Physician with a practice is currently serving 280 patients

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 22 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. 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 (Round to two decimal places as needed.) What assumption needs to be made for this analysis? The population standard deviation is known. The sample size is less than 5% of the population. The population is normally distributed. There are no assumptions needed for this analysis

Solution

A)

Note that              
              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.005          
X = sample mean =    7.6          
t(alpha/2) = critical t for the confidence interval =    2.796939505          
s = sample standard deviation =    1.5          
n = sample size =    25          
df = n - 1 =    24          
Thus,              
              
Lower bound =    6.760918149          
Upper bound =    8.439081851          
              
Thus, the confidence interval is              
              
(   6.760918149   ,   8.439081851   ) [ANSWER]

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B)

OPTION B. The population is normally distributed. [ANSWER, OPTION B]

 Physician with a practice is currently serving 280 patients. The physician would like to administer a survey to his patients to measure their satisfaction leve

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