Chapter 7

Find the moment-generating function for a chi square random variable and use it to show that \(E\left(\chi_{n}^{2}\right)=n\) and \(\operatorname{Var}\left(\chi_{n}^{2}\right)=2 n\).

Is it believable that the numbers 65,30 , and 55 are a random sample of size 3 from a normal distribution with \(\mu=50\) and \(\sigma=10\) ? Answer the question by using a chi square distribution. (Hint: Let \(Z_{i}=\left(Y_{i}-50\right) / 10\) and use Theorem 7.3.1.)

Use the fact that \((n-1) S^{2} / \sigma^{2}\) is a chi square random variable with \(n-1\) df to prove that $$ \operatorname{Var}\left(S^{2}\right)=\frac{2 \sigma^{4}}{n-1} $$ (Hint: Use the fact that the variance of a chi square random variable with \(k\) df is \(2 k\).)

If \(Y\) is a chi square random variable with \(n\) degrees of freedom, the pdf of \((Y-n) / \sqrt{2 n}\) converges to \(f_{Z}(z)\) as \(n\) goes to infinity (recall Question 7.3.2). Use the asymptotic normality of \((Y-n) / \sqrt{2 n}\) to approximate the 40 th percentile of a chi square random variable with 200 degrees of freedom.

Let \(V\) and \(U\) be independent chi square random variables with 7 and 9 degrees of freedom, respectively. Is it more likely that \(\frac{V / 7}{U / 9}\) will be between (1) \(2.51\) and \(3.29\) or (2) \(3.29\) and 4.20?

If the random variable \(F\) has an \(F\) distribution with \(m\) and \(n\) degrees of freedom, show that \(1 / F\) has an \(F\) distribution with \(n\) and \(m\) degrees of freedom.

Show that as \(n \rightarrow \infty\), the pdf of a Student \(t\) random variable with \(n\) df converges to \(f_{Z}(z)\). (Hint: To show that the constant term in the pdf for \(T_{n}\) converges to \(1 / \sqrt{2 \pi}\), use Stirling's formula, $$ \left.n ! \doteq \sqrt{2 \pi n} n^{n} e^{-n}\right) $$ Also, recall that \(\lim _{n \rightarrow \infty}\left(1+\frac{a}{n}\right)^{n}=e^{a}\).

Evaluate the integral $$ \int_{0}^{\infty} \frac{1}{1+x^{2}} d x $$ using the Student \(t\) distribution.

For a Student \(t\) random variable \(T\) with \(n\) degrees of freedom and any positive integer \(k\), show that \(E\left(T^{2 k}\right)\) exists if \(2 k0, \beta>0\), and \(\alpha \beta>1\).)

Suppose a random sample of size \(n=11\) is drawn from a normal distribution with \(\mu=15.0\). For what value of \(k\) is the following true? $$ P\left(\left|\frac{\bar{Y}-15.0}{S / \sqrt{11}}\right| \geq k\right)=0.05 $$

Let \(\bar{Y}\) and \(S\) denote the sample mean and sample standard deviation, respectively, based on a set of \(n=\) 20 measurements taken from a normal distribution with \(\mu=90.6\). Find the function \(k(S)\) for which $$ P[90.6-k(S) \leq \bar{Y} \leq 90.6+k(S)]=0.99 $$

Cell phones emit radio frequency energy that is absorbed by the body when the phone is next to the ear and may be harmful. The table in the next column gives the absorption rate for a sample of twenty high-radiation cell phones. (The Federal Communication Commission sets a maximum of \(1.6\) watts per kilogram for the absorption rate of such energy.) Construct a \(90 \%\) confidence interval for the true average cell phone absorption rate. $$ \begin{array}{ll} \hline 1.54 & 1.41 \\ 1.54 & 1.40 \\ 1.49 & 1.40 \\ 1.49 & 1.39 \\ 1.48 & 1.39 \\ 1.45 & 1.39 \\ 1.44 & 1.38 \\ 1.42 & 1.38 \\ 1.41 & 1.37 \\ 1.41 & 1.33 \\ \hline \end{array} $$

