across repeated samples, not raw values inside one sample.
Idea of a sampling distribution 13
A simulation takes \(1000\) random samples of the same size from a population and records each sample mean. The results are grouped below. <table> <tr><th>Sample-mean interval</th><th>Midpoint</th><th>Frequency</th></tr> <tr><td>\(40\le\bar X<44\)</td><td>\(42\)</td><td>\(100\)</td></tr> <tr><td>\(44\le\bar X<48\)</td><td>\(46\)</td><td>\(250\)</td></tr> <tr><td>\(48\le\bar X<52\)</td><td>\(50\)</td><td>\(400\)</td></tr> <tr><td>\(52\le\bar X<56\)</td><td>\(54\)</td><td>\(200\)</td></tr> <tr><td>\(56\le\bar X<60\)</td><td>\(58\)</td><td>\(50\)</td></tr> </table> a) Estimate the probability that a new sample mean is at least \(52\). b) Use the interval midpoints to estimate the mean of the simulated sampling distribution.

Hints

- Treat the simulation frequencies as empirical probabilities. - Identify every interval included in the requested event. - For the center, weight each representative value by how often its interval occurred.

Solution

1. There are \(200+50=250\) simulated means at least \(52\), so the estimated probability is \(\frac{250}{1000}=0.25\). 2. The midpoint-weighted mean is \(\frac{100(42)+250(46)+400(50)+200(54)+50(58)}{1000}=49.4\).

Answer

a) Approximately \(0.25\). b) Approximately \(49.4\).
Idea of a sampling distribution 14
Two computer simulations use the same population of students: - Simulation A takes \(300\) random samples of \(5\) students each. - Simulation B takes \(100\) random samples of \(20\) students each. Both record one sample mean each time. Which simulation puts more values in its histogram? Which should have the less spread-out sample means? Explain why these are different questions.

Hints

- For the histogram, count the number of means the computer records. - For spread, think about how many students contribute to each average. - The number of averages drawn is different from the number of students per average.

Solution

1. Simulation A records \(300\) means; B records \(100\) means, so A's histogram contains more plotted values. 2. Each mean in B averages \(20\) students rather than \(5\), so B's sampling distribution is less spread out under independent random sampling. 3. The number of repetitions changes the amount of simulated data; the number in each sample changes the variability of the mean.

Answer

A has more plotted values (\(300\) versus \(100\)). B should have less spread among its sample means (size \(20\) versus \(5\)). Repetitions control the number of plotted means; sample size affects their spread.
Idea of a sampling distribution 15
A population contains two values, \(1\) and \(3\). Two values are selected independently with replacement, and their mean is \(\bar X\). Construct the probability distribution of \(\bar X\).

Hints

- Make a list of all ordered outcomes of the two selections. - Calculate a mean for each possible sample. - Different samples can give the same mean.

Solution

1. The four equally likely ordered samples are \((1,1)\), \((1,3)\), \((3,1)\), and \((3,3)\). 2. Their means are \(1,2,2,3\), respectively. 3. Therefore, the probabilities are \(P(\bar X=1)=\frac{1}{4}\), \(P(\bar X=2)=\frac{1}{2}\), and \(P(\bar X=3)=\frac{1}{4}\).

Answer

<table> <tr><th>\(\bar x\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th></tr> <tr><th>\(P(\bar X=\bar x)\)</th><td>\(\frac{1}{4}\)</td><td>\(\frac{1}{2}\)</td><td>\(\frac{1}{4}\)</td></tr> </table>
Idea of a sampling distribution 16
In a sample of \(8\) students, \(\hat p\) is the fraction who walk to school. a) Can \(\hat p=0.375\)? If so, how many students walk? b) Can \(\hat p=0.30\)? Explain.

Hints

- Write sample proportion as number of successes divided by sample size. - Work backward to a count and check whether that count makes sense.

Solution

1. A sample proportion must equal a whole-number count divided by \(8\). 2. Since \(8(0.375)=3\), the first value is possible with \(3\) students walking. 3. Since \(8(0.30)=2.4\) is not a whole-number count, the second value is impossible.

Answer

a) Yes. \(3\) students walk. b) No. The count would have to be \(2.4\) students, which is not possible.
Idea of a sampling distribution 17
A population consists of the numbers \(0\) and \(10\). Two numbers are sampled either with replacement or without replacement. Let \(\bar X\) be the sample mean. a) Find \(P(\bar X=5)\) for sampling with replacement. b) Find \(P(\bar X=5)\) for sampling without replacement. c) Which method gives a more variable sample mean?

Hints

- Work out which repeated values are allowed under each method. - List the possible sample means before assigning probabilities. - One method forces the two different population values to be selected.

