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Sampling distribution of a difference in means

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Difficulty: 1 – 5

54769412

A researcher will take independent simple random samples without replacement from two populations. Population 2 contains \(620\) individuals. What is the largest sample size \(n_2\) that can be taken from Population 2 while satisfying the \(10\%\) condition for the sampling distribution of \(\bar{x}_1-\bar{x}_2\)? Explain why this condition is checked separately for each population.

Hints

- Apply the finite-population percentage limit to the population named in the question. - The two samples do not share a common population size. - Think about which sampling fraction controls dependence within each sample.

Solution

1. For Population 2, the \(10\%\) condition requires \(n_2\le0.10(620)\). 2. Thus, \(n_2\le62\), so the largest allowable sample size is \(62\). 3. The condition is checked separately because each sample is drawn from its own finite population, and each population's sampling fraction determines whether that sample can be treated as approximately independent.

Answer

The largest sample size is \(n_2=62\). Each population has its own sampling fraction, so the \(10\%\) condition must be verified separately for the two samples.

54770012

Repeated independent random sampling from two populations produces a sampling distribution of \(\bar{x}_1-\bar{x}_2\) centered at \(-4\). Population 1 has mean \(18\). What is the mean of Population 2? Explain how the center of the sampling distribution determines your answer.

Hints

- The sampling distribution of a difference is centered at the corresponding difference in population means. - Preserve the order “Population 1 minus Population 2.” - Solve the resulting one-step equation for the unknown population mean.

Solution

1. The mean of the sampling distribution of \(\bar{x}_1-\bar{x}_2\) is \(\mu_1-\mu_2\). 2. Therefore, \(18-\mu_2=-4\). 3. Solving gives \(\mu_2=22\).

Answer

Population 2 has mean \(22\).

54776612

For two independent random samples, suppose the sampling distribution of \(\bar{x}_1-\bar{x}_2\) has mean \(\delta\) and standard deviation \(\tau\). Describe the mean and standard deviation of the sampling distribution of \(\bar{x}_2-\bar{x}_1\). Explain what changes and what stays the same when the subtraction order is reversed.

Hints

- Write the reversed difference as a simple transformation of the original difference. - Think separately about how multiplying by \(-1\) affects a mean and a standard deviation. - The amount of sampling variability does not depend on which group is written first.

Solution

1. Reversing the subtraction gives \(\bar{x}_2-\bar{x}_1=-(\bar{x}_1-\bar{x}_2)\). 2. Multiplying a random variable by \(-1\) changes the sign of its mean, so the new sampling-distribution mean is \(-\delta\). 3. Multiplying by \(-1\) does not change standard deviation, so the new sampling-distribution standard deviation remains \(\tau\).

Answer

The sampling distribution of \(\bar{x}_2-\bar{x}_1\) has mean \(-\delta\) and standard deviation \(\tau\). Reversing the subtraction changes the sign of the center but not the spread.

54764012

Two independent populations of sensor response times are approximately normal. Population A has mean \(82\,\text{ms}\) and standard deviation \(12\,\text{ms}\); population B has mean \(78\,\text{ms}\) and standard deviation \(14\,\text{ms}\). Independent random samples of sizes \(36\) and \(49\), respectively, are taken. For \(\bar{x}_A-\bar{x}_B\), find the mean and standard deviation of the sampling distribution and find \(P(\bar{x}_A-\bar{x}_B>7\,\text{ms})\).

Hints

- Treat the two sample means as independent random quantities before considering their difference. - Think separately about how centers combine and how independent variability combines. - Once the distribution of the difference is described, compare \(7\,\text{ms}\) with that distribution.

Solution

1. Because the populations are approximately normal and the samples are independent, the sampling distribution of \(\bar{x}_A-\bar{x}_B\) is normal. 2. Its mean is \(82-78=4\,\text{ms}\). 3. Its standard deviation is \(\sqrt{\frac{12^2}{36}+\frac{14^2}{49}}=\sqrt{8}\approx2.83\,\text{ms}\). 4. For a difference of \(7\,\text{ms}\), \(z=\frac{7-4}{\sqrt{8}}\approx1.06\). 5. Thus, \(P(\bar{x}_A-\bar{x}_B>7)\approx0.1444\).

Answer

The sampling distribution has mean \(4\,\text{ms}\) and standard deviation \(\sqrt{8}\,\text{ms}\approx2.83\,\text{ms}\). The probability is \(P(\bar{x}_A-\bar{x}_B>7)\approx0.1444\).

54764612

Two independent populations have standard deviations \(20\) and \(15\). A study originally plans independent random samples with \(n_1=25\) and \(n_2=36\). A revised plan quadruples both sample sizes to \(n_1=100\) and \(n_2=144\), while the populations stay the same. Compare the mean and standard deviation of the sampling distribution of \(\bar{x}_1-\bar{x}_2\) under the two plans. Explain the effect of quadrupling both sample sizes.

Hints

- Changing sample sizes affects sampling variability but not the population means being compared. - Examine how each sample size enters the contribution from its population's variability. - Compare the two spreads multiplicatively rather than only by subtraction.

Solution

1. The mean of \(\bar{x}_1-\bar{x}_2\) is \(\mu_1-\mu_2\) under either sampling plan, so the center is unchanged. 2. Under the original plan, the standard deviation is \(\sqrt{\frac{20^2}{25}+\frac{15^2}{36}}=\sqrt{22.25}\approx4.72\). 3. Under the revised plan, the standard deviation is \(\sqrt{\frac{20^2}{100}+\frac{15^2}{144}}=\sqrt{5.5625}\approx2.36\). 4. Quadrupling both sample sizes cuts the standard deviation of the difference in sample means in half.

Answer

The sampling-distribution mean remains \(\mu_1-\mu_2\). The standard deviation decreases from approximately \(4.72\) to \(2.36\), so quadrupling both sample sizes halves the spread.

54766412

Population A is normally distributed with mean \(30\) and standard deviation \(5\). Population B is normally distributed with mean \(27\) and standard deviation \(4\). Independent random samples of sizes \(9\) and \(16\), respectively, are selected. Find \(P(\bar{x}_A-\bar{x}_B<1)\).

Hints

- Normal population distributions allow exact normal modeling of the sample means even with small samples. - Combine the two population centers according to the subtraction order. - Independent variability from the two sample means contributes to the spread of their difference.

Solution

1. Because both populations are normal and the samples are independent, \(\bar{x}_A-\bar{x}_B\) has a normal sampling distribution. 2. The mean is \(30-27=3\). 3. The standard deviation is \(\sqrt{\frac{5^2}{9}+\frac{4^2}{16}}\approx1.944\). 4. For a difference of \(1\), \(z=\frac{1-3}{1.944}\approx-1.03\). 5. Therefore, \(P(\bar{x}_A-\bar{x}_B<1)\approx0.1517\).

Answer

\(P(\bar{x}_A-\bar{x}_B<1)\approx0.1517\).