The following table lists the typical cost of repairing the bumper of a moderately priced midsize car damaged by a corner collision at \(3 \mathrm{mph}\). Use these observations to construct a \(95 \%\) confidence interval for \(\mu\), the true average repair cost for all such automobiles with similar damage. The sample standard deviation for these data is $$s=\$ 369.02$$. $$ \begin{array}{llll} \hline & \text { Repair } & & \text { Repair } \\ \text { Make/Model } & \text { Cost } & \text { Make/Model } & \text { Cost } \\\ \hline \text { Hyundai Sonata } & \$ 1019 & \text { Honda Accord } & \$ 1461 \\\ \text { Nissan Altima } & \$ 1090 & \text { Volkswagen Jetta } & \$ 1525 \\ \text { Mitsubishi Galant } & \$ 1109 & \text { Toyota Camry } & \$ 1670 \\ \text { Saturn AURA } & \$ 1235 & \text { Chevrolet Malibu } & \$ 1685 \\ \text { Subaru Legacy } & \$ 1275 & \text { Volkswagen Passat } & \$ 1783 \\ \text { Pontiac G6 } & \$ 1361 & \text { Nissan Maxima } & \$ 1787 \\ \text { Mazda 6 } & \$ 1437 & \text { Ford Fusion } & \$ 1889 \\ \text { Volvo S40 } & \$ 1446 & \text { Chrysler Sebring } & \$ 2484 \\ \hline \end{array} $$

How long does it take to fly from Atlanta to New York's LaGuardia airport? There are many components of the time elapsed, but one of the more stable measurements is the actual in-air time. For a sample of eighty-three flights between these destinations on Fridays in October, the time in minutes \((y)\) gave the following results: $$ \sum_{i=1}^{83} y_{i}=8622 \quad \text { and } \quad \sum_{i=1}^{83} y_{i}^{2}=899,750 $$ Find a \(99 \%\) confidence interval for the average flight time.

If a normally distributed sample of size \(n=16\) produces a \(95 \%\) confidence interval for \(\mu\) that ranges from \(44.7\) to \(49.9\), what are the values of \(\bar{y}\) and \(s\) ?

Two samples, each of size \(n\), are taken from a normal distribution with unknown mean \(\mu\) and unknown standard deviation \(\sigma .\) A \(90 \%\) confidence interval for \(\mu\) is constructed with the first sample, and a \(95 \%\) confidence interval for \(\mu\) is constructed with the second. Will the \(95 \%\) confidence interval necessarily be longer than the \(90 \%\) confidence interval? Explain.

Revenues reported last week from nine boutiques franchised by an international clothier averaged $$\$ 59,540$$ with a standard deviation of $$\$ 6860$$. Based on those figures, in what range might the company expect to find the average revenue of all of its boutiques?

What "confidence" is associated with each of the following random intervals? Assume that the \(Y_{i}\) 's are normally distributed. (a) \(\left[\bar{Y}-2.0930\left(\frac{S}{\sqrt{20}}\right), \bar{Y}+2.0930\left(\frac{S}{\sqrt{20}}\right)\right]\) (b) \(\left[\bar{Y}-1.345\left(\frac{S}{\sqrt{15}}\right), \bar{Y}+1.345\left(\frac{S}{\sqrt{15}}\right)\right]\) (c) \(\left[\bar{Y}-1.7056\left(\frac{S}{\sqrt{27}}\right), \bar{Y}+2.7787\left(\frac{S}{\sqrt{27}}\right)\right]\) (d) \(\left[-\infty, \bar{Y}+1.7247\left(\frac{S}{\sqrt{21}}\right)\right]\)

Recall Case Study 5.3.1. Assess the credibility of the theory that Etruscans were native Italians by testing an appropriate \(H_{0}\) against a two-sided \(H_{1}\). Set \(\alpha\) equal to \(0.05\). Use \(143.8 \mathrm{~mm}\) and \(6.0 \mathrm{~mm}\) for \(\bar{y}\) and \(s\), respectively, and let \(\mu_{0}=132.4\). Do these data appear to satisfy the distribution assumption made by the \(t\) test? Explain.

MBAs \(R\) Us advertises that its program increases a person's score on the GMAT by an average of forty points. As a way of checking the validity of that claim, a consumer watchdog group hired fifteen students to take both the review course and the GMAT. Prior to starting the course, the fifteen students were given a diagnostic test that predicted how well they would do on the GMAT in the absence of any special training. The following table gives each student's actual GMAT score minus his or her predicted score. Set up and carry out an appropriate hypothesis test. Use the \(0.05\) level of significance. $$ \begin{array}{lcc} \hline \text { Subject } & y_{i}=\text { act. GMAT }-\text { pre. GMAT } & y_{i}^{2} \\ \hline \text { SA } & 35 & 1225 \\ \text { LG } & 37 & 1369 \\ \text { SH } & 33 & 1089 \\ \text { KN } & 34 & 1156 \\ \text { DF } & 38 & 1444 \\ \text { SH } & 40 & 1600 \\ \text { ML } & 35 & 1225 \\ \text { JG } & 36 & 1296 \\ \text { KH } & 38 & 1444 \\ \text { HS } & 33 & 1089 \\ \text { LL } & 28 & 784 \\ \text { CE } & 34 & 1156 \\ \text { KK } & 47 & 2209 \\ \text { CW } & 42 & 1764 \\ \text { DP } & 46 & 2116 \\ \hline \end{array} $$