Solution

1. With replacement, the four equally likely ordered samples are \((0,0)\), \((0,10)\), \((10,0)\), and \((10,10)\). Two have mean \(5\), so the probability is \(\frac{2}{4}=\frac{1}{2}\). 2. Without replacement, the only two selected values must be \(0\) and \(10\), so the mean is always \(5\) and its probability is \(1\). 3. The with-replacement mean can vary among \(0,5,10\); the without-replacement mean is constant. With replacement is more variable.

Answer

a) \(\frac{1}{2}\). b) \(1\). c) Sampling with replacement gives the more variable mean.
Idea of a sampling distribution 18
A researcher records three battery runtimes: \(4\), \(5\), and \(9\) hours. One bootstrap resample is \((4,4,9)\). Is this a valid bootstrap resample? Find its mean and explain why the repeated value is allowed.

Hints

- Compare the resample's length with the original sample's length. - Check whether all values were drawn from the original observations. - Think about what sampling with replacement permits.

Solution

1. A bootstrap resample draws three observations from the original data with replacement. 2. Every entry in \((4,4,9)\) comes from the original list. The repeated \(4\) is allowed because each draw can select a value again. 3. The resampled mean is \(\frac{4+4+9}{3}=\frac{17}{3}\approx5.67\) hours.

Answer

Yes. Its mean is \(\frac{17}{3}\) hours. The \(4\) may occur twice because bootstrap sampling uses replacement.
Idea of a sampling distribution 19
One histogram shows the heights of individual plants in a greenhouse. The other shows mean heights from many repeated random samples of \(25\) plants. Identify which panel displays individual heights and which displays sample means. Explain what one recorded value stands for in each panel.
Figure for problem 563501

Hints

- Look for the display with the wider spread of values. - A sample mean combines many observations into one number. - Remember that the two horizontal axes measure heights, but the values come from different units of observation.

Solution

1. Panel a) has a much wider spread in individual heights. 2. Panel b) is tightly concentrated around a similar center, as expected when each plotted value averages \(25\) individual heights. 3. One value in a) is one plant's height; one value in b) is the average height of a whole sample of \(25\) plants.

Answer

a) Individual plant heights; each value belongs to one plant. b) Sample means; each value is an average of \(25\) plants.
Idea of a sampling distribution 20
A Monte Carlo simulation estimates an event probability using \(400\) independent trials and observes \(265\) successes. a) Find the estimated probability. b) Estimate the Monte Carlo standard error using \(\sqrt{\hat p(1-\hat p)/n}\). Round the standard error to four decimal places. c) Give an approximate interval of two standard errors on either side of the estimate. Round the interval endpoints to three decimal places, and interpret the interval as simulation uncertainty rather than uncertainty in the model itself.

Hints

- The simulation estimate is the observed fraction of successes. - Use that fraction and the number of trials in the stated standard-error formula. - For the interval, move two estimated standard errors in each direction from the estimate before rounding. - Separate Monte Carlo variability from uncertainty about whether the probability model itself is appropriate.

Solution

1. \(\hat p=\frac{265}{400}=0.6625\). 2. The estimated standard error is \(\sqrt{\frac{0.6625\cdot0.3375}{400}}\approx0.0236\). 3. Two standard errors give \(0.6625\pm2(0.02364285\ldots)\), with endpoints approximately \(0.6152143\) and \(0.7097857\). 4. Rounded to three decimal places, the interval is \((0.615,0.710)\). 5. This interval describes random Monte Carlo variation from using a finite number of simulated trials, assuming the simulation model is fixed.

Answer

a) \(\hat p=0.6625\). b) \(0.0236\). c) \((0.615,0.710)\). It describes simulation-to-simulation variation from using \(400\) trials, not uncertainty about the fixed probability model.
Idea of a sampling distribution 21
A sample of six delivery times, in minutes, is \(18,21,24,25,29,35\). A bootstrap trial samples six values from this list with replacement and calculates the sample mean. Describe how to use many bootstrap trials to estimate the probability that a resampled mean exceeds \(27\) minutes. Out of \(4000\) trials, \(932\) means exceeded \(27\). Give the estimate.

Hints

- A resample should have the same size as the original sample. - What feature distinguishes resampling with replacement from selecting a new subset? - Use the fraction of resampled means meeting the condition.

Solution

1. For each trial, draw six times independently from the observed list, allowing the same value to be selected more than once. 2. Calculate the mean of the six selected times and record whether it exceeds \(27\,\text{min}\). 3. Repeat many times. The estimated probability is the proportion of resampled means above \(27\,\text{min}\): \(\frac{932}{4000}=0.233\).

Answer

Sample six values with replacement from the given times, calculate each sample mean, and count how many means exceed \(27\,\text{min}\). The estimate is \(\frac{932}{4000}=0.233\), or \(23.3\%\).
Idea of a sampling distribution 22
Nine independent measurements have the same distribution. Their average \(\overline{X}\) has mean \(12\) and standard deviation \(1.5\). Find the mean and standard deviation of one measurement.

