Free math worksheets

- The condition compares the sample size with the size of the population from which it is drawn. - Express the sample as a fraction of the population and impose the stated percentage limit. - Check the proposed population separately after finding the threshold.

1. The \(10\%\) condition requires the population size \(N\) to be at least \(10\) times the sample size. 2. With \(n=75\), the minimum population size is \(N=10(75)=750\). 3. A population of \(600\) does not satisfy the condition because \(75\) is \(12.5\%\) of \(600\), which exceeds \(10\%\).

The minimum population size is \(750\). A population of \(600\) does not satisfy the \(10\%\) condition because the sample would be more than \(10\%\) of the population.

- Use the same population spread for all three sampling plans. - Compare the square roots of the three sample sizes rather than the sample sizes directly. - Look for a simple multiplicative pattern in the resulting sampling spreads.

1. For \(n=16\), the standard deviation is \(\frac{24}{\sqrt{16}}=6\). 2. For \(n=64\), the standard deviation is \(\frac{24}{\sqrt{64}}=3\). 3. For \(n=144\), the standard deviation is \(\frac{24}{\sqrt{144}}=2\). 4. The spread decreases according to the inverse square root of sample size: larger samples produce more concentrated sample means.

The standard deviations are \(6\), \(3\), and \(2\), respectively. Sampling variability decreases as \(1/\sqrt{n}\).

- Write out what a sample mean contains when there is only one observation. - Compare that statistic directly with the sampled value. - Then use the usual center and spread relationships at \(n=1\).

1. When \(n=1\), the sample mean is just the single sampled observation. 2. Therefore, the sampling distribution of \(\bar{x}\) has the same shape as the population, so it is strongly right-skewed. 3. Its mean is the population mean \(\mu\). 4. Its standard deviation is the population standard deviation \(\sigma\), since \(\sigma/\sqrt{1}=\sigma\).

For \(n=1\), the sampling distribution of \(\bar{x}\) is the population distribution itself: it has the same strongly right-skewed shape, mean \(\mu\), and standard deviation \(\sigma\).

- Ask what values can appear in any sample from this population. - An average cannot vary if none of its component observations vary. - A distribution with only one possible value has no spread.

1. Every sampled observation equals \(7\). 2. Therefore, every sample mean also equals \(7\), regardless of the sample size. 3. The sampling distribution is concentrated entirely at \(7\), with mean \(7\) and standard deviation \(0\).

\(\bar{x}=7\) for every possible sample. Its sampling distribution has mean \(7\) and standard deviation \(0\).

- Compare the location of the two distributions before comparing their widths. - Increasing sample size does not change the population mean being estimated. - Think about how the standard error of a sample mean changes with \(n\).

1. Both empirical sampling distributions are centered near \(50\), as expected because the sample mean is centered at the population mean. 2. Panel b) is much more concentrated around \(50\) than panel a). 3. Since the standard deviation of \(\bar{x}\) decreases as sample size increases, panel b) corresponds to \(n=36\), while panel a) corresponds to \(n=4\).

Panel b) corresponds to \(n=36\). Both distributions are centered near \(50\), but the larger sample size produces the smaller spread.

- Think about what happens to the center when many sample means are taken from the same population. - Consider how averaging \(64\) observations changes the spread compared with the spread of individual observations. - After describing the sampling distribution, translate the probability question into a standardized distance from its center.

1. Because the source population is approximately normal, the sampling distribution of \(\bar{x}\) is also approximately normal. The random sample is less than \(10\%\) of the population, supporting the usual independence approximation. 2. Its mean is \(\mu_{\bar{x}}=18.4\,\text{s}\). 3. Its standard deviation is \(\sigma_{\bar{x}}=\frac{4.8}{\sqrt{64}}=0.6\,\text{s}\). 4. The standardized value for \(19.5\,\text{s}\) is \(z=\frac{19.5-18.4}{0.6}\approx1.83\). 5. Therefore, \(P(\bar{x}>19.5)\approx0.0334\).

The sampling distribution has mean \(18.4\,\text{s}\) and standard deviation \(0.6\,\text{s}\), and \(P(\bar{x}>19.5)\approx0.0334\).

- The center of a sample-mean distribution does not depend on the sample size. - Focus on how the spread of sample means changes as the sample size grows. - Solve the spread requirement for the smallest allowable whole-number sample size, then check that the sampling fraction is small enough.

1. The sampling distribution has mean \(\mu_{\bar{x}}=240\,\text{h}\). 2. Require \(\frac{36}{\sqrt{n}}\le4.5\). 3. This gives \(\sqrt{n}\ge8\), so \(n\ge64\). 4. The smallest whole-number sample size is \(64\). Because \(64\) is less than \(10\%\) of the population, the usual sampling-standard-deviation formula is appropriate.

The smallest sample size is \(n=64\), and the sampling-distribution mean is \(240\,\text{h}\).

- First decide whether the sample size and stated population shape support an approximately normal model for sample means. - Find the center and sampling spread before translating the two endpoints. - A between-probability requires the area bounded by both standardized endpoints.

1. With \(n=49\) and no extreme population skewness or outliers, the central limit theorem supports modeling the sampling distribution of \(\bar{x}\) as approximately normal. 2. Its mean is \(12\,\text{min}\), and its standard deviation is \(\frac{7}{\sqrt{49}}=1\,\text{min}\). 3. The lower endpoint has \(z=\frac{10-12}{1}=-2\), and the upper endpoint has \(z=\frac{13.5-12}{1}=1.5\). 4. Therefore, \(P(10<\bar{x}<13.5)=P(-2<Z<1.5)\approx0.9104\).

\(P(10<\bar{x}<13.5)\approx0.9104\).

