A study was designed to compare two energy drink commercials. Each participant w

Question

A study was designed to compare two energy drink commercials. Each participant was shown the commercials in random order and asked to select the better one. Commercial A was selected by 45 out of 100 women and 80 out of 140 men. Is there evidence that commercial preferences different between genders? Be are sure to check the necessary conditions. If it is desirable to estimate the proportion of women who prefer Commercial B with plusminus 0.08 precision and 96% confidence, what sample size would you recommend?

Explanation / Answer

(a)

Data:    

n1 = 100   

n2 = 140   

p1 = 0.45   

p2 = 0.571428571   

Hypotheses:    

Ho: p1 = p2    

Ha: p1 p2    

Decision Rule:    

= 0.05   

Lower Critical z- score =   -1.959963985

Upper Critical z- score = 1.959963985

Reject Ho if |z| >   1.959963985

Test Statistic:    

Average proportion, p = (n1p1 + n2p2)/(n1 + n2) = (100 * 0.45 + 140 * 0.571428571428571)/(100 + 140) = 0.520833333

q = 1 - p = 1 - 0.520833333333333 = 0.479166667

SE = [pq * {(1/n1) + (1/n2)}] = (0.520833333333333 * 0.479166666666667 * ((1/100) + (1/140))) = 0.065408515

z = (p1 - p2)/SE = (0.45 - 0.571428571428571)/0.0654085148090355 = -1.856464281

p- value = 0.06338741   

Decision (in terms of the hypotheses):    

Since 1.856464281 < 1.959963985 we fail to reject Ho

Conclusion (in terms of the problem):    

There is no sufficient evidence that p1 p2

(b) z- score for 96% confidence is z = 2.0537

E = 0.08, p = 0.55, q = 0.45

N = (z/E)^2 * pq

N = (2.0537/0.08)^2 * 0.55 * 0.45 = 164.

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