54770612

The sampling distribution of \(\bar{x}_1-\bar{x}_2\) for two independent random samples is approximately normal with mean \(5\) and standard deviation \(2\). Find the \(90\)th percentile of this sampling distribution and interpret it in repeated-sampling terms.

Hints

- A percentile is a cutoff with a specified proportion of the distribution below it. - Start on the standardized normal scale, then return to the scale of mean differences. - The interpretation should refer to repeated pairs of samples, not individual observations.

Solution

1. The \(90\)th percentile of the standard normal distribution is approximately \(z=1.282\). 2. Converting back gives \(5+1.282(2)\approx7.56\). 3. About \(90\%\) of repeated independent sample pairs would produce a value of \(\bar{x}_1-\bar{x}_2\) below approximately \(7.56\).

Answer

The \(90\)th percentile is approximately \(7.56\). About \(90\%\) of repeated independent sample pairs would have \(\bar{x}_1-\bar{x}_2<7.56\).

54771812

Population A has mean \(28\) minutes and standard deviation \(10\) minutes. Population B has mean \(24\) minutes and standard deviation \(8\) minutes. Independent random samples of sizes \(25\) and \(16\), respectively, are taken from large populations. Find the mean and standard deviation of the sampling distribution of \(\bar{x}_A-\bar{x}_B\). Interpret the standard deviation in repeated-sampling terms.

Hints

- The center of the difference follows the same subtraction order as the two population means. - Independent sample means contribute separate sources of variability. - Interpret the final spread as variation from sample pair to sample pair, not variation among individual people.

Solution

1. The sampling-distribution mean is \(28-24=4\) minutes. 2. The standard deviation is \(\sqrt{\frac{10^2}{25}+\frac{8^2}{16}}=\sqrt{8}\approx2.83\) minutes. 3. In repeated independent sampling with these sample sizes, values of \(\bar{x}_A-\bar{x}_B\) typically vary by about \(2.83\) minutes from the population mean difference of \(4\) minutes.

Answer

Mean: \(4\) minutes. Standard deviation: \(\sqrt{8}\approx2.83\) minutes. Across repeated pairs of samples, the sample-mean difference typically varies by about \(2.83\) minutes around the population mean difference.

54773612

Two independent populations are approximately normal. Population A has mean \(64\) and standard deviation \(8\), and Population B has mean \(59\) and standard deviation \(5\). Independent random samples of sizes \(25\) and \(100\), respectively, are selected from large populations. A student claims that the sampling distribution of \(\bar{x}_A-\bar{x}_B\) has mean \(64+59=123\) and standard deviation \(\frac{8}{\sqrt{25}}-\frac{5}{\sqrt{100}}=1.1\). Correct both quantities and explain why the student's operations are wrong.

Hints

- Treat the center and the variability as two separate questions. - Keep the subtraction order when combining the population means. - For independent random quantities, think about how their variances combine rather than subtracting their standard deviations.

Solution

1. The mean of a difference equals the difference of the means, so \(\mu_{\bar{x}_A-\bar{x}_B}=64-59=5\). 2. For independent sample means, variances add even when the random variables are subtracted. 3. The standard deviation is \(\sqrt{\frac{8^2}{25}+\frac{5^2}{100}}=\sqrt{2.81}\approx1.676\). 4. Therefore, the sampling distribution is normal with mean \(5\) and standard deviation approximately \(1.676\).

Answer

The correct mean is \(5\), and the correct standard deviation is approximately \(1.676\). Centers subtract, but independent variances add before taking the square root.

54774812

A researcher plans independent simple random samples without replacement from two finite populations. Population 1 contains \(600\) individuals and the sample size is \(50\). Population 2 contains \(700\) individuals and the sample size is \(80\). Can the usual standard-deviation formula for the sampling distribution of \(\bar{x}_1-\bar{x}_2\) be used without a finite-population correction? Explain which population, if any, causes a problem.

Hints

- Check the sampling fraction separately for each population. - The usual spread formula assumes observations within each sample are approximately independent. - One failed population condition is enough to prevent using the formula as stated.

Solution

1. For sampling without replacement, each sample should be no more than \(10\%\) of its population when using the usual independence-based standard-deviation formula. 2. For Population 1, \(50/600\approx0.083\), so the \(10\%\) condition is satisfied. 3. For Population 2, \(80/700\approx0.114\), so the \(10\%\) condition is not satisfied. 4. Therefore, the usual formula \(\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}\) is not justified without accounting for the finite-population effect in Population 2.

Answer

No. Population 1 satisfies the \(10\%\) condition, but Population 2 does not because \(80>0.10(700)=70\).

54776012

Two independent random samples are taken from large populations. Sample 1 has size \(15\) and is strongly right-skewed with a clear outlier. Sample 2 has size \(50\) and shows no unusual features. Neither population is known to be normal. Is it reasonable to model the sampling distribution of \(\bar{x}_1-\bar{x}_2\) as approximately normal based only on this information? Explain.

Hints

- Check the sampling behavior of each sample mean separately before combining them. - One sample is large, but the other must still have enough shape or sample-size support. - Ask whether both components of the difference have sampling distributions that can be treated as approximately normal.

Solution

1. The sample mean from Sample 2 is supported by a large sample size of \(50\), so its sampling distribution can be modeled as approximately normal. 2. Sample 1 has only \(15\) observations and shows strong skewness and an outlier, with no population-normality information. 3. Therefore, the sampling distribution of \(\bar{x}_1\) is not justified as approximately normal. 4. Since the difference includes this poorly behaved sample mean, the information given does not justify modeling \(\bar{x}_1-\bar{x}_2\) as approximately normal.

Answer

No. The large second sample is not enough to compensate for the small, strongly skewed first sample with an outlier. Approximate normality of \(\bar{x}_1-\bar{x}_2\) is not justified from the information given.

54777812

Independent random samples are taken from two populations with means \(\mu_1\) and \(\mu_2\). The first sample has size \(1000\), and the second has size \(10\). A student claims that the sampling distribution of \(\bar{x}_1-\bar{x}_2\) should be centered closer to \(\mu_1\) because the first sample is much larger. Evaluate the claim.

Hints

- Recall the expected value of each sample mean separately. - Subtract those expected values in the same order as the statistic. - Sample size changes precision, but ask whether it changes unbiasedness.

Solution

1. The expected value of \(\bar{x}_1\) is \(\mu_1\), regardless of the first sample size. 2. The expected value of \(\bar{x}_2\) is \(\mu_2\), regardless of the second sample size. 3. Therefore, the sampling distribution of \(\bar{x}_1-\bar{x}_2\) is centered at \(\mu_1-\mu_2\). 4. Unequal sample sizes affect the spread of the sampling distribution, not the weighting of its center.

Answer

The claim is incorrect. The sampling distribution is centered at \(\mu_1-\mu_2\). The very different sample sizes affect sampling variability, not the expected difference.