- Compare the rule for the variability of individual observations with the rule for an average of many observations. - Check whether increasing the sample size by a factor of \(36\) should reduce spread by the same factor or by a square-root factor. - Revisit the distinction between variance and standard deviation when combining independent observations.

1. The standard deviation of a sample mean is found by dividing the population standard deviation by the square root of the sample size, not by the sample size itself. The sample is less than \(10\%\) of the population, so the usual formula applies. 2. The correct standard deviation is \(\sigma_{\bar{x}}=\frac{18}{\sqrt{36}}=3\). 3. Averaging \(36\) observations reduces variability by a factor of \(\sqrt{36}=6\), not by a factor of \(36\). 4. Thus, \(0.5\) understates the sampling variability of \(\bar{x}\).

Jordan divided by \(n\) instead of \(\sqrt{n}\). The correct sampling-distribution standard deviation is \(3\), not \(0.5\).

- Compare the population spread with the square root of the sample size in each study. - The center of a sample-mean distribution follows the population mean. - Look for proportional changes in the numerator and denominator of the sampling-spread expression.

1. Because the populations have the same mean, the two sampling distributions have the same mean. The stated population sizes support the usual independence approximation for both sample means. 2. Study A has sampling-distribution standard deviation \(\frac{12}{\sqrt{25}}=2.4\). 3. Study B has sampling-distribution standard deviation \(\frac{24}{\sqrt{100}}=2.4\). 4. Study B doubles the population standard deviation but quadruples the sample size, so the square root of the sample size also doubles. The two effects cancel in the standard deviation of \(\bar{x}\).

The sampling distributions have the same mean and the same standard deviation, \(2.4\). In Study B, doubling the population standard deviation is exactly offset by doubling \(\sqrt{n}\) when the sample size is quadrupled.

- One sampling-distribution parameter is inherited directly from the population. - The other parameter changes by a square-root sample-size factor. - Work backward from the sample-mean spread to recover the population spread.

1. The mean of the sampling distribution of \(\bar{x}\) equals the population mean, so \(\mu=145\). 2. For independent observations, the sampling-distribution standard deviation satisfies \(2.2=\frac{\sigma}{\sqrt{81}}\). 3. Since \(\sqrt{81}=9\), \(\sigma=2.2(9)=19.8\).

The population mean is \(145\), and the population standard deviation is \(19.8\).

- Model the hourly sample mean, not the weight of a single unit. - The two alert limits are equally far from the sampling-distribution center. - Convert the alert rule into two tail events in the standardized sampling distribution.

1. Because the individual weights are normal and the sampled units are independent, the sampling distribution of \(\bar{x}\) is normal with mean \(500\,\text{g}\) and standard deviation \(\frac{40}{\sqrt{100}}=4\,\text{g}\). 2. The alert limits are \(8\,\text{g}\) below and above the mean, corresponding to \(z=-2\) and \(z=2\). 3. The alert probability is \(P(Z<-2)+P(Z>2)\approx0.0455\).

The probability of triggering an alert is approximately \(0.0455\).

- Sample size is not the only route to a normal sampling distribution. - Determine the sampling-distribution spread before standardizing the interval endpoints. - The requested event is centered symmetrically around the population mean.

1. Because the source population is normal and the observations are independent, the sampling distribution of \(\bar{x}\) is exactly normal for any sample size. 2. Its mean is \(10\), and its standard deviation is \(\frac{2}{\sqrt{5}}\approx0.894\). 3. The standardized endpoints are approximately \(-1.118\) and \(1.118\). 4. Therefore, \(P(9<\bar{x}<11)\approx0.7364\).

\(P(9<\bar{x}<11)\approx0.7364\). The normal model is exact because the source population is normal and the sampled observations are independent.

- Compare the sample size with a fixed percentage of the finite population. - Sampling without replacement creates dependence when the sampling fraction becomes large. - The issue here is the sampling design, not the shape of the population distribution.

1. Ten percent of the population is \(0.10(500)=50\). 2. The sample size is \(80\), which is greater than \(50\). 3. Equivalently, the sample is \(16\%\) of the population. 4. Therefore, the \(10\%\) condition fails, so this guideline does not support the usual independence approximation or the unadjusted standard-deviation formula for sampling without replacement.

The \(10\%\) condition fails because \(80>50\), so the sample is too large a fraction of the population for the usual independence approximation and unadjusted formula to be justified by that guideline.

- Treat common sample-size cutoffs as context-dependent guidelines, not automatic laws. - Consider how severe the source-population skewness is in this problem. - The central limit theorem describes convergence, but the speed of convergence depends on population shape.

1. A sample size of at least \(30\) often supports an approximately normal model for the sampling distribution of a sample mean when the population is moderately nonnormal. 2. However, an extremely skewed population may require a sample size much larger than \(30\) before the normal approximation is accurate. 3. Therefore, \(n=30\) does not guarantee approximate normality in this setting. 4. More information or a substantially larger sample would be needed before confidently using the normal approximation.

No. The \(n\ge30\) guideline is not an absolute guarantee. For an extremely skewed population, a much larger sample may be needed before the sampling distribution of \(\bar{x}\) is approximately normal.

- Find the spread of the sampling distribution separately for each sample size. - The interval “within \(2\)” is the same in original units but represents different standardized distances in the two distributions. - Compare how concentration around the population mean changes as \(n\) increases.

1. For \(n=25\), the standard deviation of \(\bar{x}\) is \(\frac{10}{\sqrt{25}}=2\). Being within \(2\) units of the mean corresponds to \(-1<Z<1\), so the probability is approximately \(0.6827\). 2. For \(n=100\), the standard deviation of \(\bar{x}\) is \(\frac{10}{\sqrt{100}}=1\). Being within \(2\) units corresponds to \(-2<Z<2\), so the probability is approximately \(0.9545\). 3. The larger sample size produces a narrower sampling distribution, so sample means are more likely to fall close to the population mean.

