Question

Use the standard normal distribution or the​ t-distribution to construct a​ 95% confidence interval for the population mean. Justify your decision. If neither distribution can be​ used, explain why. Interpret the results. In a random sample of 17 patients at a​ hospital's minor emergency​ department, the mean waiting time before seeing a medical professional was 14 minutes and the standard deviation was 9 minutes. Assume the waiting times are not normally distributed. Which distribution should be used to construct the confidence​ interval? A. Use a​ t-distribution because nless than30 and sigma is known. B. Use a normal distribution because nless than30 and sigma is known. C. Use a normal distribution because nless than30 and sigma is unknown. D. Use a​ t-distribution because nless than30 and sigma is unknown. E. Cannot use the standard normal distribution or the​ t-distribution because sigma is​ unknown, n less than 30​, and the times are not normally distributed. Your answer is correct. Select the correct choice below​ and, if​ necessary, fill in any answer boxes to complete your choice. A. The​ 95% confidence interval is ​( nothing​, nothing​). ​(Round to one decimal place as​ needed.) B. Neither distribution can be used to construct the confidence interval.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 14

sample standard deviation = s = 9

sample size = n = 17

Degrees of freedom = df = n - 1 = 17 - 1 = 16

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,16 = 2.120

D. Use a​ t-distribution because n is less than30 and sigma is unknown .

Margin of error = E = t_/2,df * (s /n)

= 2.120 * (9 / 16)

= 4.6

The 95% confidence interval estimate of the population mean is,

- E < < + E

14 - 4.6 < < 14 + 4.6

9.4 < < 18.6

(9.4 , 18.6)

The​ 95% confidence interval is : (9.4 , 18.6)