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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 25 patients had an average satisfaction score of 8.1 on a scale of 1-10. The sample standard deviation was 1.21
a. Construct a 99% confidence interval to estimate the average satisfaction score for the physician's practice.( round to two decimals)
ps-EVERYONE keeps getting this wrong and its really frustrating because I cant get it either :/
Solution :
Given that,
Point estimate = sample mean =
= 8.1
sample standard deviation = s = 1.21
sample size = n = 25
Degrees of freedom = df = n - 1 = 24
At 99% confidence level the t is ,
t
/2,df = t0.005,24= 2.797
Margin of error = E = t/2,df
* (s /
n)
= 2.797 * ( 1.21/
25)
= 0.68
The 99% confidence interval estimate of the population mean is,
- E <
<
+ E
8.1 - 0.68 <
< 8.1 + 0.68
7.42 <
< 8.78
( 7.91 , 8.29 )
Question Help A physician with a practice is currently serving 280 patients. The physician would like...