# Probabilities, Confidence Intervals and Binomial Random Variables

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8.50 Refer to Exercise 8.25, vhere we reported on the results of Harris polls conducted in 1983 and 1992. The 1992 results summarized the responses of n ? 1251 individuals, whereas we assumed that the results for the 1983 poll involved n = 1250 respondents. The data are summarized in the accompanying table for your convenience.

Health Issue 1983 Survey 1992 Survey

Ate recommended amount of fibrous food .59 .53

Avoided fat .55 .51

Avoided excess salt .53 .46

Used scatbelts .19 .70

Used smoke detectors .67 .90

a Construct a 98% confidence interval for the difference in the proportions of Americans who ate the recommended amounts of fibrous foods in 1983 and in 1992.

b Based upon the interval you obtained in (a), do you think that the proportion of Americans who ate (he recommended amount of fibrous foods decreased between 1983 and 1992? Explain.

8.54 Alice Chang and her colleagues (A. Chang, T. L. Rosenthal, R. H. Bryant, R. Fl. Rosenthal. R. M. Heidlage, and B. K. Fritzler. 'Companng High School and College Students' Leisure in- Wrests and Stress Ratings," Behavioral Research Therapy 31(2), 1993) tested 559 high school students using a leisure-interest checklist (LIC) survey. The means and standard deviations for each of the seven LIC factor scales are given in the following table for n = 252 male students and n = 307 female students.

a Give a 95% confidence interval for the difference in the means for male and female students for the cultural activity scale. Interpret this interval.

8.58 Let Y be a binomial random variable with parameter p. Find the sample size necessary to 1

estimate p to within .05 with probability .95 in the following situations:

a if p is thought to be approximately .9.

b if no information about p is known (usc p = .5 in estimating the variance of ).

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Probabilities, Confidence Intervals and Binomial Random Variables are investigated. The solution is detailed and well presented. The response received a rating of "5/5" from the student who originally posted the question.

###### Education

- BSc , Wuhan Univ. China
- MA, Shandong Univ.

###### Recent Feedback

- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
- "excellent work"
- "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
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