A manufacturer of pipe for laying underground electrical cables is concerned about the pipe's rate of corrosion and whether a special coating may retard that rate. As a way of measuring corrosion, the manufacturer examines a short length of pipe and records the depth of the maximum pit. The manufacturer's tests have shown that in a year's time in the particular kind of soil the manufacturer must deal with, the average depth of the maximum pit in a foot of pipe is \(0.0042\) inch. To see whether that average can be reduced, ten pipes are coated with a new plastic and buried in the same soil. After one year, the following maximum pit depths are recorded (in inches): \(0.0039,0.0041,0.0038,0.0044,0.0040,0.0036\), \(0.0034,0.0036,0.0046\), and \(0.0036\). Given that the sample standard deviation for these ten measurements is \(0.000383\) inch, can it be concluded at the \(\alpha=0.05\) level of significance that the plastic coating is beneficial?

In athletic contests, a wide-spread conviction exists that the home team has an advantage. However, one explanation for this is that a team schedules some much weaker opponents to play at home. To avoid this bias, a study of college football games considered only games between teams ranked in the top twenty- five. For three hundred seventeen such games, the margin of victory \((y=\) home team score-visitors score) was recorded. For these data, \(\bar{y}=4.57\) and \(s=18.29\). Does this study confirm the existence of a home field advantage? Test \(H_{0}: \mu=0\) versus \(H_{1}: \mu>0\) at the \(0.05\) level of significance.

The first analysis done in Example 7.4.2 (using all \(n=12\) banks with \(\bar{y}=58.667\) ) failed to reject \(H_{0}: \mu=62\) at the \(\alpha=0.05\) level. Had \(\mu_{0}\) been, say, \(61.7\) or \(58.6\), the same conclusion would have been reached. What do we call the entire set of \(\mu_{0}\) 's for which \(H_{0}: \mu=\mu_{0}\) would not be rejected at the \(\alpha=0.05\) level?

Explain why the distribution of \(t\) ratios calculated from small samples drawn from the exponential pdf, \(f_{Y}(y)=e^{-y}, y \geq 0\), will be skewed to the left [recall Figure 7.4.6(b)]. (Hint: What does the shape of \(f_{Y}(y)\) imply about the possibility of each \(y_{i}\) being close to 0 ? If the entire sample did consist of \(y_{i}\) 's close to 0 , what value would the \(t\) ratio have?)

Suppose one hundred samples of size \(n=3\) are taken from each of the pdfs (1) \(f_{Y}(y)=2 y, \quad 0 \leq y \leq 1\) and (2) \(f_{Y}(y)=4 y^{3}, \quad 0 \leq y \leq 1\) and for each set of three observations, the ratio $$ \frac{\bar{y}-\mu}{s / \sqrt{3}} $$ is calculated, where \(\mu\) is the expected value of the particular pdf being sampled. How would you expect the distributions of the two sets of ratios to be different? How would they be similar? Be as specific as possible.

Suppose that random samples of size \(n\) are drawn from the uniform pdf, \(f_{Y}(y)=1,0 \leq y \leq 1\). For each sample, the ratio \(t=\frac{\bar{y}-0.5}{s / \sqrt{n}}\) is calculated. Parts (b) and (d) of Figure 7.4.6 suggest that the pdf of \(t\) will become increasingly similar to \(f_{T_{n-1}}(t)\) as \(n\) increases. To which pdf is \(f_{T_{n-1}}(t)\), itself, converging as \(n\) increases?

