According to a hospital administrator,historical records over the past 10 years have shown that 20 percent of the major surgery patients are dissatisfied with after-surgery care in the hospital.A scientific poll based on 400 hospital patients has just been conducted.Sixty-four patients indicated that they were dissatisfied with the after-surgery care.What are the mean and the standard deviation of the sampling distribution of ?
A)16% and .034%
B)20% and 1.83%
C)20% and 2%
D)20% and .034%
E)20% and 16%

Question

According to a hospital administrator,historical records over the past 10 years have shown that 20 percent of the major surgery patients are dissatisfied with after-surgery care in the hospital.A scientific poll based on 400 hospital patients has just been conducted.Sixty-four patients indicated that they were dissatisfied with the after-surgery care.What are the mean and the standard deviation of the sampling distribution of  ?
A)16% and .034% 
B)20% and 1.83% 
C)20% and 2% 
D)20% and .034% 
E)20% and 16%

Quizplus · Accepted Answer

The population proportion of dissatisfied patients is 20%. The sample size is 400, and the number of dissatisfied patients is 64. The sample proportion of dissatisfied patients is:

p̂ = 64/400 = 0.16

The mean of the sampling distribution of proportions is the same as the population proportion:

μ = 20% = 0.20

The standard deviation of the sampling distribution of proportions is:

σ = √[(p(1-p))/n] = √[(0.20*0.8)/400] ≈ 0.02 = 2%

Therefore, the mean and standard deviation of the sampling distribution of proportions are 20% and 2%, respectively. So, the answer is C.

According to a Hospital Administrator,historical Records Over the Past 10