Hints

- Relate the center of an average to the center of its components. - Work backward from the way averaging changes spread. - Use the number of independent measurements in the scaling factor.

Solution

1. An average has the same mean as each identically distributed component, so \(\mu_X=12\). 2. For nine independent measurements, \(\sigma_{\overline{X}}=\frac{\sigma_X}{\sqrt{9}}\). 3. Therefore, \(1.5=\frac{\sigma_X}{3}\), so \(\sigma_X=4.5\).

Answer

One measurement has mean \(12\) and standard deviation \(4.5\).
Idea of a sampling distribution 23
Independent instrument readings have standard deviation \(3.2\) units. A technician will average \(n\) readings. Find the smallest integer \(n\) for which the standard deviation of the average is at most \(0.8\) units.

Hints

- Express the spread of an average in terms of the number of independent readings. - Translate “at most” into an inequality. - Check that the final value is an integer count and satisfies the requirement.

Solution

1. The standard deviation of the average is \(\frac{3.2}{\sqrt{n}}\). 2. Require \(\frac{3.2}{\sqrt{n}}\le0.8\), which gives \(\sqrt{n}\ge4\). 3. Therefore, \(n\ge16\), so the smallest integer is \(16\).

Answer

\(n=16\) readings.
Idea of a sampling distribution 24
A population consists of \(2,5,9\). Two values are sampled independently with replacement. Let \(R\) be the range of the two sampled values. a) Construct the sampling distribution of \(R\). b) Find \(E(R)\).

Hints

- Treat ordered samples as equally likely because sampling is with replacement. - Group samples according to the statistic rather than the raw pair. - Use the completed sampling distribution for the long-run average range.

Solution

1. There are nine equally likely ordered samples. The three samples with matching values have range \(0\). 2. The pairs using \(2\) and \(5\) give range \(3\) in two orders; those using \(5\) and \(9\) give range \(4\) in two orders; those using \(2\) and \(9\) give range \(7\) in two orders. 3. The expected range is \(0(3/9)+3(2/9)+4(2/9)+7(2/9)=\frac{28}{9}\approx3.11\).

Answer

a) <table> <tr><th>\(r\)</th><th>\(0\)</th><th>\(3\)</th><th>\(4\)</th><th>\(7\)</th></tr> <tr><th>\(P(R=r)\)</th><td>\(\frac{1}{3}\)</td><td>\(\frac{2}{9}\)</td><td>\(\frac{2}{9}\)</td><td>\(\frac{2}{9}\)</td></tr> </table> b) \(E(R)=\frac{28}{9}\approx3.11\).
Idea of a sampling distribution 25
In a population, the proportion with a certain characteristic is \(p=0.40\). A random sample of \(3\) independent individuals is selected, and \(\hat p\) is the sample proportion with the characteristic. a) Construct the sampling distribution of \(\hat p\). b) Find its mean and standard deviation.

Hints

- List every possible count of individuals with the characteristic. - Convert each count to a sample proportion. - Use the resulting distribution to summarize its center and spread.

Solution

1. The number with the characteristic can be \(0,1,2,3\), so \(\hat p\) can be \(0,\frac{1}{3},\frac{2}{3},1\). 2. The corresponding probabilities are \((0.60)^3=0.216\), \(3(0.40)(0.60)^2=0.432\), \(3(0.40)^2(0.60)=0.288\), and \((0.40)^3=0.064\). 3. The sampling-distribution mean is \(E(\hat p)=0.40\). 4. Its standard deviation is \(\sqrt{\frac{0.40(0.60)}{3}}=\frac{\sqrt2}{5}\approx0.2828\).

Answer

a) <table> <tr><th>\(\hat p\)</th><th>\(0\)</th><th>\(\frac{1}{3}\)</th><th>\(\frac{2}{3}\)</th><th>\(1\)</th></tr> <tr><th>Probability</th><td>\(0.216\)</td><td>\(0.432\)</td><td>\(0.288\)</td><td>\(0.064\)</td></tr> </table> b) Mean \(0.40\); standard deviation \(\frac{\sqrt2}{5}\approx0.2828\).
Idea of a sampling distribution 26
A finite population consists of \(10,20,30,40\). A simple random sample of size \(2\) is selected without replacement. Let \(M\) be the smaller sampled value. a) Construct the sampling distribution of \(M\). b) Find \(E(M)\).

Hints

- List the possible samples under the stated sampling method. - Compute the requested statistic for each sample. - Group identical statistic values before finding the expected value.