54778412

Every value in Population A is exactly \(10\), so \(\mu_A=10\) and \(\sigma_A=0\). Population B has mean \(\mu_B\) and standard deviation \(6\). Independent random samples are taken, with any positive sample size from Population A and a sample size of \(36\) from Population B. Describe the mean and standard deviation of the sampling distribution of \(\bar{x}_A-\bar{x}_B\).

Hints

- Consider what the sample mean from a population with no variability can be. - Find the center using the population means in the stated subtraction order. - In the spread formula, one population's variance contribution is zero.

Solution

1. Every possible sample mean from Population A equals \(10\), so that component contributes no sampling variability. 2. The sampling-distribution mean is \(10-\mu_B\). 3. The standard deviation is \(\sqrt{\frac{0^2}{n_A}+\frac{6^2}{36}}=1\). 4. Thus, all sampling variability in the difference comes from Population B.

Answer

The sampling distribution has mean \(10-\mu_B\) and standard deviation \(1\).

54780212

Let \(\bar{x}_1\) and \(\bar{x}_2\) be independent sample means with sampling-distribution variances \(V_1\) and \(V_2\). Compare the sampling-distribution variances of \(\bar{x}_1-\bar{x}_2\) and \(\bar{x}_1+\bar{x}_2\). Explain why subtraction does not make the variance contributions subtract.

Hints

- Rewrite a difference as addition of a negative random variable. - Think about what multiplying a random variable by \(-1\) does to its variance. - Independence determines how the two variance contributions combine.

Solution

1. For independent random variables, the variance of a sum is the sum of the variances: \(\operatorname{Var}(\bar{x}_1+\bar{x}_2)=V_1+V_2\). 2. For the difference, \(\operatorname{Var}(\bar{x}_1-\bar{x}_2)=\operatorname{Var}(\bar{x}_1)+\operatorname{Var}(-\bar{x}_2)\). 3. Multiplying by \(-1\) does not change variance, so \(\operatorname{Var}(-\bar{x}_2)=V_2\). 4. Therefore, \(\operatorname{Var}(\bar{x}_1-\bar{x}_2)=V_1+V_2\), the same variance as the sum.

Answer

Both sampling distributions have variance \(V_1+V_2\). Subtraction changes the center but not the fact that independent variance contributions add.

54780812

Two independent populations both have population standard deviation \(10\). Independent random samples have sizes \(n_1=25\) and \(n_2=100\). A student calculates the standard deviation of \(\bar{x}_1-\bar{x}_2\) as \(10/\sqrt{25+100}\). Explain the error and find the correct standard deviation.

Hints

- Do not combine the sample sizes before accounting for each sample mean's variability. - Write one variance term for each independent sample mean. - Add the variance terms, then take the square root.

Solution

1. Each sample mean has its own variance contribution based on its own sample size. 2. The correct variance is \(\frac{10^2}{25}+\frac{10^2}{100}=4+1=5\). 3. Therefore, the standard deviation is \(\sqrt{5}\approx2.236\). 4. Combining the sample sizes before dividing incorrectly treats the two independent sample means as though they came from one pooled sample.

Answer

The correct standard deviation is \(\sqrt{5}\approx2.236\). The two variance contributions must be calculated separately and then added.

54782012

Let \(D=\bar{x}_1-\bar{x}_2\) have an approximately normal sampling distribution with mean \(\delta\) and standard deviation \(\tau\). An analyst reports the transformed statistic \(S=5+2D\). Describe the sampling distribution of \(S\).

Hints

- Treat the reported statistic as a linear transformation of the original mean difference. - Constants added to a random variable affect its center but not its spread. - Multiplying a random variable scales both its center contribution and its standard deviation.

Solution

1. Adding \(5\) and multiplying by \(2\) form a linear transformation of the approximately normal statistic \(D\), so \(S\) is also approximately normal. 2. Its mean is \(5+2\delta\). 3. Adding \(5\) does not change spread, while multiplying by \(2\) doubles standard deviation. 4. Therefore, the standard deviation of \(S\) is \(2\tau\).

Answer

\(S\) is approximately normal with mean \(5+2\delta\) and standard deviation \(2\tau\).

54783812

Population A has mean \(30\) and standard deviation \(5\). Population B has mean \(27\) and standard deviation \(8\). Independent random samples of sizes \(25\) and \(64\), respectively, are repeatedly taken. A simulation of \(5000\) values of \(\bar{x}_A-\bar{x}_B\) has mean \(2.99\) and standard deviation \(1.42\). Compare the simulation results with the theoretical center and spread of the sampling distribution. Are the simulation results plausible?

Hints

- Find what the difference of sample means should be centered around. - Account for the sampling variability contributed by each independent sample. - A simulation with finitely many repetitions should be close to, but not exactly equal to, theoretical values.

Solution

1. The theoretical mean is \(\mu_A-\mu_B=30-27=3\). 2. The theoretical standard deviation is \(\sqrt{\frac{5^2}{25}+\frac{8^2}{64}}=\sqrt{2}\approx1.414\). 3. The simulation mean \(2.99\) is very close to \(3\), and the simulation standard deviation \(1.42\) is very close to \(1.414\). 4. Small discrepancies are expected from a finite simulation, so the reported simulation is plausible.

Answer

The theoretical center is \(3\) and the theoretical standard deviation is approximately \(1.414\). The simulated values \(2.99\) and \(1.42\) are close, so the simulation results are plausible.

54784412

Independent random samples have sizes \(40\) and \(50\). The sample standard deviations are \(7\) and \(9\), respectively. A student says the sampling distribution of \(\bar{x}_1-\bar{x}_2\) therefore has standard deviation \(\sqrt{7^2/40+9^2/50}\). Explain what is wrong with that statement and identify the quantity the student actually calculated.

Hints

- Distinguish population parameters from statistics computed from the samples. - Ask which spread values appear in the theoretical sampling-distribution formula. - Sample standard deviations can estimate an unknown sampling spread.

Solution

1. The theoretical standard deviation of \(\bar{x}_1-\bar{x}_2\) depends on the population standard deviations \(\sigma_1\) and \(\sigma_2\). 2. Because those population standard deviations are unknown, the exact theoretical sampling-distribution standard deviation is also unknown. 3. Replacing \(\sigma_1\) and \(\sigma_2\) with the sample standard deviations gives the estimated standard error \(\sqrt{\frac{7^2}{40}+\frac{9^2}{50}}\approx1.687\).

Answer

The student calculated an estimated standard error, approximately \(1.687\), not the exact theoretical standard deviation of the sampling distribution. The latter uses the unknown population standard deviations.

54785612

Two populations have the same mean and the same standard deviation \(12\). Independent random samples of size \(36\) are taken from each population. A student says that because the population means are equal, \(\bar{x}_1-\bar{x}_2\) should always equal \(0\). Correct the student’s reasoning by giving the center and standard deviation of the sampling distribution of \(\bar{x}_1-\bar{x}_2\).

Hints

- Distinguish the expected difference from the difference observed in a particular pair of samples. - Each sample mean has sampling variability even when the populations have the same center. - Independent variability from the two samples does not cancel automatically.

