Question

A physician with a practice is currently serving 280 patients. The physician would like to administer a survey to his patients to measure their satisfaction level with his practice. A random sample of 22 patients had an average satisfaction score of 8.3 on a scale of​ 1-10. The sample standard deviation was 1.3 . Complete parts a and b below. a. Construct a​ 99% confidence interval to estimate the average satisfaction score for the​ physician's practice. The​ 99% confidence interval to estimate the average satisfaction score is left parenthesis nothing comma nothing right parenthesis . ​(Round to two decimal places as​ needed.)

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 8.3

sample standard deviation = s = 1.3

sample size = n = 22

Degrees of freedom = df = n - 1 = 22 - 1 = 21

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,21 = 2.831

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

= 2.831 * ( 1.3 / 22)

= 0.78

The 99% confidence interval estimate of the population mean is,

- E < < + E

8.3 - 0.78 < < 8.3 + 0.78

7.52 < < 9.08

(7.52 , 9.08)