Assume you plan to construct a 95% confidence interval. The numbers of online ap

Question

Assume you plan to construct a 95% confidence interval. The numbers of online applications from simple random samples of college applications for 2003 and for the current year are given below. Find (a) the margin of error E, and (b) the 95% confidence interval. a. The margin of error is b. Let p1 be the proportion of online applications for 2003 and p_2 be the proportion of online applications for the current year. Find the 95% confidence interval estimate of the difference in proportions, p_1 - p_2.

Explanation / Answer

p1 = 13/34

p2 = 16/29

s1 = sqrt(13/34 * 21/34 *1/34) = 0.0833

s2 = sqrt(16/29 * 13/29 * 1/29) = 0.09235

s = sqrt(s1^2 + s2^2)

= 0.124368

margin of error = z-score * s

= 1.96 * 0.124368

= 0.24376

b)

p1 - p2 = -167/986

lower bound = -167/986 - 0.24376 =- 0.4131

upper bound = -167/986 + 0.24376 = 0.0744

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