Solution

1. Equal population means imply that the sampling distribution is centered at \(\mu_1-\mu_2=0\). 2. The two sample means still vary independently from sample to sample. 3. The standard deviation of their difference is \(\sqrt{\frac{12^2}{36}+\frac{12^2}{36}}=\sqrt{8}\approx2.828\). 4. Thus, a difference of \(0\) is the center of the sampling distribution, not the only possible sample difference.

Answer

The sampling distribution is centered at \(0\) and has standard deviation \(\sqrt{8}\approx2.828\). Equal population means do not make two independent sample means identical in every sample.

54787412

The sampling distribution of \(\bar{x}_A-\bar{x}_B\) is approximately normal with mean \(-4\) and standard deviation \(1.5\). One pair of independent samples produces \(\bar{x}_A-\bar{x}_B=-1\). Standardize the observed difference relative to its sampling distribution and interpret the resulting z-score.

Hints

- A standardized value measures distance from the center of the relevant sampling distribution. - Do not replace the stated center with zero just because the statistic is a difference. - Express the observed deviation in units of the sampling-distribution spread.

Solution

1. Standardize relative to the sampling-distribution center \(-4\), not relative to \(0\). 2. The z-score is \(z=\frac{-1-(-4)}{1.5}=2\). 3. The observed sample-mean difference is therefore \(2\) sampling-distribution standard deviations above the true population mean difference.

Answer

\(z=2\). The observed difference \(-1\) is \(2\) standard deviations above the sampling-distribution mean of \(-4\).

54788012

The sampling distribution of \(D=\bar{x}_A-\bar{x}_B\) is approximately normal with mean \(5\) and standard deviation \(2\). Find the probability that an observed sample-mean difference is positive but smaller than the true population mean difference; that is, find \(P(0<D<5)\).

Hints

- Center the standardization at the true population mean difference. - One endpoint is exactly the center of the sampling distribution. - Convert both endpoints to standardized values before finding the area between them.

Solution

1. Standardizing the lower endpoint gives \(z=(0-5)/2=-2.5\). 2. The upper endpoint \(D=5\) is the center of the sampling distribution, so its z-score is \(0\). 3. Therefore, \(P(0<D<5)=P(-2.5<Z<0)\approx0.4938\).

Answer

\(P(0<D<5)\approx0.4938\).

54789212

The sampling distribution of \(D=\bar{x}_A-\bar{x}_B\) is approximately normal with mean \(3\) and standard deviation \(2\). What is the probability that the observed sample-mean difference has the same positive sign as the true population mean difference?

Hints

- Translate “same positive sign” into an inequality for the sample-mean difference. - Standardize the zero boundary relative to the true center of the sampling distribution. - The desired event is the area to the right of that boundary.

Solution

1. Having the same positive sign means \(D>0\). 2. Standardizing the boundary gives \(z=(0-3)/2=-1.5\). 3. Therefore, \(P(D>0)=P(Z>-1.5)\approx0.9332\).

Answer

The probability is approximately \(0.9332\).

54790412

Two independent samples are taken from populations A and B, but the same miscalibrated instrument adds \(2\) units to every measurement in both samples. Let the true population means be \(\mu_A\) and \(\mu_B\). How does this common additive bias affect the center of the sampling distribution of the observed difference \(\bar{x}_A-\bar{x}_B\)?

Hints

- Write the expected observed mean for each group after the common offset is added. - Keep the subtraction order visible. - A shift shared equally by both groups may disappear when one mean is subtracted from the other.

Solution

1. The observed sample mean for A is centered at \(\mu_A+2\), and the observed sample mean for B is centered at \(\mu_B+2\). 2. The observed mean difference is therefore centered at \((\mu_A+2)-(\mu_B+2)\). 3. The common additive bias cancels, leaving center \(\mu_A-\mu_B\). 4. Thus, this particular shared calibration offset biases each group mean but not their difference.

Answer

The center remains \(\mu_A-\mu_B\). Adding the same fixed \(2\)-unit bias to both groups cancels in the difference of sample means.

54791012

Two independent sample means have sampling-distribution standard deviations \(3\) and \(4\). A student says the standard deviation of their difference should be \(1\) because \(4-3=1\). Find the correct standard deviation of the difference and explain why the spreads do not subtract.

Hints

- Work with variances when combining independent sources of random variability. - The minus sign in a difference affects the center, but squaring removes that sign in the variance calculation. - Convert back to a standard deviation only after combining the variances.

Solution

1. For independent sample means, sampling variances add when the means are added or subtracted. 2. The variance of the difference is \(3^2+4^2=9+16=25\). 3. Therefore, the standard deviation of the difference is \(\sqrt{25}=5\). 4. Subtracting the point estimates does not imply subtracting their independent sampling variability.

Answer

The standard deviation is \(5\), not \(1\).

54791612

Two independent sample means come from continuous populations, and the sampling distribution of \(D=\bar{x}_A-\bar{x}_B\) is continuous and centered at \(0\). What is \(P(D=0)\)? Explain why a distribution can be centered at \(0\) even though the probability of observing exactly \(0\) is \(0\).

Hints

- Recall how probabilities work for a single point in a continuous distribution. - Distinguish the center of a distribution from probability concentrated at one exact value. - Think in terms of intervals around \(0\), not just the point itself.

Solution

1. A continuous random variable assigns probability \(0\) to any single exact value. 2. Therefore, \(P(D=0)=0\). 3. Being centered at \(0\) describes the location or expected value of the entire sampling distribution, not the probability mass at the single point \(0\). 4. Values arbitrarily close to \(0\) can have substantial total probability even though the exact point has probability \(0\).

Answer

\(P(D=0)=0\). The sampling distribution can still be centered at \(0\) because a continuous distribution’s center is not required to have positive point probability.

54792212

For independent samples, the variance of \(D=\bar{x}_A-\bar{x}_B\) is currently \(18\). Both sample sizes are then doubled while the population standard deviations stay the same. What happens to the variance and standard deviation of \(D\)?

Hints

- Sample size appears in the denominator of each variance contribution. - Combine the effect on variances before converting to standard deviation. - Halving a variance does not halve its standard deviation.

Solution

1. Each component variance has the form \(\sigma^2/n\). 2. Doubling both sample sizes halves each component variance, so the total variance also halves from \(18\) to \(9\). 3. The standard deviation changes from \(\sqrt{18}\) to \(\sqrt{9}=3\). 4. Thus, doubling both sample sizes divides the standard deviation by \(\sqrt{2}\), not by \(2\).

Answer

The variance decreases from \(18\) to \(9\). The standard deviation decreases from \(\sqrt{18}\) to \(3\), a factor of \(1/\sqrt{2}\).

54792812

Two independent sample means come from normal populations with the same population mean but possibly different standard deviations and different sample sizes. What is \(P(\bar{x}_A-\bar{x}_B>0)\)? Explain why unequal sample sizes and unequal spreads do not change this probability.