Solution

1. The six equally likely samples are \(\{10,20\},\{10,30\},\{10,40\},\{20,30\},\{20,40\},\{30,40\}\). 2. Their minimum values are \(10,10,10,20,20,30\). 3. Thus, the probabilities of \(10,20,30\) are \(3/6,2/6,1/6\). 4. The expected minimum is \(10(1/2)+20(1/3)+30(1/6)=\frac{50}{3}\approx16.67\).

Answer

a) <table> <tr><th>\(m\)</th><th>\(10\)</th><th>\(20\)</th><th>\(30\)</th></tr> <tr><th>\(P(M=m)\)</th><td>\(\frac{1}{2}\)</td><td>\(\frac{1}{3}\)</td><td>\(\frac{1}{6}\)</td></tr> </table> b) \(E(M)=\frac{50}{3}\approx16.67\).
Idea of a sampling distribution 27
A population consists of \(0,10,20\). A sample of size \(2\) is selected in two different ways: method A samples with replacement, and method B samples without replacement. Let \(\bar X_A\) and \(\bar X_B\) be the sample means. a) Construct both sampling distributions. b) Compare their means and variances.

Hints

- Keep the two sampling mechanisms separate when listing possible samples. - Group repeated sample means and assign probabilities from the appropriate sample space. - Compare both center and spread after constructing the distributions.

Solution

1. With replacement, the nine ordered samples produce means \(0,5,10,15,20\) with probabilities \(1/9,2/9,3/9,2/9,1/9\). 2. Without replacement, the three equally likely samples produce means \(5,10,15\), each with probability \(1/3\). 3. Both sampling distributions have mean \(10\). 4. The with-replacement variance is \(\frac{100}{3}\), while the without-replacement variance is \(\frac{50}{3}\). Sampling without replacement produces less variability here.

Answer

a) Method A: <table> <tr><th>\(\bar x_A\)</th><th>\(0\)</th><th>\(5\)</th><th>\(10\)</th><th>\(15\)</th><th>\(20\)</th></tr> <tr><th>Probability</th><td>\(\frac{1}{9}\)</td><td>\(\frac{2}{9}\)</td><td>\(\frac{1}{3}\)</td><td>\(\frac{2}{9}\)</td><td>\(\frac{1}{9}\)</td></tr> </table> Method B: <table> <tr><th>\(\bar x_B\)</th><th>\(5\)</th><th>\(10\)</th><th>\(15\)</th></tr> <tr><th>Probability</th><td>\(\frac{1}{3}\)</td><td>\(\frac{1}{3}\)</td><td>\(\frac{1}{3}\)</td></tr> </table> b) Both means are \(10\). The variances are \(\frac{100}{3}\) with replacement and \(\frac{50}{3}\) without replacement.
Idea of a sampling distribution 28
Three histograms come from a study of commute times. They show (in some order) all commuters in a population, the individual times in one random sample of \(50\) commuters, and the means of \(5000\) random samples of \(50\) commuters each. a) Which panel shows the sample means? Explain how you know. b) Can you tell for certain which of the other two panels shows the population and which shows the single sample? Explain. c) Describe what one data value represents in each kind of distribution. Why do sample means vary less than individual commute times?
Figure for problem 547550

Hints

- Look for the plot whose values vary over the smallest range. - Could a random sample have a shape close to the population it came from? - Decide whether one data value stands for one person or one entire group of people.

Solution

1. Panel c) has a narrower, roughly symmetric distribution centered near the other panels' centers. It shows the means from repeated samples. 2. Panels a) and b) both show individual commute times with nearly the same shape and spread. The plots alone do not identify which is the population and which is one sample. 3. A value in the population or one-sample histogram is one person's commute time. A value in panel c) is the mean commute time from an entire sample of \(50\) people. 4. Averaging many individuals tends to reduce the effect of high and low individual times, making the distribution of sample means less spread out.

Answer

a) Panel c), because its values are much less spread out and form a roughly symmetric mound. b) No. Panels a) and b) are too similar to distinguish confidently from the histograms alone. c) The first two kinds of plot show individual commutes; panel c) shows one mean per sample. Sample means have less variation because individual highs and lows tend to balance when averaged.
Idea of a sampling distribution 29
A population consists of \(1,2,5,8\), with population mean \(4\). Two values are sampled independently with replacement, and \(\bar X\) is the sample mean. a) Find \(P(\bar X=4)\). b) The mean of the sampling distribution of \(\bar X\) is \(4\). Explain why this does not contradict part a).

Hints

- Translate the required sample mean into a required sum of the two observations. - Check all possible population-value pairs. - Distinguish a distribution's center from the values in its support.

Solution

1. For \(\bar X=4\), the two sampled values would need to sum to \(8\). 2. No ordered pair from \(\{1,2,5,8\}\) has sum \(8\), so \(P(\bar X=4)=0\). 3. The mean of a distribution is a weighted balance point and does not have to be one of the distribution's possible values. 4. Averaging all \(16\) equally likely sample means gives \(4\), even though no individual sample mean equals \(4\).

