A local library is evaluating the possibility of expanding its hours of operation.The library staff want to estimate the mean number of persons who patronize the library during the peak weekday hours from 3:30 P.M.to 5:00 P.M.They randomly sample 12 weekdays and find an average attendance of 220.15, with a standard deviation of 22.4.What is a 90% confidence interval for the true mean weekly patronage during the peak hours?
A) (208.627, 231.673)
B) (209.514, 230.787)
C) (179.920, 260.380)
D) (208.537, 231.763)

Question

Quizplus · Accepted Answer

To calculate the confidence interval for the true mean weekly patronage during the peak hours with 90% confidence, we can use the following formula:

CI = x̄ ± t(α/2, n-1) * s/√n

Where x̄ is the sample mean, t(α/2, n-1) is the t-score with (n-1) degrees of freedom and α = 1- 0.90 (since we want a 90% confidence level), s is the sample standard deviation, and n is the sample size.

Plugging in the values given in the problem, we get:

CI = 220.15 ± t(0.05, 11) * 22.4/√12

From the t-distribution table, we can find that t(0.05, 11) = 1.796.

Plugging this value in the above equation, we get:

CI = 220.15 ± 1.796 * 22.4/√12
CI = (208.537, 231.763)

Therefore, the best choice is D) (208.537, 231.763).

A Local Library Is Evaluating the Possibility of Expanding Its