Hints

- Determine the center of the mean-difference sampling distribution first. - Use the fact that both source populations are normal. - Spread affects how far values vary from the center, but symmetry determines the probability of being on either side of the center.

Solution

1. Because the population means are equal, the sampling distribution of \(\bar{x}_A-\bar{x}_B\) is centered at \(0\). 2. Each sample mean is normally distributed, so their independent difference is also normal. 3. A normal distribution centered at \(0\) is symmetric, so half of its probability lies above \(0\). 4. Unequal sample sizes or population standard deviations change the spread of the difference but not its center or symmetry.

Answer

\(P(\bar{x}_A-\bar{x}_B>0)=0.5\).

54793412

Independent random samples are repeatedly taken from populations with means \(\mu_A\) and \(\mu_B\). Both sample sizes increase without bound while the population standard deviations remain fixed. Describe what happens to the sampling distribution of \(\bar{x}_A-\bar{x}_B\). A student says it must collapse toward \(0\); explain the error.

Hints

- Track the center and spread separately as the sample sizes grow. - Increasing sample size changes the variance terms but not the population means. - A distribution can become very narrow around a nonzero center.

Solution

1. The sampling distribution is centered at \(\mu_A-\mu_B\) for every pair of sample sizes. 2. Its variance is \(\frac{\sigma_A^2}{n_A}+\frac{\sigma_B^2}{n_B}\). 3. As both sample sizes increase without bound, both variance terms approach \(0\), so the sampling distribution becomes increasingly concentrated. 4. It collapses around the true population mean difference \(\mu_A-\mu_B\), not around \(0\) unless the two population means are actually equal.

Answer

The sampling distribution becomes concentrated at \(\mu_A-\mu_B\) as both sample sizes grow. Its spread approaches \(0\), but its center does not move to \(0\) unless \(\mu_A=\mu_B\).

54765212

Two independent populations have standard deviations \(8\) and \(12\). A random sample of size \(16\) will be taken from the first population. The sample size from the second population has not yet been chosen. What is the smallest integer value of \(n_2\) that makes the standard deviation of the sampling distribution of \(\bar{x}_1-\bar{x}_2\) no more than \(3\)?

Hints

- Write the sampling variability of the difference using the contributions from both independent samples. - Only one sample size is unknown, so isolate the term that contains it. - Because sample size must be a whole number, check which side of the cutoff the neighboring integers fall on.

Solution

1. The sampling-distribution standard deviation is \(\sqrt{\frac{8^2}{16}+\frac{12^2}{n_2}}\). 2. Require \(\sqrt{4+\frac{144}{n_2}}\le3\). 3. Squaring gives \(4+\frac{144}{n_2}\le9\), so \(\frac{144}{n_2}\le5\) and \(n_2\ge28.8\). 4. The smallest integer is \(n_2=29\). With \(n_2=29\), the standard deviation is approximately \(2.994\), while \(n_2=28\) gives approximately \(3.024\).

Answer

The smallest sample size is \(n_2=29\).

54765812

Forty volunteers participate in an experiment comparing two study environments. The researchers randomly assign \(20\) volunteers to a quiet room and \(20\) to a room with background conversation, then record each person's score on the same timed task. The volunteers were not randomly sampled from a larger population. For the sampling distribution of the difference in treatment-group sample means, explain which randomization condition is satisfied, whether a \(10\%\) population condition is needed for this randomized experiment, and what limitation remains when interpreting the comparison.

Hints

- Distinguish how participants entered the study from how treatments were assigned after they entered. - Conditions for random sampling without replacement do not play the same role in a randomized experiment. - Think separately about causal conclusions and conclusions about a wider population.

Solution

1. Random assignment to the two treatments satisfies the randomization condition for comparing the treatment groups. 2. For a randomized experiment, the \(10\%\) condition used for sampling without replacement is not required. 3. Random assignment supports a cause-and-effect comparison between the two study environments for the volunteers in the experiment, provided other design requirements are met. 4. Because the volunteers were not randomly sampled from a broader population, the results cannot automatically be generalized to all students or another larger population.

Answer

Random assignment satisfies the relevant randomization condition, and the \(10\%\) sampling condition is not needed for this randomized experiment. The design can support a causal comparison for the participants, but the lack of random sampling limits generalization to a broader population.

54767012

Two independent populations both have standard deviation \(10\). Plan A uses independent random samples with \(n_1=n_2=25\). Plan B increases only the first sample size to \(n_1=100\) while keeping \(n_2=25\). Find the standard deviation of the sampling distribution of \(\bar{x}_1-\bar{x}_2\) under each plan. Explain why increasing only one sample size by a factor of \(4\) does not cut the overall standard deviation in half.

Hints

- The difference contains sampling variability from two independent sample means. - Changing one sample size affects only one of those two contributions. - Compare variances before taking square roots to see why the reduction is not proportional to the first sample size alone.

Solution

1. Under Plan A, the standard deviation is \(\sqrt{\frac{10^2}{25}+\frac{10^2}{25}}=\sqrt{8}\approx2.83\). 2. Under Plan B, the standard deviation is \(\sqrt{\frac{10^2}{100}+\frac{10^2}{25}}=\sqrt{5}\approx2.24\). 3. Quadrupling the first sample size cuts only the first sample mean's variance contribution from \(4\) to \(1\). 4. The second sample still contributes variance \(4\), so the total variance falls from \(8\) to \(5\), not to \(2\). Therefore, the overall standard deviation is not halved.

Answer

Plan A: standard deviation \(\sqrt{8}\approx2.83\). Plan B: standard deviation \(\sqrt{5}\approx2.24\). Only one source of sampling variability is reduced, so the total standard deviation does not fall by one-half.

54767612

Two independent populations are compared. Population A is normally distributed, and a random sample of \(8\) observations is taken from it. Population B is strongly nonnormal but not extremely skewed, and a random sample of \(40\) observations is taken from it. Both populations are large enough for the \(10\%\) condition. Explain why the sampling distribution of \(\bar{x}_A-\bar{x}_B\) can be modeled as approximately normal even though one sample size is only \(8\).

Hints

- Check the shape condition for each sample mean separately before combining them. - A small sample is not automatically a problem when its source population already has the needed shape. - The larger sample can smooth the nonnormal shape of the other population at the sample-mean level.

Solution

1. Because Population A is normal, the sampling distribution of \(\bar{x}_A\) is normal for any sample size, including \(n_A=8\). 2. Population B is nonnormal, but \(n_B=40\) is large enough for the central limit theorem to support an approximately normal model for \(\bar{x}_B\), assuming the population is not extremely skewed. 3. The two random samples are independent and satisfy the population-size condition. 4. The difference of an independent normal variable and an approximately normal variable is approximately normal, so \(\bar{x}_A-\bar{x}_B\) can be modeled as approximately normal.

Answer

The small sample from Population A is acceptable because that population is normal, making \(\bar{x}_A\) normal even for \(n_A=8\). The larger sample from Population B makes \(\bar{x}_B\) approximately normal by the central limit theorem. With independent samples, their difference is therefore approximately normal.