Answer

a) \(P(\bar X=4)=0\). b) A distribution's mean need not be an attainable value; it is the weighted average of all possible sample-mean values.
Idea of a sampling distribution 30
A population consists of \(1,4,6\). Two values are sampled independently with replacement. Let \(M\) be the larger sampled value. a) Construct the sampling distribution of \(M\). b) Find \(E(M)\).

Hints

- List or classify the equally likely ordered samples. - Count how many samples produce each possible maximum. - Use the resulting distribution to calculate its expected value.

Solution

1. There are nine equally likely ordered samples. Only \((1,1)\) has maximum \(1\), so its probability is \(1/9\). 2. The samples \((1,4),(4,1),(4,4)\) have maximum \(4\), so the probability is \(3/9\). 3. The remaining five samples have maximum \(6\), so the probability is \(5/9\). 4. The expected maximum is \(1(1/9)+4(3/9)+6(5/9)=\frac{43}{9}\approx4.78\).

Answer

a) <table> <tr><th>\(m\)</th><th>\(1\)</th><th>\(4\)</th><th>\(6\)</th></tr> <tr><th>\(P(M=m)\)</th><td>\(\frac{1}{9}\)</td><td>\(\frac{1}{3}\)</td><td>\(\frac{5}{9}\)</td></tr> </table> b) \(E(M)=\frac{43}{9}\approx4.78\).
Idea of a sampling distribution 31
A city wants to estimate average household water use. Method A takes a simple random sample of \(100\) households. Method B randomly selects \(20\) blocks and then randomly selects \(5\) households from each chosen block. Both methods collect \(100\) observations. Households on the same block tend to have similar water use. Which method's sample mean would generally have the more variable sampling distribution? Explain.

Hints

- Compare the dependence among observations under the two designs. - Ask whether every additional household contributes equally new information. - Sampling-distribution spread depends on the sampling method as well as the number of observations.

Solution

1. Method A spreads observations across the city and produces more nearly independent information. 2. In Method B, households within the same block tend to be similar, so several observations may repeat much of the same information. 3. The effective amount of independent information is therefore smaller for Method B. 4. Method B's sample mean would generally have the more variable sampling distribution.

Answer

Method B would generally be more variable because within-block similarity reduces the effective independent information despite the same nominal sample size.
Idea of a sampling distribution 32
A website estimates a population proportion by posting an open poll and using the proportion among volunteers. The site repeats this procedure on many days and calls the resulting distribution a sampling distribution. Explain why the repeated values can form a distribution but why standard simple-random-sample formulas may not describe its center or spread.

Hints

- Separate repeated outcomes from the method that generates them. - Check whether every population member has a controlled chance to enter the sample. - Sampling-distribution formulas depend on the sampling design, not only the sample size.

Solution

1. Repeating the volunteer procedure produces a distribution of the resulting proportions, so repeated-sample variation can be observed. 2. The respondents are self-selected rather than randomly sampled from the population. 3. The respondent composition can systematically differ from the population and can vary according to factors not represented in simple random sampling. 4. Therefore, the usual simple-random-sample center and standard-error formulas are not justified for this procedure.

Answer

The repeated volunteer proportions do have a distribution, but it is generated by a self-selection mechanism. It need not be centered at the population proportion or have the spread predicted for a simple random sample.
Idea of a sampling distribution 33
A company has two equally large departments with very different average commute times. Method A takes a simple random sample of \(100\) employees from the whole company. Method B randomly samples \(50\) employees from each department and combines the results using equal weights. Commute times are relatively similar within each department. Which method should generally produce a less variable sampling distribution for the estimated company mean? Explain.

Hints

- Identify the major source of differences in the population. - Compare whether each design fixes or randomizes the representation of the two groups. - Sampling designs can reduce variation without changing the total sample size.

Solution

1. Method B guarantees balanced representation of both equally large departments. 2. Because commute times are relatively homogeneous within each department but differ between departments, stratifying removes department-composition variation from sample to sample. 3. Method A can randomly overrepresent one department, adding variability. 4. Therefore, Method B should generally produce the less variable sampling distribution.

Answer

Method B should generally be less variable because stratification controls the sample's department composition and uses the within-department similarity.
Idea of a sampling distribution 34
A population has three equally common categories: A, B, and C. Two independent observations are selected with replacement. Let \(D\) be the number of different categories represented in the sample. Construct the sampling distribution of \(D\), and find its mean and standard deviation.

Hints

- List the ordered pairs of category labels. - Separate pairs with matching labels from pairs with different labels. - Summarize the resulting two-value distribution.

