According to a report on sleep deprivation by the Centers for Disease Control an

Question

According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 8.2%, while this proportion is 8.9% for Oregon residents. These data are based on independently selected simple random samples of 11870 California and 4115 Oregon residents. Calculate a 80% confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived and interpret it in context of the data.

a) Calculate the following (round each to two decimal places):

n1p1 =

n1(1-p1) =

n2p2 =

n2(1-p2) =

c) What is the sample difference in proportions?

d) What is the critical value?

e) What is the standard error of the point estimate?

f) What is the confidence interval?

Explanation / Answer

here p1=0.082

and n1=11870

hence n1p1=973.34

and n1(1-p1)=10896.66

n2=4115

p2=0.089

n2p2=366.235

n2(1-p2)=3748.765

c) sample differnce in proportions =p1-p2 =-0.007

d) critical value z=-/+1.2816

e)std error=(p1(1-p1)/n1+p2(1-p2)/n2)1/2 =0.0051

f) confidence interval =sample diffierence in proportion -/+ z*std error =-0.01354 ; -0.0005

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