According to a Scripps Survey Research Center poll, 15% of men and9% of women in

Question

According to a Scripps Survey Research Center poll, 15% of men and9% of women in the United States are left-handed. Suppose thatthese percentages are based on random samples of 1200 men and 1080women. (from [Mann 2007])
a. Construct a 99% confidence interval for the difference betweenthe two population proportions.
b. At the 1% significance level, can you conclude that thepercentage of left-handed men in the US is higher than thepercentage of left-handed women?
c. Repeat the test of part b using a p-value approach.

Explanation / Answer

Men: p1=0.15, n1=1200 Women: p2=0.09, n2=1080 (a) =0.01, Z(0.005)=2.58 (check normal table) The 99% CI is (p1-p2) ± Z*([p1*(1-p1)/n1] + [p2*(1-p2)/n2]) --> (0.15-0.09) ± 2.58*sqrt(0.15*(1-0.15)/1200 +0.09*(1-0.09)/1080) --> (0.025, 0.094) (b) The test hypothesis is Ho:p1p2 The test statistic is Z=(p1-p2) /([p1*(1-p1)/n1] + [p2*(1-p2)/n2]) =(0.15-0.09) /sqrt(0.15*(1-0.15)/1200 + 0.09*(1-0.09)/1080) =4.45 =0.01, Z(0.005)=2.58 (check normal table) Since Z=4.45>2.58, we reject Ho. So you can conclude that thepercentage of left-handed men in the US is higher than thepercentage of left-handed women (c) Since the p-value P(Z>4.45) ˜0 < =0.01, wereject Ho.

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