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MME 524 F' 99 Test 2

(20 points) A study directed by Dr. Michael Gaziano, a heart specialist at the Harvard Medical School-affiliated Brigham and Women's Hospital in Boston, suggests that tea may reduce the chance of a heart attack by 44 percent. Gaziano's study examined 340 men and women who had suffered heart attacks and matched them by age, sex and neighborhood with people who had never had a heart attack. It then investigated their coffee- and tea-drinking habits over the course of a year. Gaziano thinks that the reason for the reduced incidence of heart attacks among tea drinkers is that tea contains powerful amounts of flavonoids, natural substances that make blood cells less prone to clots, which can cause heart attacks. The study was adjusted for factors that could have skewed the results, such as smoking, exercise, alcohol intake and family history of heart trouble. Total calories consumed, intake of fatty foods and body mass index which compares the girth of people of different heights to determine obesity was about the same across the board.
What kind of a study was this? Be as precise as you can, and justify your answer.

ANS: A retrospective observational study, since subjects are classified into groups after the fact, and causal factors sought. (10 points)

Based on this study, is it valid to conclude that drinking tea causes a reduction in the chance of a heart attack? Why or why not?

ANS: No, because, even though the researchers tried to adjust for extraneous factors, this is not a controlled experiment. (10 points)

(10 points) Figure 1 displays four frequency histograms, each constructed from a different data set of 200 observations. Each has mean 8.42. Which histogram, that graphing Y1 (upper left), Y2 (lower left), Y3 (upper right) or Y4 (lower right), has quartiles Q1=8.09, Q2=9.00, Q3=9.44? Justify your answer.

Figure 1: Four frequency histograms for problem 2.
\psfig {file=/math/mathlab/sas/sasmath/ma2611/e99/t2q2.eps,

ANS: Y3, because of the left skewness. (10 points)

(30 points) The proper operation of a nuclear power plant depends on the reactor feedwater system. If the concentrations of metals, such as iron or copper, are too high, reactor performance can be adversely affected. In order to monitor metal concentrations, measurements are taken each day. A recent set of copper concentration measurements, one taken on each of 15 successive days, had a mean of 0.2631 ppb and a standard deviation of 0.0129 ppb. It is desired to perform statistical inference for the process that produced these measurements.

If it is desired to estimate the process mean, what assumptions about the process should be checked?

ANS: Stationarity and normality. (5 points)

Assume you have checked the assumptions in (a), and there is no evidence that the assumptions do not hold.

Obtain a point estimate for the process mean copper concentration.

ANS: $\overline{y}=0.2631$. (5 points)

Obtain a 95% confidence interval for the mean. Tell what ``95% confidence'' means here.

ANS: $0.2631\pm \frac{0.0129}{\sqrt{15}}t{14,0.975}=
0.2631\pm \frac{0.0129}{\sqrt{15}}2.1448=(0.2560,0.2702)$. (10 points) 95% confidence means that in repeated sampling from the same population, approximately 95% of all intervals constructed in this way will contain the true population mean. (5 points)

Obtain an interval you are 95% confident will contain tomorrow's measurement.

ANS: A 95% prediction interval is $0.2631\pm 0.0129\left(\sqrt{1+\frac{1}{15}}\right)2.1448=(0.2345,0.2916)$.(5 points)

Management is interested in an interval that it can be 95% confident will contain at least 99% of the population of all possible measurements. Construct such an interval.

ANS: A 95% tolerance interval for a proportion 0.99 of the population values is $0.2631\pm (0.0129)(3.878)=(0.2131,0.3131)$. (5 points)

(10 points) A survey is to be done to estimate the proportion of a population with a certain characteristic. How large a sample is needed to be 99% confident that the estimate will be within 0.02 of the true proportion?

ANS: n=(0.5)(0.5)(2.5758)2/(0.02)2=4147. (10 points)

(15 points) The new game ``Textbook Toss'' is sweeping college campuses. To play the game, the player first selects a page of his/her textbook at random and then flips a fair coin. If the coin lands heads, the player wins the number (in cents) of the selected page (e.g., if page 348 is selected, the player wins $3.48). If the coin lands tails the player loses the number of the selected page (e.g., if page 348 is selected, the player loses $3.48).

Suppose the book used in the game has 944 pages, and let Y denote the amount won by the player on one play of ``Textbook Toss''. What is the distribution of Y?

ANS: Y has probability mass function $p_Y(y)= \frac{1}{1888}, y=-944, -943, \ldots, -2, -1, 1, 2, \ldots,
943, 944.$ (10 points)

What is the expected value of Y? Support your answer.

ANS: 0, by the symmetry of the distribution about 0. (5 points)

(15 points) Historically, a proportion 0.001 of monthly bills sent to residential customers by a utility company contains at least one error.

A random sample of 15000 residential bills from the most recent six-month billing period is taken. If the billing error rate conforms to the historical proportion, and if presence of errors in different bills are independent, what kind of distribution model describes the number of the 15000 bills with errors?

ANS: b(15000,0.001) (5 points)

Of the 15000 bills in the random sample, 25 have at least one error. Is this convincing evidence that the proportion of bills with errors has increased from 0.01? Use a large-sample approximation to help you decide.

ANS: Suppose $Y\sim b(15000,0.001)$. Then $P(Y\geq
25)=P(Y\geq 24.5)= P(\frac{Y-15}{\sqrt{15000(0.001)(0.999)}}\geq
\frac{24.5-15}{\sqrt{15000(0.001)(0.999)}})\approx P(Z\geq
2.45)=0.0071.$ This seems convincing evidence that the proportion of bills with errors has increased from 0.01. (10 points)

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