Evaluate the following probabilities: (a) \(P\left(\chi_{17}^{2} \geq 8.672\right)\) (b) \(P\left(x_{6}^{2}<10.645\right)\) (c) \(P\left(9.591 \leq \chi_{20}^{2} \leq 34.170\right)\) (d) \(P\left(\chi_{2}^{2}<9.210\right)\)

Find the value \(y\) that satisfies each of the following equations: (a) \(P\left(\chi_{9}^{2} \geq y\right)=0.99\) (b) \(P\left(\chi_{15}^{2} \leq y\right)=0.05\) (c) \(P\left(9.542 \leq x_{22}^{2} \leq y\right)=0.09\) (d) \(P\left(y \leq \chi_{31}^{2} \leq 48.232\right)=0.95\)

For what value of \(n\) is each of the following statements true? (a) \(P\left(\chi_{n}^{2} \geq 5.009\right)=0.975\) (b) \(P\left(27.204 \leq \chi_{n}^{2} \leq 30.144\right)=0.05\) (c) \(P\left(\chi_{n}^{2} \leq 19.281\right)=0.05\) (d) \(P\left(10.085 \leq x_{n}^{2} \leq 24.769\right)=0.80\)

For df values beyond the range of Appendix Table A.3, chi square cutoffs can be approximated by using a formula based on cutoffs from the standard normal pdf, \(f_{Z}(z) .\) Define \(\chi_{p, n}^{2}\) and \(z_{p}^{*}\) so that \(P\left(\chi_{n}^{2} \leq \chi_{p, n}^{2}\right)=p\) and \(P\left(Z \leq z_{p}^{*}\right)=p\), respectively. Then $$ \chi_{p, n}^{2} \doteq n\left(1-\frac{2}{9 n}+z_{p}^{*} \sqrt{\frac{2}{9 n}}\right)^{3} $$ Approximate the 95 th percentile of the chi square distribution with 200 df. That is, find the value of \(y\) for which $$ P\left(\chi_{200}^{2} \leq y\right) \doteq 0.95 $$

A random sample of size \(n=19\) is drawn from a normal distribution for which \(\sigma^{2}=12.0\). In what range are we likely to find the sample variance, \(s^{2}\) ? Answer the question by finding two numbers \(a\) and \(b\) such that $$ P\left(a \leq S^{2} \leq b\right)=0.95 $$

In Case Study 7.5.1, the \(95 \%\) confidence interval was constructed for \(\sigma\) rather than for \(\sigma^{2}\). In practice, is an experimenter more likely to focus on the standard deviation or on the variance, or do you think that both formulas in Theorem \(7.5 .1\) are likely to be used equally often? Explain.

If a \(90 \%\) confidence interval for \(\sigma^{2}\) is reported to be \((51.47,261.90)\), what is the value of the sample standard deviation?

Let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) be a random sample of size \(n\) from the pdf $$ f_{Y}(y)=\left(\frac{1}{\theta}\right) e^{-y / \theta}, \quad y>0 ; \quad \theta>0 $$ (a) Use moment-generating functions to show that the ratio \(2 n \bar{Y} / \theta\) has a chi square distribution with \(2 n\) df. (b) Use the result in part (a) to derive a \(100(1-\alpha) \%\) confidence interval for \(\theta\).

When working properly, the amounts of cement that a filling machine puts into \(25-\mathrm{kg}\) bags have a standard deviation \((\sigma)\) of \(1.0 \mathrm{~kg}\). In the next column are the weights recorded for thirty bags selected at random from a day's production. Test \(H_{0}: \sigma^{2}=1\) versus \(H_{1}: \sigma^{2}>1\) using the \(\alpha=0.05\) level of significance. Assume that the weights are normally distributed. Use the following sums: $$ \sum_{i=1}^{30} y_{i}=758.62 \text { and } \sum_{i=1}^{30} y_{i}^{2}=19,195.7938 $$

A stock analyst claims to have devised a mathematical technique for selecting high-quality mutual funds and promises that a client's portfolio will have higher average ten-year annualized returns and lower volatility; that is, a smaller standard deviation. After ten years, one of the analyst's twenty-four- stock portfolios showed an average ten-year annualized return of \(11.50 \%\) and a standard deviation of \(10.17 \%\). The benchmarks for the type of funds considered are a mean of \(10.10 \%\) and a standard deviation of \(15.67 \%\). (a) Let \(\mu\) be the mean for a twenty-four-stock portfolio selected by the analyst's method. Test at the \(0.05\) level that the portfolio beat the benchmark; that is, test \(H_{0}: \mu=10.1\) versus \(H_{1}: \mu>10.1\). (b) Let \(\sigma\) be the standard deviation for a twenty-fourstock portfolio selected by the analyst's method. Test at the \(0.05\) level that the portfolio beat the benchmark; that is, test \(H_{0}: \sigma=15.67\) versus \(H_{1}: \sigma<15.67\).