A national study found that 44% of college students engage in binge drinking (5

Question

A national study found that 44% of college students engage in binge drinking (5 drinks at a sitting for men, 4 for women). A professor surveyed a random sample of 200 students at her college.

a. What is the mean of the sampling distribution of the sample proportion?

b. What is the standard deviation of the sampling distribution of the sample proportion?

c. The professor finds that 101 of the surveyed students engage in binge drinking. Is this a surprisingly high value? Explain.

(Show work please!)

Explanation / Answer

a) mean of the sampling distribution of the sample proportion p=0.44

b) here n=200 ; standard deviation of the sampling distribution of the sample proportion =(p*(1-p)/n)1/2

=0.0351

c) probabiltiy of getitng 101 or more (101/200 =0.505 ~ 50.5%) or more surveyed students engage in binge drinking

=P(X>0.505)=1-P(X<0.505)=1-P(Z<(0.505-0.44)/0.0351)=1-P(Z<1.8519)=1-0.9680 =0.0320

as probability of getting that or more extreme value is less then 0.05 ; therefore it is unusual and surprisingly high value.

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