54768212

Two independent normal populations both have mean \(50\). Population 1 has standard deviation \(6\), and Population 2 has standard deviation \(8\). Independent random samples of sizes \(36\) and \(64\), respectively, are selected. Find the probability that the two sample means differ by less than \(2\) units in absolute value: \(P(|\bar{x}_1-\bar{x}_2|<2)\).

Hints

- Equal population means make the center of the difference especially simple. - Translate the absolute-value condition into a two-sided interval before finding probability. - Combine independent sampling variability from both means when finding the spread of their difference.

Solution

1. The sampling distribution of \(\bar{x}_1-\bar{x}_2\) is normal with mean \(50-50=0\). 2. Its standard deviation is \(\sqrt{\frac{6^2}{36}+\frac{8^2}{64}}=\sqrt{2}\approx1.414\). 3. The event \(|\bar{x}_1-\bar{x}_2|<2\) is equivalent to \(-2<\bar{x}_1-\bar{x}_2<2\). 4. The standardized endpoints are approximately \(-1.414\) and \(1.414\). 5. Therefore, the probability is approximately \(0.8427\).

Answer

\(P(|\bar{x}_1-\bar{x}_2|<2)\approx0.8427\).

54768812

Two independent populations are sampled. Population 1 has standard deviation \(12\), and \(n_1=36\). Population 2 has unknown standard deviation \(\sigma_2\), and \(n_2=25\). The sampling distribution of \(\bar{x}_1-\bar{x}_2\) has standard deviation \(3\). Find \(\sigma_2\).

Hints

- Work with variance contributions before taking the final square root. - One population's contribution to the sampling variability is completely known. - Use the stated overall spread to isolate the unknown population variance.

Solution

1. The variance of the difference in independent sample means is \(\frac{12^2}{36}+\frac{\sigma_2^2}{25}\). 2. Since the standard deviation is \(3\), the variance is \(9\), so \(4+\frac{\sigma_2^2}{25}=9\). 3. Therefore, \(\frac{\sigma_2^2}{25}=5\), giving \(\sigma_2^2=125\). 4. Thus, \(\sigma_2=\sqrt{125}\approx11.18\).

Answer

\(\sigma_2=\sqrt{125}\approx11.18\).

54771212

Two independent normal populations are compared. Population 1 has mean \(6.5\) and standard deviation \(3\), with a random sample of size \(25\). Population 2 has mean \(5.0\) and standard deviation \(2.4\), with a random sample of size \(36\). Find the probability that the sample from Population 1 has a lower sample mean than the sample from Population 2.

Hints

- Rewrite the comparison of two sample means as an inequality involving their difference. - The population mean difference determines the center of the sampling distribution. - Independent sampling variability from both groups contributes to the chance of a reversal.

Solution

1. The sampling distribution of \(\bar{x}_1-\bar{x}_2\) is normal with mean \(6.5-5.0=1.5\). 2. Its standard deviation is \(\sqrt{\frac{3^2}{25}+\frac{2.4^2}{36}}\approx0.721\). 3. Population 1 has the lower sample mean when \(\bar{x}_1-\bar{x}_2<0\). 4. The standardized value for \(0\) is \(z=\frac{0-1.5}{0.721}\approx-2.08\). 5. Therefore, the probability is approximately \(0.0188\).

Answer

\(P(\bar{x}_1<\bar{x}_2)\approx0.0188\).

54772412

Population 1 has standard deviation \(18\), and Population 2 has standard deviation \(6\). Two independent sampling plans use the same total number of observations. Plan A uses \(n_1=81\) and \(n_2=9\). Plan B uses \(n_1=9\) and \(n_2=81\). Find the standard deviation of \(\bar{x}_1-\bar{x}_2\) under each plan and explain why the plans have very different precision.

Hints

- Each population's contribution depends on both its variability and its sample size. - Compare which plan assigns the larger sample to the population with the larger standard deviation. - Equal total sample size does not imply equal variance of the difference.

Solution

1. Under Plan A, the standard deviation is \(\sqrt{\frac{18^2}{81}+\frac{6^2}{9}}=\sqrt{8}\approx2.83\). 2. Under Plan B, the standard deviation is \(\sqrt{\frac{18^2}{9}+\frac{6^2}{81}}\approx6.04\). 3. Population 1 is much more variable, so giving it the larger sample size greatly reduces its contribution to the sampling variability. 4. The same total sample size does not guarantee the same precision when variability differs substantially between populations.

Answer

Plan A has standard deviation approximately \(2.83\); Plan B has standard deviation approximately \(6.04\). Allocating more observations to the more variable population produces much greater precision here.

54773012

A university compares the time needed to complete the same online advising task under two interface versions. For Version A, the population mean completion time is \(42\,\text{s}\) with population standard deviation \(6\,\text{s}\). For Version B, the population mean is \(38\,\text{s}\) with population standard deviation \(9\,\text{s}\). Independent random samples of \(64\) users for Version A and \(81\) users for Version B are selected from large populations. Find the mean and standard deviation of the sampling distribution of \(\bar{x}_A-\bar{x}_B\), and find \(P(2<\bar{x}_A-\bar{x}_B<5)\).

Hints

- Find the center of the difference by keeping the subtraction order consistent. - Combine the two independent sources of sampling variability before standardizing. - The requested probability lies between two cutoffs, so both standardized bounds are needed.

Solution

1. The sampling distribution of \(\bar{x}_A-\bar{x}_B\) is approximately normal because both sample sizes are at least \(30\). 2. Its mean is \(42-38=4\,\text{s}\). 3. Its standard deviation is \(\sqrt{\frac{6^2}{64}+\frac{9^2}{81}}=\sqrt{1.5625}=1.25\,\text{s}\). 4. The standardized bounds are \(z=\frac{2-4}{1.25}=-1.60\) and \(z=\frac{5-4}{1.25}=0.80\). 5. Therefore, \(P(2<\bar{x}_A-\bar{x}_B<5)=P(-1.60<Z<0.80)\approx0.7333\).

Answer

The sampling distribution has mean \(4\,\text{s}\) and standard deviation \(1.25\,\text{s}\). The probability is approximately \(0.7333\).

54774212

Two independent populations have population standard deviations \(7\) and \(9\). A researcher plans to take independent random samples of the same size \(n\) from both populations. What is the minimum common sample size \(n\) needed so that the standard deviation of the sampling distribution of \(\bar{x}_1-\bar{x}_2\) is at most \(1.5\)?

Hints

- Write the sampling-distribution spread using the same unknown sample size for both groups. - Translate “at most \(1.5\)” into an inequality for that spread. - Check the neighboring whole numbers after solving the inequality.