For \(n=25\), the probability is approximately \(0.6827\). For \(n=100\), it is approximately \(0.9545\). The larger sample gives the greater probability because its sample means vary less.

- Think about what adding the same constant to every observation does to a sample mean. - Separate the effect on the center from the effect on the spread. - The usual sample-size effect on the spread of a sample mean still applies after calibration.

1. Adding \(7\) to every observation adds \(7\) to every sample mean, so \(\overline{Y}=\overline{X}+7\). 2. Therefore, the sampling-distribution mean is \(\mu+7\). 3. Adding a constant does not change variability, so the population standard deviation remains \(\sigma\). 4. Because the observations are independent, the sampling-distribution standard deviation is \(\frac{\sigma}{\sqrt{n}}\).

The sampling distribution of \(\overline{Y}\) has mean \(\mu+7\) and standard deviation \(\frac{\sigma}{\sqrt{n}}\).

- Do not confuse the shape of the population with the center of the sampling distribution. - Recall what determines the expected value of a sample mean. - The central limit theorem describes what happens as sample size grows, but it does not supply a universal cutoff for every population shape.

1. The sampling distribution of the sample mean is centered at the population mean \(\mu\), not at a population mode. 2. Because the sample is a small fraction of a very large population, its standard deviation is \(\frac{\sigma}{\sqrt{64}}=\frac{\sigma}{8}\). 3. The sampling distribution does not have to preserve the population's bimodal shape. For a finite-variance population, the central limit theorem makes the distribution of sample means increasingly normal as sample size grows, but bimodality alone does not guarantee that \(n=64\) is already large enough for an accurate normal approximation. 4. Therefore, Mei's claims about the center and the required bimodal shape are incorrect.

The statement is incorrect. The sampling distribution is centered at \(\mu\) and has standard deviation \(\frac{\sigma}{8}\). It need not be bimodal; whether \(n=64\) is large enough for an accurate normal approximation depends on more than the fact that the population is bimodal.

- Find the center and spread of the sample-mean distribution before working with the two bounds. - Standardize the lower and upper endpoints separately. - The requested probability is the area between the two standardized values.

1. Because the source population is normal and the observations are independent, the sampling distribution of \(\bar{x}\) is normal with mean \(70\,\text{min}\) and standard deviation \(\frac{12}{\sqrt{36}}=2\,\text{min}\). 2. The lower bound gives \(z=\frac{68-70}{2}=-1.00\). 3. The upper bound gives \(z=\frac{73-70}{2}=1.50\). 4. Therefore, \(P(68<\bar{x}<73)=P(-1<Z<1.5)\approx0.7745\).

The probability is approximately \(0.7745\).

- Separate random sampling variability from systematic selection bias. - Ask how participants enter each sample before comparing the sample sizes. - A larger sample can narrow a distribution without correcting a biased center.

1. A simple random sample gives each appropriate population member a chance to be selected through a random mechanism, so the sample mean is an unbiased estimator of the population mean under the sampling model. 2. Method B uses voluntary response, so people who choose to participate may differ systematically from the population. 3. Increasing the number of voluntary responses can reduce random sampling variability around the volunteer-group tendency, but it does not remove selection bias. 4. Therefore, the larger sample size in Method B does not guarantee that its sample-mean distribution is centered at the population mean.

Method A uses random sampling and is designed to produce a sample mean centered at the population mean. Method B can remain biased even with \(400\) responses because voluntary response affects the center, not just the spread.

- Distinguish the population standard deviation from the sample standard deviation. - Check whether the sample is small relative to the population before using the usual sampling-distribution formula. - The question asks for the theoretical spread of all possible sample means under the stated population.

1. The theoretical standard deviation of the sampling distribution uses the population standard deviation \(\sigma\), when it is known. 2. Because \(36\) is less than \(10\%\) of the population, the sampled observations are approximately independent for this purpose. 3. Therefore, \(\sigma_{\bar{x}}=\frac{15}{\sqrt{36}}=2.5\). 4. The value \(\frac{12}{\sqrt{36}}=2\) is an estimated standard error based on this particular sample, not the known theoretical standard deviation of the sampling distribution.

The correct sampling-distribution standard deviation is \(2.5\). The value \(2\) uses the sample standard deviation and is an estimate, not the known theoretical spread.

- Ask whether different repetitions of the stated sampling procedure can produce different samples. - Sampling variability comes from variation among possible samples. - Compare a census with the independence assumption behind the usual sample-mean spread formula.

1. Every repetition selects the entire population, so every sample contains exactly the same \(80\) values. 2. Therefore, every sample mean equals the population mean \(\mu\). 3. The sampling distribution is concentrated at the single value \(\mu\) and has standard deviation \(0\). 4. The usual \(\sigma/\sqrt{n}\) formula assumes approximately independent sampling observations; sampling the entire finite population without replacement violates that approximation completely.

The sampling distribution consists only of \(\mu\), so its mean is \(\mu\) and its standard deviation is \(0\). The usual \(\sigma/\sqrt{n}\) formula does not apply to a census drawn without replacement.

- Find each sample mean's sampling-distribution standard deviation from the population standard deviation and sample size. - Compare the fixed distance of \(4\) units with each sampling-distribution standard deviation. - Once the distances are standardized, compare the two resulting normal probabilities.