Solution

1. There are \(9\) equally likely ordered samples. 2. The \(3\) samples AA, BB, and CC give \(D=1\), so \(P(D=1)=\frac13\). 3. The other \(6\) samples give \(D=2\), so \(P(D=2)=\frac23\). 4. The mean is \(1\cdot\frac13+2\cdot\frac23=\frac53\). 5. The variance is \(\frac29\), so the standard deviation is \(\frac{\sqrt2}{3}\approx0.471\).

Answer

<table> <tr><th>\(D\)</th><th>\(1\)</th><th>\(2\)</th></tr> <tr><th>Probability</th><td>\(\frac13\)</td><td>\(\frac23\)</td></tr> </table> The mean is \(\frac53\), and the standard deviation is \(\frac{\sqrt2}{3}\approx0.471\).
Idea of a sampling distribution 35
The finite population is \(\{2,5,9,12,17\}\). A simple random sample of \(4\) values is selected without replacement, and \(\bar X\) is calculated. Construct the sampling distribution of \(\bar X\), then find its mean and standard deviation.

Hints

- Describe each sample by the value left out. - Use the population total to obtain each sample mean efficiently. - Treat the complete list of simple random samples as equally likely.

Solution

1. Each sample is determined by the one population value omitted, and the population total is \(45\). 2. Omitting \(2,5,9,12,17\) gives sample means \(10.75,10,9,8.25,7\), respectively. 3. The five samples are equally likely, so each mean has probability \(0.20\). 4. The mean of the sampling distribution is \(9\), equal to the population mean. 5. The variance is \(\frac{69}{40}\), so the standard deviation is \(\sqrt{\frac{69}{40}}\approx1.313\).

Answer

<table> <tr><th>\(\bar X\)</th><th>\(7\)</th><th>\(8.25\)</th><th>\(9\)</th><th>\(10\)</th><th>\(10.75\)</th></tr> <tr><th>Probability</th><td>\(0.20\)</td><td>\(0.20\)</td><td>\(0.20\)</td><td>\(0.20\)</td><td>\(0.20\)</td></tr> </table> Mean: \(9\). Standard deviation: \(\sqrt{\frac{69}{40}}\approx1.313\).
Idea of a sampling distribution 36
Two independent observations \(X_1\) and \(X_2\) are selected with replacement from the population \(\{0,1,2\}\), with each value equally likely. Let \(D=X_2-X_1\). Construct the sampling distribution of \(D\), and find its mean and standard deviation.

Hints

- Preserve the order of the two observations when listing samples. - Group pairs that yield the same signed difference. - Look for symmetry before carrying out all summary calculations.

Solution

1. The \(9\) ordered pairs are equally likely. 2. Differences \(-2,-1,0,1,2\) occur in \(1,2,3,2,1\) pairs, respectively. 3. The corresponding probabilities are \(\frac19,\frac29,\frac39,\frac29,\frac19\). 4. Symmetry gives mean \(0\). 5. The variance is \(\frac43\), so the standard deviation is \(\frac{2}{\sqrt3}\approx1.155\).

Answer

<table> <tr><th>\(D\)</th><th>\(-2\)</th><th>\(-1\)</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th></tr> <tr><th>Probability</th><td>\(\frac19\)</td><td>\(\frac29\)</td><td>\(\frac39\)</td><td>\(\frac29\)</td><td>\(\frac19\)</td></tr> </table> The mean is \(0\), and the standard deviation is \(\frac{2}{\sqrt3}\approx1.155\).
Idea of a sampling distribution 37
A population contains equal numbers of the values \(7\) and \(17\). Its mean is \(12\), and its standard deviation is \(5\). A sampling plan chooses one value \(X_1\) at random and pairs it with \(X_2=24-X_1\). Let \(\bar X=\frac{X_1+X_2}{2}\). a) Describe the sampling distribution of \(\bar X\). b) Compare its standard deviation with \(\frac{5}{\sqrt{2}}\), the standard deviation for the mean of two independent observations.

Hints

- List the two possible ordered pairs under the pairing rule. - Compute their means before using any standard-error formula. - Does the independence assumption hold for this sampling plan?

Solution

1. Since the population contains only \(7\) and \(17\), the pairing rule always produces another value in the population. 2. The sample mean is \(\bar X=\frac{X_1+(24-X_1)}{2}=12\) for every pair. 3. Thus \(P(\bar X=12)=1\), and the sampling distribution has standard deviation \(0\). 4. Independent sampling would give \(\frac{5}{\sqrt{2}}\approx3.536\), but that formula does not apply to this dependent pair.

Answer

a) \(\bar X\) is always \(12\), so \(P(\bar X=12)=1\). b) Its standard deviation is \(0\), compared with \(\frac{5}{\sqrt{2}}\approx3.536\) for independent observations.
Idea of a sampling distribution 38
A population value is either \(0\) with probability \(0.40\) or \(1\) with probability \(0.60\). Three independent observations are selected. Let \(M\) be the sample mode. Because the sample size is odd, there is no tie. Construct the sampling distribution of \(M\), and find its mean and standard deviation.