Solution

1. With equal sample sizes, the standard deviation of \(\bar{x}_1-\bar{x}_2\) is \(\sqrt{\frac{7^2}{n}+\frac{9^2}{n}}=\sqrt{\frac{130}{n}}\). 2. Require \(\sqrt{\frac{130}{n}}\le1.5\). 3. Squaring gives \(\frac{130}{n}\le2.25\), so \(n\ge\frac{130}{2.25}\approx57.78\). 4. The minimum whole-number sample size is \(n=58\) from each population. 5. At \(n=58\), the standard deviation is approximately \(1.497\); at \(n=57\), it is approximately \(1.510\).

Answer

The researcher needs at least \(58\) observations from each population.

54775412

Two independent random samples are taken from the same very large, approximately normal population, which has population standard deviation \(15\). The first sample has size \(25\), and the second has size \(100\). Find the standard deviation of the sampling distribution of \(\bar{x}_1-\bar{x}_2\), and approximate the probability that the two sample means differ by more than \(6\) units in absolute value.

Hints

- Since the two samples come from the same population, first determine the expected difference between their sample means. - Combine the independent sampling variability from both samples. - An absolute-difference probability requires considering both tails of the sampling distribution.

Solution

1. Because the population is approximately normal and the samples are independent, the sampling distribution of \(\bar{x}_1-\bar{x}_2\) is normal and centered at \(\mu-\mu=0\). 2. Its standard deviation is \(15\sqrt{\frac{1}{25}+\frac{1}{100}}=\sqrt{11.25}\approx3.354\). 3. A difference of \(6\) units corresponds to \(z=\frac{6}{3.354}\approx1.789\). 4. The probability of an absolute difference greater than \(6\) is \(2P(Z>1.789)\approx0.0736\).

Answer

The standard deviation is approximately \(3.354\), and \(P(|\bar{x}_1-\bar{x}_2|>6)\approx0.0736\).

54777212

Two independent populations have standard deviations \(4\) and \(12\). Independent random samples of size \(25\) are taken from each population. For the sampling distribution of \(\bar{x}_1-\bar{x}_2\), what percentage of the variance comes from Population 2? Explain what this says about which population contributes more to the sampling variability.

Hints

- Separate the two variance terms inside the standard-error expression. - Compare each term with their sum rather than comparing the standard deviations directly. - Equal sample sizes do not imply equal contributions when population variability differs.

Solution

1. Population 1 contributes variance \(\frac{4^2}{25}=0.64\). 2. Population 2 contributes variance \(\frac{12^2}{25}=5.76\). 3. The total variance is \(0.64+5.76=6.40\). 4. Population 2 contributes \(\frac{5.76}{6.40}=0.90\), or \(90\%\), of the variance. 5. Even with equal sample sizes, the much larger population standard deviation makes Population 2 the dominant source of sampling variability.

Answer

Population 2 contributes \(90\%\) of the variance of \(\bar{x}_1-\bar{x}_2\).

54779012

From one finite population, a researcher selects a simple random sample of \(40\) individuals without replacement. Those \(40\) individuals are then removed from the population, and a second simple random sample of \(40\) is selected from the remaining individuals. The researcher wants to use the standard independent-samples formula for the spread of \(\bar{x}_1-\bar{x}_2\). Explain why the two sample means are not independent under this design.

Hints

- Ask whether knowing who entered the first sample changes what can happen in the second sample. - Independence between samples requires that one sample's selection not alter the distribution of the other. - The usual difference-of-means variance formula assumes no covariance between the two sample means.

Solution

1. The composition of the second sample depends on which individuals were selected into the first sample because those individuals cannot appear again. 2. Information about the first sample therefore changes the set of values available for the second sample. 3. This creates dependence between \(\bar{x}_1\) and \(\bar{x}_2\). 4. The usual variance formula that simply adds the two independent sample-mean variances is not justified without accounting for this dependence.

Answer

The sample means are not independent because the first sample changes the population available for the second sample. The standard independent-samples spread formula therefore does not apply as stated.

54779612

Independent random samples are taken from two large populations. Population 1 has standard deviation \(12\) with sample size \(36\), and Population 2 has standard deviation \(8\) with sample size \(64\). Approximate the probability that the observed difference \(\bar{x}_1-\bar{x}_2\) exceeds the true population mean difference \(\mu_1-\mu_2\) by more than \(5\) units.

Hints

- Center the problem on the estimation error rather than on the unknown population means themselves. - Find the spread of the difference in independent sample means. - Translate “more than \(5\) above the true difference” into an upper-tail standardized event.

Solution

1. The sampling error \((\bar{x}_1-\bar{x}_2)-(\mu_1-\mu_2)\) is centered at \(0\). 2. Its standard deviation is \(\sqrt{\frac{12^2}{36}+\frac{8^2}{64}}=\sqrt{5}\approx2.236\). 3. An overstatement of \(5\) units corresponds to \(z=\frac{5}{\sqrt{5}}\approx2.236\). 4. Therefore, the probability of overstating the true difference by more than \(5\) units is \(P(Z>2.236)\approx0.0127\).

Answer

The probability is approximately \(0.0127\).

54781412

Population 1 has standard deviation \(12\), and Population 2 has standard deviation \(6\). A researcher wants the two populations to contribute equally to the variance of the sampling distribution of \(\bar{x}_1-\bar{x}_2\). If the sample from Population 2 has size \(25\), what sample size should be used for Population 1?

Hints

- Set the two variance contributions equal to one another. - Remember that standard deviations are squared in the variance terms. - A larger population spread requires a larger sample to create the same variance contribution.

Solution

1. Equal variance contributions require \(\frac{12^2}{n_1}=\frac{6^2}{25}\). 2. This gives \(\frac{144}{n_1}=\frac{36}{25}\). 3. Solving gives \(n_1=100\). 4. The population with twice the standard deviation needs four times the sample size to contribute the same sampling variance.

Answer

Population 1 should have sample size \(100\).

54783212

Two populations each have standard deviation \(10\). A study will take independent random samples totaling \(100\) observations and use \(\bar{x}_A-\bar{x}_B\). Compare the sampling-distribution standard deviations for allocations \((n_A,n_B)=(50,50)\), \((30,70)\), and \((10,90)\). Which allocation is most precise?

Hints

- Keep the total sample size fixed while comparing how each group contributes to the variability of the difference. - Compute the spread for each proposed allocation rather than comparing only the total number sampled. - Greater imbalance can make the smaller group dominate the uncertainty.

Solution

1. For independent samples, the standard deviation is \(\sqrt{\frac{10^2}{n_A}+\frac{10^2}{n_B}}\). 2. For \((50,50)\), the standard deviation is \(\sqrt{2+2}=2.000\). 3. For \((30,70)\), it is \(\sqrt{\frac{100}{30}+\frac{100}{70}}\approx2.182\). 4. For \((10,90)\), it is \(\sqrt{10+\frac{100}{90}}\approx3.333\). 5. The balanced allocation \((50,50)\) has the smallest standard deviation and is therefore the most precise of the three.

Answer

The standard deviations are \(2.000\), approximately \(2.182\), and approximately \(3.333\), respectively. The \((50,50)\) allocation is most precise.