1. For Population A, the sampling-distribution standard deviation is \(\sigma_{\bar{x}_A}=\frac{12}{\sqrt{36}}=2\). 2. For Population B, the sampling-distribution standard deviation is \(\sigma_{\bar{x}_B}=\frac{20}{\sqrt{100}}=2\). 3. Both source populations are normal, so both sample means are normally distributed. In each case, being within \(4\) units of the population mean means being within \(\frac{4}{2}=2\) sampling-distribution standard deviations of the center. 4. Therefore, for each population, \(P(|\bar{x}-\mu|<4)=P(-2<Z<2)\approx0.9545\). 5. The probabilities are equal because the different population standard deviations and sample sizes produce the same sampling-distribution standard deviation, \(2\).

For Population A, \(\sigma_{\bar{x}_A}=2\), and for Population B, \(\sigma_{\bar{x}_B}=2\). For each population, the probability that the sample mean is within \(4\) units of its population mean is approximately \(0.9545\). The probabilities are equal because the two sampling distributions have the same standard deviation.

- First find the center and spread of the sample mean on the original measurement scale. - Check that the sample is small relative to the population before using the usual spread formula. - The same conversion factor multiplies both the center and the spread.

1. Before conversion, the sampling distribution of \(\bar{x}\) has mean \(12\,\text{miles}\) and standard deviation \(\frac{3}{\sqrt{25}}=0.6\,\text{mile}\). 2. Multiplying every observation by \(1.6\) multiplies both the sample mean and its sampling-distribution standard deviation by \(1.6\). 3. The converted mean is \(12\cdot1.6=19.2\,\text{km}\). 4. The converted standard deviation is \(0.6\cdot1.6=0.96\,\text{km}\).

The sampling distribution has mean \(19.2\,\text{km}\) and standard deviation \(0.96\,\text{km}\).

- Separate the ideas of center and spread in a sampling distribution. - Ask what random sampling implies about the long-run center of \(\bar{x}\). - Larger samples affect how much sample means vary from sample to sample.

1. Under simple random sampling, the sample mean is an unbiased estimator of the population mean, so its sampling distribution is centered at \(48\) for both sample sizes. 2. Increasing the sample size does not move the center toward \(0\). 3. The sampling-distribution standard deviation decreases as the sample size increases, so the sample means for \(n=100\) are less variable than those for \(n=25\).

The claim is false. Both sampling distributions are centered at \(48\). Increasing the sample size reduces the spread of \(\bar{x}\); it does not shrink the center toward \(0\).

- Ask whether the added rows contain new random outcomes. - The usual sample-size effect on spread assumes independent observations. - Reformatting or duplicating data cannot create new information about the population mean.

1. Duplicating each measurement creates exact copies, not new independent observations from the population. 2. The duplicated dataset contains the same information about the population mean as the original \(30\) observations. 3. Therefore, the effective sample size remains \(30\); using a standard deviation proportional to \(1/\sqrt{60}\) would falsely claim more precision. 4. The sampling variability is determined by the original data-collection process, not by the number of rows after duplication.

The effective sample size is still \(30\). Duplicating rows does not create independent information and does not reduce the sampling-distribution standard deviation as a genuine sample of size \(60\) would.

- First identify the center and spread of a standardized individual value. - Then apply the sampling-distribution relationships for a sample mean. - The sample size affects the spread but not the center.

1. Standardizing the population gives \(Y\) mean \(0\) and standard deviation \(1\). 2. The sample mean \(\bar{Y}\) is therefore centered at \(0\). 3. Its sampling-distribution standard deviation is \(1/\sqrt{16}=0.25\).

The sampling distribution of \(\bar{Y}\) has mean \(0\) and standard deviation \(0.25\).

- Use the fact that an average must lie between the smallest and largest values being averaged. - Compare the true support of the statistic with the unbounded support of a normal approximation. - An approximation does not override an exact logical constraint.

1. A sample mean is an average of sampled population values. 2. If every sampled value is at most \(10\), their average cannot exceed \(10\). 3. Therefore, \(P(\bar{x}>10)=0\) exactly. 4. A normal approximation can assign extremely small probability outside the true range because a normal distribution is unbounded; such tail probability is an approximation artifact.

The exact probability is \(0\). An average of values all between \(0\) and \(10\) cannot be greater than \(10\), even if a normal approximation assigns a tiny impossible tail probability there.

- Keep variance and standard deviation formulas distinct. - You can convert the population variance to a population standard deviation first. - Check the result by squaring the sampling-distribution standard deviation.

1. The population standard deviation is \(\sqrt{81}=9\). 2. The standard deviation of \(\bar{x}\) is \(9/\sqrt{9}=3\). 3. Therefore, the variance of \(\bar{x}\) is \(3^2=9\), equivalently \(81/9=9\). 4. Ethan incorrectly divided the variance by \(\sqrt{n}\); variances scale by \(1/n\), while standard deviations scale by \(1/\sqrt{n}\).

The sampling-distribution variance is \(9\), and the sampling-distribution standard deviation is \(3\).

- Distinguish a statement about one realized sample from a statement about a sampling distribution. - Compare \(\operatorname{Var}(\bar{x})\) for the two sample sizes. - A smaller spread does not determine which of two particular random outcomes is closer to the center.

1. Random samples vary, so a particular sample of size \(10\) can happen to have a mean very close to \(\mu\), while a particular sample of size \(100\) can be farther away. 2. For independent observations with population variance \(\sigma^2\), \(\operatorname{Var}(\bar{x})=\frac{\sigma^2}{n}\). 3. Therefore, the size-\(100\) sampling distribution has one tenth the variance, and \(\frac{1}{\sqrt{10}}\) times the standard deviation, of the size-\(10\) sampling distribution. 4. This guarantees less sampling variability, not that every realized larger-sample mean is closer to \(\mu\).

A size-\(100\) sample mean is not guaranteed to be closer to \(\mu\) than a size-\(10\) sample mean. What is guaranteed is a smaller sampling variance: \(\sigma^2/100\) instead of \(\sigma^2/10\).

