Question

To estimate the mean number of visitors(per/week) to the W.I.U Main Library, a random sample of 12 weeks selected. The sample mean is found to be 2509 visitors with a sample standard deviation of 630 visitors.

(a) Estimate the mean number of visitors per week to the WIU library with a 95% confidence interval.

(b) What is the critical value and degrees of freedom in above confidence interval?

(c) What does this interval mean?

Answer 1

Answer: To estimate the mean number of visitors(per/week) to the W.I.U Main Library, a random sample of 12 weeks selected. The sample mean is found to be 2509 visitors with a sample standard deviation of 630 visitors.

Solution:

Sample mean, = 2509

Sample standard deviation, s = 630

Sample size, n = 12

(a) Estimate the mean number of visitors per week to the WIU library with a 95% confidence interval.

The critical value, t_{​​​​​}​​​​​​_{​​​​​​c} at = 0.05 and degree of freedom

DF = 12 - 1 = 11

t_{​​​​​​c}​​​​​ (0.05,11) = 2.201

CI = t_{​​​​​​c}​​​​​ * S /√n

CI = 2509 2.201 * 630 / √12

CI = 2509 400.2856

CI = (2108.7144, 2909.2856)

Therefore, the mean number of visitors per week to the WIU library with a 95% confidence interval is between 2108.7144 and 2909.2856.

(b) What is the critical value and degrees of freedom in above confidence interval?

Critical value, t_{​​​​​c} (0.05, 11)  = 2.201

Degree of freedom, df

DF = 12 - 1 = 11

(c) What does this interval mean?

This interval mean that we are 95% confident that true mean of visitors per week to the W.I.U Main Library lies between 2108.7144 and 2909.2856.