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Material Type: Assignment; Class: STATISTICAL INFERENCE; Subject: Statistics; University: University of Pennsylvania; Term: Unknown 1989;
Typology: Assignments
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Homework Assignments (#1) ( Due 9/16 )
Chapter 7: #1, 3, 7, 20, 23, 25, 30, 35, 36, and
A. At the beginning of Bill Clinton’s presidential administration he spoke on proposed economic reforms. A sample of people who heard the speech (n = 611) were asked if they favored higher taxes on all forms of energy; 43% responded “Yes”. ( Time , March 1, 1993, p. 26.) (a) Based on this sample, give a 90% confidence interval for the population proportion of “Yes” answers. (b) What is the “population” here? (c) Newspaper and newsmagazine surveys often report that with 95% confidence the margin of error of the survey is at most ±3%. Did the Time survey above meet these specifications? If not, how many respondents would Time have needed to survey in order to do so?
B. A gasoline company tested 20 one-gallon samples of gasoline produced during a day. The results were as follows: 87.5, 86.9, 86.6, 87.3, 87.9, 88.0, 86.7, 87.5, 87.2, 87.0, 88.1, 87.5, 86.5, 87.7, 88.0, 87.1, 87.0, 87.6, 87.5, 88.3. (a) Give a 95% confidence interval for the mean octane rating of the day’s production. Assume a normal population. [You may use JMPin or a hand calculator to perform the numerical computations needed here and in (b), below.] (b) Construct a normal probability plot for this data to investigate whether the assumption of normality used in part (a) seems plausible. What do you conclude? (Note: This is a relatively small sample – of size n = 20 – so any conclusion concerning normality of the population must necessarily be a very tentative one.)
C. Suppose the gasoline company in problem B performed similar tests on each of 100 days, and thus produced 100 confidence intervals – one for each of the 100 days. Let Y denote the number of intervals out of the 100 that contain the true mean octane rating
of that day’s production. What is the distribution of Y? What is the probability that Y ≥
90? Y ≥ 95? Y ≥ 99? [Partial answer: P(Y ≥ 90) = 0.972; a normal approximation that
gives a good approximation to this answer would also provide an acceptable solution to this question.]
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