- Distinguish the distribution of individual observations from the distribution of sample averages. - Each plotted value in panel b) summarizes many observations. - Think about the role of sample size in the Central Limit Theorem.

1. The individual observations inherit the population's strong right skew, which is visible in panel a). 2. Each value in panel b) is an average of \(40\) observations rather than a single observation. 3. Averaging reduces the influence of individual extreme values, and the Central Limit Theorem says the sampling distribution of the sample mean becomes more nearly normal as sample size grows under suitable conditions. 4. Therefore, a skewed population can still produce an approximately bell-shaped sampling distribution of \(\bar{x}\) for a sufficiently large sample size.

Panel b) is more nearly bell-shaped because it is a distribution of averages of \(40\) observations. The Central Limit Theorem allows the sampling distribution of \(\bar{x}\) to be approximately normal even when the population of individual values is skewed.

- Compare the new and old standard errors as a ratio. - Under the stated independent-observation model, standard error changes with the square root of sample size. - Because the target is an upper bound on standard error, the final sample size must be rounded upward.

1. For independent observations from the same population, the standard error of a sample mean is proportional to \(1/\sqrt n\). 2. The ratio of the new standard error to the old standard error is \(\sqrt{\frac{50}{n}}\). 3. Require \(\sqrt{\frac{50}{n}}\le0.70\). 4. Squaring gives \(\frac{50}{n}\le0.49\), so \(n\ge\frac{50}{0.49}\approx102.04\). 5. The smallest integer sample size meeting the requirement is \(103\).

\(n=103\).

- The sampling mechanism does not shift the center of the sample mean. - Use the finite-population correction because the sample is a substantial fraction of the population. - Compare the corrected spread with the value that would apply under independent sampling.

1. The sampling-distribution mean is the population mean, \(50\). 2. The finite-population standard deviation is \(\frac{12}{\sqrt{90}}\sqrt{\frac{300-90}{300-1}}=\frac{12}{\sqrt{90}}\sqrt{\frac{210}{299}}\approx1.060\). 3. Ignoring the finite population gives \(\frac{12}{\sqrt{90}}\approx1.265\). 4. Sampling \(30\%\) of the population without replacement removes substantial uncertainty, so the corrected spread is smaller.

a) \(50\). b) \(\frac{12}{\sqrt{90}}\sqrt{\frac{210}{299}}\approx1.060\). c) \(\frac{12}{\sqrt{90}}\approx1.265\). The uncorrected value is larger because it ignores the reduced variability from sampling a large fraction without replacement.

- Use the first sampling distribution to recover the population spread. - Keep that population parameter fixed when changing the sample size. - Solve the second standard-error relationship for the new sample size.

1. Since \(4=\frac{\sigma}{\sqrt{49}}\), the population standard deviation is \(\sigma=4(7)=28\). 2. For a target standard deviation of \(2\), require \(2=\frac{28}{\sqrt n}\). 3. Thus, \(\sqrt n=14\), so \(n=196\).

a) \(\sigma=28\). b) \(n=196\).

- Compare the two observed spreads as a ratio. - Think about how sampling spread changes when sample size is multiplied. - Changing sample size affects precision, not the population quantity being estimated.

1. For independent samples from the same population, the standard deviation of \(\bar X\) is inversely proportional to \(\sqrt n\). 2. Reducing the standard deviation from \(6\) to \(3\) is a factor of \(2\), so the sample size must increase by a factor of \(2^2=4\). 3. Study B therefore uses approximately \(25\cdot4=100\) observations per sample. 4. Both sampling distributions are centered at the same population mean.

a) Approximately \(100\) observations. b) Their centers should be approximately equal because both estimate the same population mean.

- The population shape affects whether a small-sample mean can be modeled normally. - Separate the questions of center, spread, and shape; sample size influences them differently. - A large random sample can make the distribution of sample means much more regular than the original population distribution.

1. For \(n=16\), the sampling-distribution mean is \(\$48\) and the standard deviation is \(\frac{30}{\sqrt{16}}=\$7.50\). 2. Because the population is strongly right-skewed and \(n=16\) is small, an approximately normal sampling distribution is not guaranteed. 3. For \(n=100\), the sampling-distribution mean is \(\$48\) and the standard deviation is \(\frac{30}{\sqrt{100}}=\$3.00\). 4. With \(n=100\), the central limit theorem supports modeling the sampling distribution of \(\bar{x}\) as approximately normal, provided the population is not so extremely skewed that an even larger sample is needed. 5. Both sample sizes satisfy the \(10\%\) condition because the population is much larger than either sample.

For \(n=16\): mean \(\$48\), standard deviation \(\$7.50\), and normality is not guaranteed because the population is strongly skewed and the sample is small. For \(n=100\): mean \(\$48\), standard deviation \(\$3.00\), and the sampling distribution is approximately normal by the central limit theorem, barring extreme skewness.

- Describe the sampling distribution before looking for the requested cutoff. - The question asks for a percentile of sample means, not a percentile of individual job sizes. - Convert the desired cumulative probability into a standardized location and then return to page units.

1. The sampling distribution of \(\bar{x}\) is normal because the population is normal. The sample is less than \(10\%\) of the population, so the usual independence approximation applies. 2. Its mean is \(120\,\text{pages}\), and its standard deviation is \(\frac{24}{\sqrt{36}}=4\,\text{pages}\). 3. The \(95\)th percentile of the standard normal distribution is approximately \(1.645\). 4. Therefore, \(c=120+1.645(4)\approx126.58\,\text{pages}\). 5. About \(95\%\) of simple random samples of \(36\) jobs have a sample mean print-job size below about \(126.58\,\text{pages}\).

\(c\approx126.58\,\text{pages}\). About \(95\%\) of simple random samples of \(36\) print jobs have \(\bar{x}<126.58\,\text{pages}\).

