For this problem, find the lower and upper limit. A. Political polls typically s

Question

For this problem, find the lower and upper limit.

A. Political polls typically sample randomly from the U.S population to investigate the percentage of voters who favor some candidate or issue. Suppose that one such poll asks voters how they feel about the President's handling of foreign affairs. The results show that 575 out of the 1280 people polled say they either "approve" or "strongly approve" of the President's handling of foreign affairs. Based on the results of this sample, find a 95% confidence interval estimate for the proportion of the entire voter population who "approve" or "strongly approve" of the President's handling of foreign affairs.

B. The U.S. Department of Transportation requires tire manufacturers to provide tire performance information on the sidewall of the tire so that a prospective customer can be better informed when making a purchasing decision. One very important measure of tire performance is the tread wear index, which indicates the tire's resistance to tread wear compared with a tire graded with a base of 100. This means that a tire with a grade of 200 should last twice as long, on average, as a tire graded with a base of 100. Suppose that a consumer organization wants to estimate the actual tread wear index of a brand name of tires graded 200 that are produced by a certain manufacturer. A random sample of 18 of these tires produced a sample mean tread wear index of 195.3 and a sample standard deviation of 21.4. Assuming that the population of tread wear indices is normally distributed, compute a 95% confidence interval estimate for the population mean tread wear index for tires produced by this manufacturer under this brand name. Place your answer, rounded to 2 decimal places, in the blanks

C. A manufacturer of computer paper has a production process that operates continuously throughout an entire production shift. The paper is expected to have a mean length of 11 inches and the standard deviation of the length is inch. At periodic intervals, a sample is selected to determine whether the mean paper length is still equal to 11 inches or whether something has gone wrong in the production process to change the mean length of the paper produced. You select a random sample of 100 sheets, and the mean paper length is found to be 10.9980 inches. Construct a 95% confidence interval estimate for the population mean paper length.

D. A random sample of 56 credit card holders showed that 41 regularly paid their credit card bills on time. Find a 95% confidence interval for p, the proportion of all credit card holders who pay their credit card bills on time. Place your answer, rounded to 4 decimal places, in the blanks

B. The U.S. Department of Transportation requires tire manufacturers to provide tire performance information on the sidewall of the tire so that a prospective customer can be better informed when making a purchasing decision. One very important measure of tire performance is the tread wear index, which indicates the tire's resistance to tread wear compared with a tire graded with a base of 100. This means that a tire with a grade of 200 should last twice as long, on average, as a tire graded with a base of 100. Suppose that a consumer organization wants to estimate the actual tread wear index of a brand name of tires graded 200 that are produced by a certain manufacturer. A random sample of 18 of these tires produced a sample mean tread wear index of 195.3 and a sample standard deviation of 21.4. Assuming that the population of tread wear indices is normally distributed, compute a 95% confidence interval estimate for the population mean tread wear index for tires produced by this manufacturer under this brand name. Place your answer, rounded to 2 decimal places, in the blanks

Explanation / Answer

Solving question 1. Please repost rest

n = 1280

people in favor,x = 575

proportion in favor , p = 575/1280 =0.4492

Standard error of the mean = SEM = x(N-x)/N3 = 0.014

= (1-CL)/2 = 0.025

Z for 95% confidence interval is = 1.960

Hence, the confidence interval is ( 0.4492- 1.96*0.014, 0.4492+1.96*0.014)=(0.422,0.476)

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