Hints

- Determine how many occurrences are needed for a value to be the mode. - Group samples according to the number of \(1\) values. - Once the two probabilities are known, treat the statistic as a two-value random variable.

Solution

1. The mode is \(1\) when at least two of the three observations equal \(1\). 2. The probability is \(3(0.60)^2(0.40)+(0.60)^3=0.648\). 3. Therefore, \(P(M=0)=1-0.648=0.352\). 4. Since \(M\) takes values \(0\) and \(1\), its mean is \(0.648\). 5. Its standard deviation is \(\sqrt{0.648(0.352)}=\frac{6\sqrt{99}}{125}\approx0.4776\).

Answer

<table> <tr><th>\(M\)</th><th>\(0\)</th><th>\(1\)</th></tr> <tr><th>Probability</th><td>\(0.352\)</td><td>\(0.648\)</td></tr> </table> Mean: \(0.648\). Standard deviation: \(\frac{6\sqrt{99}}{125}\approx0.4776\).
Idea of a sampling distribution 39
A population rating is equally likely to be any integer from \(1\) through \(6\). Two ratings are sampled independently with replacement, and \(\bar X\) is their mean. a) Construct the sampling distribution of \(\bar X\). b) Find its mean and standard deviation. c) Find \(P(\bar X\ge5)\).

Hints

- Organize ordered pairs by their sum. - Convert each possible sum into a sample mean. - Use the frequency pattern to calculate the requested summaries and tail probability.

Solution

1. The \(36\) ordered samples are equally likely. The possible means \(1,1.5,2,2.5,3,3.5,4,4.5,5,5.5,6\) occur with frequencies \(1,2,3,4,5,6,5,4,3,2,1\). 2. The sampling-distribution mean is \(3.5\). 3. The variance is \(\frac{35}{24}\), so the standard deviation is \(\sqrt{\frac{35}{24}}\approx1.2076\). 4. Means at least \(5\) occur in \(3+2+1=6\) of the \(36\) samples, so the probability is \(1/6\).

Answer

a) <table> <tr><th>\(\bar x\)</th><th>\(1\)</th><th>\(1.5\)</th><th>\(2\)</th><th>\(2.5\)</th><th>\(3\)</th><th>\(3.5\)</th><th>\(4\)</th><th>\(4.5\)</th><th>\(5\)</th><th>\(5.5\)</th><th>\(6\)</th></tr> <tr><th>Probability</th><td>\(\frac{1}{36}\)</td><td>\(\frac{2}{36}\)</td><td>\(\frac{3}{36}\)</td><td>\(\frac{4}{36}\)</td><td>\(\frac{5}{36}\)</td><td>\(\frac{6}{36}\)</td><td>\(\frac{5}{36}\)</td><td>\(\frac{4}{36}\)</td><td>\(\frac{3}{36}\)</td><td>\(\frac{2}{36}\)</td><td>\(\frac{1}{36}\)</td></tr> </table> b) Mean \(3.5\); standard deviation \(\sqrt{35/24}\approx1.2076\). c) \(P(\bar X\ge5)=\frac{1}{6}\).
Idea of a sampling distribution 40
A population consists of \(0,1,2\), each equally likely. Three observations are sampled independently with replacement. Let \(M\) be the sample median. a) Construct the sampling distribution of \(M\). b) Find \(E(M)\).

Hints

- Classify samples by how many observations fall at each extreme. - Use symmetry to avoid repeating equivalent counting. - Assign the remaining samples to the middle median value.

Solution

1. There are \(27\) equally likely ordered samples. The median is \(0\) when at least two observations are \(0\), which occurs in \(7\) samples. 2. By symmetry, the median is \(2\) in \(7\) samples. 3. The remaining \(13\) samples have median \(1\). 4. Thus, \(E(M)=0(7/27)+1(13/27)+2(7/27)=1\).

Answer

a) <table> <tr><th>\(m\)</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th></tr> <tr><th>\(P(M=m)\)</th><td>\(\frac{7}{27}\)</td><td>\(\frac{13}{27}\)</td><td>\(\frac{7}{27}\)</td></tr> </table> b) \(E(M)=1\).
Idea of a sampling distribution 41
A finite population consists of \(0,2,4,10\). A simple random sample of size \(3\) is selected without replacement. Let \(D\) be the sample median minus the sample mean. Construct the sampling distribution of \(D\), and find its mean and standard deviation.

Hints

- List samples by the one population value omitted. - For each sample, calculate both center measures before taking their difference. - Treat the resulting statistic values as an equally likely distribution.

