Question

A recent poll determined that 1/4 of the people support a particular candidate. Following a major advertising campaign, a new survey showed that 57 out of 192 people surveyed supported the candidate. Do the data indicate a significant change in attitude? Test with alpha set at .05. For full points indicate in the box below, 1) null hypothesis, 2)standard deviation or standard error, 3) z, 4) fail or fail to reject null hypothesis, and 5) interpretation or conclusion.

Answer 1

Let p be the probability that people support a particular candidate. Here p = 1/4 = 0.25, Now the Variance = p*(1-p) = 0.25*(1-0.25) = 0.1875. So standard deviation sd = sqrt(variance) = sqrt(0.1875) = 0.433.

Sample size = 192, out of which 57 supported the candidate, thus p0 = 57/192 = 0.297.

1) Null Hypothesis H0: p0 = p

2) Here we already calculated the population standard deviation which is 0.433. Thus, do not calculate standard error.

3) Z-value = p0-p/Standard Deviation = (0.297-0.25)/0.433 = 0.109

4) We fail to reject the null hypothesis as Z-value is smaller than Z-critical at 0.05 alpha which is 1.96.

5) Thus, the advertisement campaign did a good job to confirm that p is 0.25.

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