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Quiz: Ch 9 Quiz: One Sample Confidence Intervals dy This Question: 1 pt tio Subm

ID: 3072937 • Letter: Q

Question

Quiz: Ch 9 Quiz: One Sample Confidence Intervals dy This Question: 1 pt tio Submit Quiz 4 of 10 (4 complete) This Quiz: 10 pts possible A poll conducted in 1972 asked 1005 people, "During the past year, about how many books, either hardcover or paperback, did you read either all or part of the way through?" Results of the survey indicated that x 18.7 books and s 19 7 books a Construct a 95% confidence nterval or the mean number of books read either all or part of duringthe preceding year interpret the interval b) Compare these results to a recent survey of 1005 people The results of the survey indicated that-12 5 books and s : 15 8 books A 95% confidence interval for this survey is (11.52, 13 48) Were people reading more in 1972 than they are today? n s Click the icon to view the table of areas under the t-distribution an (a) The 95% confidence interval for 19726 OD n (Use ascending order Round to two decimal places as needed) th Interpret the confidence interval Which of the following is correct? A. The probability that the interval contains the mean number of books read is. we are 95% confident that the mean number of books read n 1972 s n the interval The probability that the interval does not contain the mean number of books read is 95% 95% O c We are 95% confident that the mean number of books read in 1972 is not in the interval. D ch (b) Were people reading more in 1972 than they are today? O A. No, people were reading less in 1972 than they are reading today O B. Yes, people were reading more in 1972 than they are reading today O C. People were reading approximately the same number of books in 1972 as they are today Click to select your answer(s).

Explanation / Answer

a)

n = 1005

mean = 18.7

z-value of 95% CI = 1.9600

std. dev. = 19.7

SE = std.dev./sqrt(n) = 19.7/sqrt(1005) = 0.62142

ME = z*SE = 1.96*0.62142 = 1.2180

Lower Limit = Mean - ME = 18.7 - 1.2180 = 17.48204

Upper Limit = Mean + ME = 18.7 + 1.2180 = 19.91796

95% CI (17.48 , 19.92 )

Option B

b)

Yes, people were were reading more in 1972 than the are today