A physician with a practice is currently serving 280 patients. The physician wou

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 21 patients had an average satisfaction score of 8.4 on a scale of 1-10. The sample standard deviation was 1.3. Complete parts a and b below. a. Construct a 99 percentage confidence interval to estimate the average satisfaction score for the physician's practice. The 99 percentage confidence interval to estimate the average satisfaction score is (Round to two decimal places as needed.)

Explanation / Answer

Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=8.4
Standard deviation( sd )=1.3
Sample Size(n)=21
Confidence Interval = [ 8.4 ± t a/2 ( 1.3/ Sqrt ( 21) ) ]
= [ 8.4 - 2.845 * (0.284) , 8.4 + 2.845 * (0.284) ]
= [ 7.593,9.207 ]

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