- Do not use the same spread for an individual observation and for an average of \(25\) observations. - Compare how many standard deviations \(74\) lies above the center in each distribution. - The shared mean does not imply the two distributions have the same concentration around that mean.

1. For an individual value, \(z=\frac{74-70}{10}=0.4\), so \(P(X>74)\approx0.3446\). 2. The sample is less than \(10\%\) of the population, so the usual independence approximation applies. For the sample mean, the standard deviation is \(\frac{10}{\sqrt{25}}=2\). 3. Thus, \(z=\frac{74-70}{2}=2\) for \(\bar{x}\), so \(P(\bar{x}>74)\approx0.0228\). 4. Both distributions are centered at \(70\), but sample means are much less variable than individual observations, making a mean above \(74\) much less common.

\(P(X>74)\approx0.3446\), while \(P(\bar{x}>74)\approx0.0228\). Sample means have the same center as individual values but a much smaller standard deviation, so values far from the mean are less likely for \(\bar{x}\).

- Express the spread of possible sample means in terms of the unknown sample size. - Think about the standardized cutoff that captures the middle \(95\%\) of a normal distribution. - After solving for the sample size, check which way a noninteger result must be rounded to preserve the probability requirement.

1. Because the population distribution is normal and the observations are independent, the sampling distribution of \(\bar{x}\) is normal with standard deviation \(\frac{4.2}{\sqrt{n}}\). 2. To place the middle \(95\%\) of sample means within \(0.7\,\text{min}\) of the population mean, require \(0.7\ge1.960\left(\frac{4.2}{\sqrt{n}}\right)\). 3. Solving gives \(n\ge\left(\frac{1.960\cdot4.2}{0.7}\right)^2\approx138.30\). 4. The sample size must be a whole number and must meet the requirement, so round up to \(n=139\). 5. For \(n=139\), the probability is approximately \(0.9506\), while \(n=138\) gives approximately \(0.9498\).

The minimum sample size is \(139\).

- First describe the distribution of a sample mean for samples of size \(25\). - Convert the cutoff of \(55\,\text{s}\) into a probability for one sample mean. - Use that probability to find an expected count out of \(300\) repeated samples.

1. Because the source population is normal and each sample consists of independent observations, the sampling distribution of \(\bar{x}\) is normal with mean \(52\,\text{s}\) and standard deviation \(\frac{10}{\sqrt{25}}=2\,\text{s}\). 2. For \(55\,\text{s}\), the standardized value is \(z=\frac{55-52}{2}=1.50\). 3. The probability that one sample mean exceeds \(55\,\text{s}\) is \(P(Z>1.50)\approx0.0668\). 4. Across \(300\) independent samples, the expected number is \(300\cdot0.0668\approx20.0\).

About \(20\) of the \(300\) sample means would be expected to exceed \(55\,\text{s}\).

- The center is the same for individual observations and sample means, but their spreads are not. - Standardize \(118\,\text{s}\) once using the population standard deviation and once using the standard deviation of \(\bar{x}\). - Compare the resulting standardized distances before comparing the tail probabilities.

1. For one individual time, \(z=\frac{118-100}{12}=1.50\), so \(P(X>118)\approx0.0668\). 2. Because the source population is normal and the sampled times are independent, the sampling distribution of \(\bar{x}\) is normal with mean \(100\,\text{s}\) and standard deviation \(\frac{12}{\sqrt{16}}=3\,\text{s}\). 3. For the sample mean, \(z=\frac{118-100}{3}=6\), so \(P(\bar{x}>118)\approx9.87\times10^{-10}\). 4. The same numerical cutoff is far more extreme for a sample mean because sample means vary much less than individual observations.

For an individual time, the flag probability is approximately \(0.0668\). For a sample mean of \(16\) times, the flag probability is approximately \(9.87\times10^{-10}\). The cutoff is therefore vastly more extreme for the sample mean.

- Convert the condition on the total into an equivalent condition on the sample mean. - Use both the sampling fraction and the stated population shape when deciding whether the usual sampling model is reasonable. - Standardize the mean cutoff using the spread of sample means, not the spread of individual households.

1. A total above \(2000\,\text{lb}\) for \(100\) households is equivalent to a sample mean above \(\frac{2000}{100}=20\,\text{lb}\). 2. The sample is less than \(10\%\) of the population, and with \(n=100\) and no extreme shape features, the central limit theorem supports modeling the sampling distribution of \(\bar{x}\) as approximately normal. 3. The sampling distribution has mean \(18\,\text{lb}\) and standard deviation \(\frac{12}{\sqrt{100}}=1.2\,\text{lb}\). 4. The standardized value is \(z=\frac{20-18}{1.2}\approx1.67\). 5. Therefore, \(P(\bar{x}>20)\approx0.0478\).

The probability that the sampled households total more than \(2000\,\text{lb}\) is approximately \(0.0478\).

- Look at how the observations were selected, not only how many were recorded. - Ask whether two observations from the same selected unit could be related. - The usual sample-mean spread formula relies on independence among the sampled values.

1. The formula \(\sigma/\sqrt{n}\) assumes the sampled observations are independent or approximately independent. 2. People from the same household may have related measurements, so the \(40\) observations are clustered rather than independently sampled individuals. 3. Randomly selecting households does not make the four observations within a household independent of one another. 4. Therefore, treating the data as \(40\) independent observations can understate the sampling variability of the mean.

The formula \(\sigma/\sqrt{40}\) is not justified from the stated design because observations within the same household may be dependent. The effective sampling structure is clustered, not a simple random sample of \(40\) independent people.

- Find the center and standard deviation of the uniform population first. - Use the sample size to determine the spread of sample means. - The requested interval is symmetric around the population mean, so the standardized bounds have equal magnitude.

