Question

A physician with a practice is currently serving 290 patients. The physician would like to administer a survey to his patients to measure their satisfaction level with his practice. A random sample of 21 patients had an average satisfaction score of 7.9 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,.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 7.9

sample standard deviation = s = 1.3

sample size = n = 21

Degrees of freedom = df = n - 1 = 21 - 1 = 20

At 99% confidence level the t is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

t _/2,df = t_0.005,20 = 2.845

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

= 2.845 * (1.3 / 21)

= 0.8

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

- E < < + E

7.9 - 0.8 < < 7.9 + 0.8

7.1 < < 8.7

(7.1 , 8.7)

The​ 99% confidence interval to estimate the average satisfaction score is :
(7.7 , 8.7)