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 20 patients had an average satisfaction score of 7.6 on a scale of​ 1-10. The sample standard deviation was 1.5. 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 (_,_) .

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 7.6

sample standard deviation = s = 1.5

sample size = n = 20

Degrees of freedom = df = n - 1 = 20 - 1 = 19

At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

t/2,df = t0.005,19 = 2.861

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

= 2.861 * (1.5 / 20)

Margin of error = E = 0.96

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

  ± E

= 7.6    ± 0.96

= ( 6.64, 8.56 )