Many high school students take the AP tests in different subject areas. In 2007,

Question

Many high school students take the AP tests in different subject areas. In 2007, of the 211,693 students who took the calculus AB exam 102,598 of them were female and 109,095 of them were male ("AP exam scores," 2013). Estimate the difference in proportion of female students taking the calculus AB exam versus male students taking the calculus AB exam using a 90% confidence level.

(i) Enter the level of significance ? used for this test:

(ii) Determine pˆfemale exam takers and pˆmale exam takers

(iii)  Determine Z score corresponding to desired confidence level

(iv) Determine error bound of the proportion:

(v) Let "p1" refer to proportion of female students taking the calculus AB exam. Let "p2" represent proportion of male students taking the calculus AB exam. Determine the confidence interval of the proportion difference p1 – p2:

(vi) Using the confidence interval, select the correct description of the result of the survey:

A. We estimate with 90% confidence that the true proportion of female students taking the calculus AB exam is anywhere from 2.70% to 3.42% higher than the proportion of male students taking the calculus AB exam.

B. We estimate with 90% confidence that the true proportion of male students taking the calculus AB exam is anywhere from 2.70% to 3.42% lower than the proportion of female students taking the calculus AB exam.

C. We estimate with 90% confidence that the true mean population of female students taking the calculus AB exam is anywhere from 2.70% to 3.42% higher than the true mean population of male students taking the calculus AB exam.

D. We estimate with 90% confidence that the true proportion of female students taking the calculus AB exam is anywhere from 2.70% to 3.42% lower than the proportion of male students taking the calculus AB exam.

Explanation / Answer

p1 = 102598/211693 = 0.484

p2 = 109095/211693 = 0.515

Since the null hypothesis states that P1=P2, we use a pooled sample proportion (p) to compute the standard error of the sampling distribution.
p = (p1 * n1 + p2 * n2) / (n1 + n2)

where p1 is the sample proportion from population 1, p2 is the sample proportion from population 2, n1 is the size of sample 1, and n2 is the size of sample 2.

p = (p1 * n1 + p2 * n2) / (n1 + n2)
= (0.484*211693 + 0.515*211693)/(211693+211693) = 0.499

Compute the standard error (SE) of the sampling distribution difference between two proportions.
SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

where p is the pooled sample proportion, n1 is the size of sample 1, and n2 is the size of sample 2.

SE = sqrt( 0.499 * ( 1 - 0.499 ) * [ (1/211693) + (1/211693) ])
= 0.0015

The test statistic is a z-score (z) defined by the following equation.
z = (0.484 - 0.515) /0.0015

where p1 is the proportion from sample 1, p2 is the proportion from sample 2, and SE is the standard error of the sampling distribution.

z = (0.014-0.018)/0.0023 = -20.6

now check the z table for the p value as
0.0000
as the p value is less than 0.1 , hence we reject the null hypothesis

Please note that we can answer only 4 subparts of a question at a time , as per the answering guidelines

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