1. For a uniform distribution from \(58\) to \(82\), the population mean is \(\mu=70\), and the population standard deviation is \(\sigma=\frac{82-58}{\sqrt{12}}=4\sqrt{3}\approx6.928\). 2. Since \(n=36\), the sampling distribution of \(\bar{x}\) is approximately normal with standard deviation \(\frac{4\sqrt{3}}{\sqrt{36}}=\frac{2\sqrt{3}}{3}\approx1.155\). 3. The endpoints \(68\) and \(72\) are each \(2\) units from the mean, corresponding to standardized values approximately \(-1.732\) and \(1.732\). 4. Therefore, \(P(68<\bar{x}<72)\approx0.9167\).

The probability is approximately \(0.9167\).

- Decide which plan produces independent draws and therefore uses the unadjusted sample-mean spread. - For the without-replacement plan, apply the supplied finite-population factor after computing \(\sigma/\sqrt{n}\). - Keep the radicals exact while simplifying each expression.

1. With replacement, the draws are independent, so the sampling-distribution standard deviation is \(\frac{10}{\sqrt{30}}\). 2. Without replacement, the finite-population factor is \(\sqrt{\frac{50-30}{50-1}}=\sqrt{\frac{20}{49}}\). 3. The without-replacement standard deviation is \(\frac{10}{\sqrt{30}}\sqrt{\frac{20}{49}}=\frac{20}{7\sqrt{6}}\). 4. Sampling without replacement produces less variability because selected values are not replaced.

a) \(\frac{10}{\sqrt{30}}\) b) \(\frac{20}{7\sqrt{6}}\) The without-replacement sampling distribution is less variable.

- First find the probability for a single sample mean. - It may be easier to work with the complementary event that every sample mean stays at or below the cutoff. - Use the independence of the repeated samples when combining the single-sample probability.

1. For one sample, \(\bar{x}\) is normal with mean \(50\) and standard deviation \(10/\sqrt{25}=2\). 2. The cutoff \(54\) corresponds to \(z=(54-50)/2=2\), so \(P(\bar{x}>54)\approx0.02275\). 3. The probability that none of the \(20\) independent sample means exceeds \(54\) is \((1-0.02275)^{20}\approx0.63112\). 4. Therefore, the probability that at least one exceeds \(54\) is \(1-0.63112\approx0.36888\), which rounds to \(0.3689\).

The probability is approximately \(0.3689\).

- First find the sampling distribution of one sample mean. - Then treat the independent sample means as the values being averaged at the second stage. - Averaging independent quantities reduces spread while preserving their common center.

1. Each sample mean is centered at \(80\) and has standard deviation \(12/\sqrt{25}=2.4\). 2. The \(16\) sample means are independent because they come from independent random samples. 3. Averaging the \(16\) sample means keeps the center at \(80\). 4. The standard deviation of their average is \(2.4/\sqrt{16}=0.6\), equivalent to \(12/\sqrt{400}\).

The sampling distribution of \(G\) has mean \(80\) and standard deviation \(0.6\).

- List the possible numbers of \(10\)s that can appear in a sample of size \(3\). - Convert each count into a possible sample mean. - The expected value of a distribution does not have to be one of its possible outcomes.

1. If the sample contains \(k\) values equal to \(10\), then \(\bar{x}=10k/3\), so the possible sample means are \(0,\frac{10}{3},\frac{20}{3},10\). 2. Thus, \(5\) is not a possible realized value of \(\bar{x}\). 3. The population mean is \(0(0.5)+10(0.5)=5\). 4. The sample mean is unbiased, so the mean of its sampling distribution is \(5\), even though that center need not itself be a possible outcome.

The possible sample means are \(0,\frac{10}{3},\frac{20}{3},10\), so \(5\) cannot occur. Nevertheless, the sampling distribution is centered at \(E(\bar{x})=\mu=5\).

- Find the expected value of one observation under the actual selection mechanism. - The sample mean reflects the distribution being sampled by the procedure, not merely the listed population composition. - Unbiasedness depends on how the sample is selected.

1. Under the biased sampling method, one selected observation has expected value \(0(0.2)+10(0.8)=8\). 2. The mean of a fixed number of independent draws from this selection mechanism also has expected value \(8\). 3. Therefore, the sampling distribution of \(\bar{x}\) is centered at \(8\), not at the population mean \(5\). 4. The usual unbiasedness of the sample mean depends on a sampling process that represents the population rather than systematically oversampling high values.

The sampling distribution is centered at \(8\). The biased sampling method oversamples the value \(10\), so \(\bar{x}\) is not centered at the population mean \(5\).

- Separate the center of a sampling distribution from its spread. - A simple random sample can remain representative even when it is a large fraction of the population. - The \(10\%\) condition is tied to dependence from sampling without replacement.

1. A simple random sample without replacement still gives a sample mean whose expected value is the population mean. 2. Therefore, \(\bar{x}\) remains an unbiased estimator of \(\mu\) even though the sampling fraction is large. 3. The failed \(10\%\) condition affects the approximation that treats sampled observations as independent and uses the usual \(\sigma/\sqrt{n}\) spread formula without a finite-population adjustment. 4. Thus, the issue is sampling variability and dependence, not bias in the center.

The sample mean is still centered at the population mean. Failing the \(10\%\) condition affects the usual independence and standard-error approximation, not the unbiasedness of \(\bar{x}\).

- Compare the center of the histogram with the marked population mean. - Bias concerns the center of an estimator's sampling distribution. - Separate the effect of sample size on variability from the effect of sampling design on bias.

