A study was conducted to determine if there was a difference in the humor conten

Question

A study was conducted to determine if there was a difference in the humor content in British and American trade magazine advertisements. In an independent random sample of 270 American trade magazine advertisements, 56 were humorous. An independent random sample of 203 British trade magazines contained 52 humorous ads. Do these data provide evidence at the .05 significance level that there is a difference in the proportion of humorous ads in British versus American trade magazines?

State the null and alternate hypotheses. Ho: American (Click to select)=<> British H1: American (Click to select)=<> British

Make the decision rule. (Negative amounts should be indicated by a minus sign. Round your answers to 2 decimal places.) Reject Ho if z is < _____or z >______ . Evaluate the test statistic. (Negative amounts should be indicated by a minus sign. Round your answer to 2 decimal places.) Test statistic

What is the decision regarding the null hypothesis? Ho is rejected or not rejected. The proportions could not or could be the same.

Explanation / Answer

p-hat A = 56/270 = 0.2074 (American)
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p-hat B = 52/203 = 0.2562 (British)
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Ho: pa - pb = 0
Ha: pa - pb # 0
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pbar = (x1 + x2) / (n1 + n2) = (56 + 52) / (270 + 203) = 0.2283qbar = 1 – pbar = 1 – 0.2283 = 0.7717

At = 0.05 for a two-tailed test, zcrit = ±1.96ztest = [ (p1’ – p2’) – (p1 – p2) ] / [(pbar)(qbar)(1/n1 + 1/n2)]

ztest = [ (0.2074 – 0.2562) – 0 ] / [(0.2283)(0.7717)(1/270 + 1/203)]

ztest = -1.2515Since ztest > zcrit, the null hypothesis cannot be rejected.There is not sufficient evidence to support the claim that there is a difference in the proportion of humorous advertisements in British vs American trade magazines.

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