Solution

1. The four equally likely samples are obtained by omitting one population value. 2. Omitting \(0,2,4,10\) gives \(D=-\frac43,-\frac23,-2,0\), respectively. 3. Each value has probability \(\frac14\). 4. The mean is \(-1\). 5. The variance is \(\frac59\), so the standard deviation is \(\frac{\sqrt5}{3}\approx0.745\).

Answer

<table> <tr><th>\(D\)</th><th>\(-2\)</th><th>\(-\frac43\)</th><th>\(-\frac23\)</th><th>\(0\)</th></tr> <tr><th>Probability</th><td>\(\frac14\)</td><td>\(\frac14\)</td><td>\(\frac14\)</td><td>\(\frac14\)</td></tr> </table> The mean is \(-1\), and the standard deviation is \(\frac{\sqrt5}{3}\approx0.745\).
Idea of a sampling distribution 42
A population has mean \(\mu\) and standard deviation \(\sigma\). Before taking a sample, a fair coin is flipped. If it lands heads, one observation is selected; if it lands tails, four independent observations are selected. Let \(M\) be the mean of the observations selected. Find the mean and standard deviation of the sampling distribution of \(M\).

Hints

- Consider the two possible sample sizes separately. - Compare the centers of the two conditional sampling distributions. - Combine variances only after accounting for how often each sampling plan is used.

Solution

1. Conditional on a sample size of \(1\), the sample mean has mean \(\mu\) and variance \(\sigma^2\). 2. Conditional on a sample size of \(4\), the sample mean has mean \(\mu\) and variance \(\frac{\sigma^2}{4}\). 3. Both conditional means equal \(\mu\), so the unconditional mean is \(\mu\). 4. The unconditional variance is \(\frac12\sigma^2+\frac12\cdot\frac{\sigma^2}{4}=\frac{5\sigma^2}{8}\). 5. The standard deviation is \(\sigma\sqrt{\frac58}\).

Answer

The sampling distribution has mean \(\mu\) and standard deviation \(\sigma\sqrt{\frac58}\).
Idea of a sampling distribution 43
A systematic sample selects every second item from the production sequence shown. The starting position is chosen uniformly at random from the first two positions. Let \(\bar X\) be the sample mean. a) Construct the sampling distribution of \(\bar X\). b) Compare its center and shape with what would be expected from simple random samples of \(5\) items.
Figure for problem 547583

Hints

- Follow each possible random starting position through the positions shown. - Do not assume that a fairly large sample fraction guarantees a small spread when the sampling interval matches a population pattern. - Compare whether each method mixes the two item values within a sample.

Solution

1. Starting at position \(1\) selects positions \(1,3,5,7,9\), whose values are all \(0\), so \(\bar X=0\). 2. Starting at position \(2\) selects positions \(2,4,6,8,10\), whose values are all \(10\), so \(\bar X=10\). 3. Each start has probability \(0.5\), so the sampling distribution places probability \(0.5\) at each of \(0\) and \(10\). 4. Its mean is \(5\), equal to the population mean, but it is extremely spread out and two-point rather than concentrated near \(5\). 5. Simple random samples of \(5\) items would usually contain a mix of both values and have means closer to \(5\).

Answer

a) \(P(\bar X=0)=0.5\) and \(P(\bar X=10)=0.5\). b) The center is \(5\), but this systematic design produces a two-point, highly variable distribution. Simple random sampling would produce means concentrated more closely around \(5\).
Idea of a sampling distribution 44
A population consists of \(0,2,\) and \(6\), each equally likely. Two observations are selected independently with replacement. Define the sample midrange \(R\) as the average of the smaller and larger observations. Construct the sampling distribution of \(R\), and find its mean and standard deviation.

Hints

- List ordered samples because the two selections are independent. - Several different samples can produce the same midpoint of the extremes. - Combine equal statistic values before finding the numerical summaries.

Solution

1. The \(9\) ordered samples are equally likely. 2. The midrange values \(0,1,2,3,4,6\) occur in \(1,2,1,2,2,1\) ordered samples, respectively. 3. Their probabilities are therefore \(\frac19,\frac29,\frac19,\frac29,\frac29,\frac19\). 4. The mean is \(\frac83\). 5. The variance is \(\frac{28}{9}\), so the standard deviation is \(\frac{2\sqrt7}{3}\approx1.764\).

Answer

<table> <tr><th>\(R\)</th><th>\(0\)</th><th>\(1\)</th><th>\(2\)</th><th>\(3\)</th><th>\(4\)</th><th>\(6\)</th></tr> <tr><th>Probability</th><td>\(\frac19\)</td><td>\(\frac29\)</td><td>\(\frac19\)</td><td>\(\frac29\)</td><td>\(\frac29\)</td><td>\(\frac19\)</td></tr> </table> The mean is \(\frac83\), and the standard deviation is \(\frac{2\sqrt7}{3}\approx1.764\).

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