1. The empirical sampling distribution is centered near \(102\), noticeably to the right of the reference line at \(100\). 2. That systematic shift suggests the sampling method produces a biased estimator of the population mean. 3. Increasing sample size can reduce sampling variability, but it does not by itself remove bias caused by a flawed sampling method. 4. Therefore, larger samples taken with the same biased method could produce a narrower distribution still centered away from \(100\).

The plot suggests upward bias because the sampling distribution is centered above \(100\). Larger samples can reduce spread, but they do not necessarily correct bias from the sampling method.

- List the ordered pairs that can occur when sampling with replacement. - Different ordered pairs can produce the same sample mean, so combine equal outcomes when assigning probabilities. - Check the sampling-distribution center and spread against the population mean and the sample-mean spread rule.

1. There are \(9\) equally likely ordered samples of size \(2\). 2. The possible sample means and probabilities are \(P(\bar{x}=2)=\frac{1}{9}\), \(P(\bar{x}=4)=\frac{2}{9}\), \(P(\bar{x}=6)=\frac{3}{9}\), \(P(\bar{x}=8)=\frac{2}{9}\), and \(P(\bar{x}=10)=\frac{1}{9}\). 3. The mean of this sampling distribution is \(6\), equal to the population mean. 4. The population variance is \(\frac{(2-6)^2+(6-6)^2+(10-6)^2}{3}=\frac{32}{3}\), so the standard deviation of \(\bar{x}\) is \(\sqrt{\frac{32/3}{2}}=\frac{4}{\sqrt{3}}\approx2.309\).

\(P(\bar{x}=2)=\frac{1}{9}\), \(P(\bar{x}=4)=\frac{2}{9}\), \(P(\bar{x}=6)=\frac{3}{9}\), \(P(\bar{x}=8)=\frac{2}{9}\), and \(P(\bar{x}=10)=\frac{1}{9}\). The sampling distribution has mean \(6\) and standard deviation \(\frac{4}{\sqrt{3}}\approx2.309\).

- List all equally likely subsets of two values from the population. - Several different samples can have the same sample mean, so combine their probabilities. - Because the sample is a large fraction of this tiny population, sampling without replacement reduces the spread compared with independent draws.

1. The \(6\) equally likely samples are the unordered pairs from the four population values. 2. Their sample means are \(3,5,7,7,9,11\). 3. Thus, \(P(\bar{x}=3)=\frac{1}{6}\), \(P(\bar{x}=5)=\frac{1}{6}\), \(P(\bar{x}=7)=\frac{2}{6}\), \(P(\bar{x}=9)=\frac{1}{6}\), and \(P(\bar{x}=11)=\frac{1}{6}\). 4. The sampling-distribution mean is \(7\), equal to the population mean. 5. The population variance is \(20\). Including the finite-population correction for sampling without replacement, the sampling-distribution variance is \(\frac{20}{2}\cdot\frac{4-2}{4-1}=\frac{20}{3}\). 6. Therefore, the exact sampling-distribution standard deviation is \(\sqrt{\frac{20}{3}}\).

\(P(\bar{x}=3)=\frac{1}{6}\), \(P(\bar{x}=5)=\frac{1}{6}\), \(P(\bar{x}=7)=\frac{1}{3}\), \(P(\bar{x}=9)=\frac{1}{6}\), and \(P(\bar{x}=11)=\frac{1}{6}\). The mean is \(7\), and the standard deviation is \(\sqrt{\frac{20}{3}}\).

- Treat the two possible sample sizes as two different sampling distributions first. - Their centers are the same, but their spreads are not. - When equally mixing distributions with the same center, compare their variances rather than averaging their sample sizes.

1. The sampling-distribution standard deviation must be calculated for each actual sample size, not from the average sample size. 2. For \(n=25\), the standard deviation is \(20/\sqrt{25}=4\). For \(n=100\), it is \(20/\sqrt{100}=2\). 3. Both component distributions have the same mean \(100\), so the variance of the equally weighted mixture is the average of their variances: \(\frac{4^2+2^2}{2}=10\). 4. The combined distribution therefore has standard deviation \(\sqrt{10}\), not \(20/\sqrt{62.5}\).

Averaging the sample sizes does not produce the correct sampling variability. The combined distribution has standard deviation \(\sqrt{10}\).

- Each sample mean should contribute in proportion to how many observations it represents. - After writing the combined statistic as a weighted sum, track how independent sources of variability combine. - Check whether the result agrees with what you would expect from one sample containing all \(100\) observations.

1. The two sample means should not receive equal weight because the second sample contains three times as many observations as the first. 2. The combined mean is \(\bar{x}=\frac{25}{100}\bar{x}_1+\frac{75}{100}\bar{x}_2=0.25\bar{x}_1+0.75\bar{x}_2\). 3. Its sampling-distribution mean is \(0.25(60)+0.75(60)=60\). 4. Because the two samples are independent, its variance is \((0.25)^2\frac{10^2}{25}+(0.75)^2\frac{10^2}{75}=1\). 5. Therefore, the sampling-distribution standard deviation is \(1\), the same as for a single random sample of size \(100\).

The correct combined mean is \(0.25\bar{x}_1+0.75\bar{x}_2\). Its sampling distribution has mean \(60\) and standard deviation \(1\).

- First identify the sampling-distribution spread for samples of size \(25\). - Convert the stated lower-tail probability to the supplied standardized location. - Work backward from the standardized sample-mean value to the unknown center.

1. The sampling distribution of \(\bar{x}\) is normal with mean \(\mu\) and standard deviation \(10/\sqrt{25}=2\). 2. The given probability means that \(47\) has standardized value approximately \(-0.842\). 3. Therefore, \(\frac{47-\mu}{2}=-0.842\). 4. Solving gives \(47-\mu=-1.684\), so \(\mu\approx48.684\), which rounds to \(48.68\).

The population mean is approximately \(48.68\).