Question
In: Statistics and Probability
A physician with a practice is currently serving 300 patients. The physician would like to administer...
A physician with a practice is currently serving 300 patients. The physician would like to administer a survey to his patients to measure their satisfaction level with his practice. A random sample of 24 patients had an average satisfaction score of 8.1 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 (____,____)
Solutions
Expert Solution
Solution :
Given that,
= 8.1
s = 1.5
n = 24
Degrees of freedom = df = n - 1 = 24- 1 = 23
At 99% confidence level the t is ,
= 1 - 99% = 1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
t
/2,df = t0.005,23 =2.807
Margin of error = E = t/2,df
* (s /
= 2.807 * (1.5 /
= 0.86
Margin of error = 0.86
The 99% confidence interval estimate of the population mean is,
- E < 
<
+ E
8.1 - 0.86 <
< 8.1 + 0.86
7.24 <
< 8.96
(7.24, 8.96 )
orchestra answered 2 years agoRelated Solutions
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