54785012

Two surveys estimate a mean from the same city. Survey A samples \(100\) residents and Survey B samples \(100\) residents, but \(30\) residents appear in both samples. An analyst wants to use the usual independent-samples formula for the standard deviation of \(\bar{x}_A-\bar{x}_B\). Explain why the overlap is a problem for that formula.

Hints

- Check whether the two sample statistics are built from completely separate sets of observations. - Shared observations can make two statistics move together. - The usual spread formula relies on independence between the two sample means.

Solution

1. The usual formula for the spread of \(\bar{x}_A-\bar{x}_B\) assumes that the two sample means are independent. 2. Because \(30\) residents contribute to both sample means, the two statistics share observations and can be associated. 3. When the sample means are not independent, their covariance affects the variance of the difference. 4. Therefore, simply adding the two separate sampling variances is not justified for these overlapping samples.

Answer

The independent-samples formula is not justified because the two sample means share \(30\) observations and are therefore not independent. The overlap creates covariance that the usual formula ignores.

54786212

For two independent sample means, the sampling-distribution standard deviation of \(\bar{x}_A\) is \(2.4\), while the standard deviation of \(\bar{x}_A-\bar{x}_B\) is \(3.0\). Find the sampling-distribution standard deviation of \(\bar{x}_B\).

Hints

- Work with variances rather than subtracting standard deviations directly. - Independence determines how the two sources of sampling variability combine. - One component can be recovered from the total variance and the other known component.

Solution

1. Independence gives \(\operatorname{Var}(\bar{x}_A-\bar{x}_B)=\operatorname{Var}(\bar{x}_A)+\operatorname{Var}(\bar{x}_B)\). 2. Therefore, \(3.0^2=2.4^2+\sigma_{\bar{x}_B}^2\). 3. Thus, \(\sigma_{\bar{x}_B}=\sqrt{9-5.76}=\sqrt{3.24}=1.8\).

Answer

The sampling-distribution standard deviation of \(\bar{x}_B\) is \(1.8\).

54786812

A simulation of the sampling distribution of \(\bar{x}_A-\bar{x}_B\) is approximately normal. About \(95\%\) of the simulated differences lie between \(3.08\) and \(10.92\). Using the normal approximation that a central \(95\%\) interval extends about \(1.96\) standard deviations from the mean, estimate the center and standard deviation of this sampling distribution.

Hints

- Use symmetry to locate the center of the reported central interval. - Measure the distance from that center to one endpoint. - Relate that distance to the stated number of standard deviations.

Solution

1. The center is the midpoint of the symmetric interval: \(\frac{3.08+10.92}{2}=7.00\). 2. The distance from the center to either endpoint is \(10.92-7.00=3.92\). 3. If this distance represents \(1.96\) standard deviations, then the standard deviation is \(3.92/1.96=2.00\). 4. Thus, the simulated sampling distribution is centered near \(7\) with standard deviation near \(2\).

Answer

The estimated center is \(7\), and the estimated sampling-distribution standard deviation is \(2\).

54788612

The sampling distribution of \(\bar{x}_A-\bar{x}_B\) is approximately normal with mean \(4\) and standard deviation \(1.5\). Give an interval that will contain about \(95\%\) of future values of \(\bar{x}_A-\bar{x}_B\). Then explain why this is not a \(95\%\) confidence interval for the population mean difference.

Hints

- Center the interval on the mean of the sampling distribution. - Use the sampling-distribution standard deviation to describe future statistics. - Distinguish a distribution of statistics from an interval that estimates a fixed unknown parameter.

Solution

1. About \(95\%\) of a normal distribution lies within \(1.96\) standard deviations of its mean. 2. The endpoints are \(4\pm1.96(1.5)=4\pm2.94\), giving approximately \((1.06,6.94)\). 3. This interval describes the random values of the statistic \(\bar{x}_A-\bar{x}_B\) across future samples when the population parameters are treated as fixed and known. 4. A confidence interval instead starts with one observed sample and uses it to estimate the unknown population mean difference.

Answer

About \(95\%\) of future sample-mean differences will lie in approximately \((1.06,6.94)\). This is a sampling-distribution prediction interval for the statistic, not a confidence interval estimating an unknown parameter from one sample.

54789812

Independent samples compare two population means. Population A has standard deviation \(6\), and Population B has standard deviation \(8\). The sample size from B is fixed at \(16\), while the sample size from A can be made arbitrarily large. As \(n_A\) becomes extremely large, what value does the standard deviation of \(\bar{x}_A-\bar{x}_B\) approach? Explain why it does not approach \(0\).

Hints

- Separate the two groups’ contributions to the variance of the difference. - Consider what happens to each contribution as only one sample size grows. - A fixed sample continues to contribute sampling uncertainty.

Solution

1. The sampling-distribution standard deviation is \(\sqrt{\frac{6^2}{n_A}+\frac{8^2}{16}}\). 2. As \(n_A\) becomes extremely large, \(36/n_A\) approaches \(0\). 3. The contribution from Population B remains \(64/16=4\). 4. Therefore, the standard deviation approaches \(\sqrt{4}=2\), not \(0\).

Answer

The standard deviation approaches \(2\). Making only the A sample arbitrarily large cannot remove the sampling variability coming from the fixed B sample.

54782612

Population A consists of the values \(8\) and \(12\), each equally likely. Population B consists of the values \(3\) and \(9\), each equally likely. Independently take a sample of size \(2\) with replacement from each population. Find the exact sampling distribution of \(\bar{x}_A-\bar{x}_B\), and then find \(P(\bar{x}_A-\bar{x}_B>4)\).

Hints

- First determine all possible sample means for each population and how likely each one is. - The two samples are independent, so combine the probabilities from the two separate sampling distributions. - Values of the difference can occur in more than one way, so combine probabilities for repeated outcomes.

Solution

1. For Population A, \(\bar{x}_A\) can be \(8,10,12\) with probabilities \(\frac14,\frac12,\frac14\), respectively. 2. For Population B, \(\bar{x}_B\) can be \(3,6,9\) with probabilities \(\frac14,\frac12,\frac14\), respectively. 3. Combining the independent sampling distributions gives \(\bar{x}_A-\bar{x}_B=-1,1,2,3,4,5,6,7,9\) with probabilities \(\frac1{16},\frac2{16},\frac2{16},\frac1{16},\frac4{16},\frac1{16},\frac2{16},\frac2{16},\frac1{16}\), respectively. 4. The values greater than \(4\) are \(5,6,7,9\), whose probabilities sum to \(\frac{1+2+2+1}{16}=\frac6{16}=0.375\).

Answer

The exact distribution is \(-1: \frac1{16},\ 1: \frac2{16},\ 2: \frac2{16},\ 3: \frac1{16},\ 4: \frac4{16},\ 5: \frac1{16},\ 6: \frac2{16},\ 7: \frac2{16},\ 9: \frac1{16}\). Also, \(P(\bar{x}_A-\bar{x}_B>4)=\frac6{16}=0.375\).

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