I. Suppose we want a 93% confidence interval for the average amount spent on boo
ID: 3315492 • Letter: I
Question
I. Suppose we want a 93% confidence interval for the average amount spent on books by freshmen in their first year at a major university. The interval is to have a margin of error of $2, and the amount spent has a normal distribution with a standard deviation = $30. The number of observations required is closest to: A) 28. B) 30 C) 739 D) 865 2. Scores on the SAT Mathematics test(SAT-M) are believed to be normally distributed with mean The scores of a random sample of seven students who recently took the exam are 550, 620, 480, 570, 690, 750, and 500. A 90% confidence interval for based on these data is: A)1521.73, 666.85) B)(532.86, 655.71) C)(S66.86, 621.71) D)(502.91,685.67). 3. Nine airlines are selected at random. For each airline, we record the current fee for checking a single bag. The average for these 9 airlines is x = $25. Assume that the current fee follows a normal distribution with unknown mean and standard deviation =$6. A 90% confidence interval for is: A) $25 ± $6.00 B) $25 ± $3.29 C) $25 ± $9.87 $25 ± $3.92 4 A researcher is interested in the size of the current balance of credit card holders. To estimate this, he obtains the size of the current balance of a random sample of 25 credit card holders, A 90% confidence interval for the mean current balance of credit card holders is found to be S662-72 ± $44.70, which of the following would produce a confidence interval with a smaller margin of error than this 90% confidence interval? A) Obtain the balances of only five credit card holders rather than 25, because five are likely to be more uniform than 25. Obtain the balances of 100 credit card holders rather than 25, Compute a 99% confidence interval rather than a 90% confidence interval. The increase in confidence indicates that we have a better interval None of the above. 8) C) S. You measure the weights of a random sample of 400 male workers in the automotive industry. The sample mean is 176.2 bs. Suppose that the weights of male workers in the automotive industry follow a normal distribution with unknown mean and standard deviation . 11.1 lbs. A 95% confidence interval for is: A) [15444, 19796). D) (175.29, 177.11). 8 (157.94, 194.46) CI 175,11, 177.29) 6. Researchers are studying sales at car dealerships in two locations. The researchers are going to compute independent 90% confidence intervals for the mean sales for the car dealerships at each location. The probability that at least one of the intervals will cover the true mean yields at that location is: A) 081 B)0.19 0.99 D) 0.95Explanation / Answer
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Z/2 at 0.07% LOS is = 1.812 ( From Standard Normal Table )
Standard Deviation ( S.D) = 30
ME =2
n = ( 1.812*30/2) ^2
= (54.36/2 ) ^2
= 738.752 ~ 739
ANSWER:
Option C.