In an article in the Journal of Advertising, Weinberger and Spotts compare the use of humor...

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In an article in the Journal of Advertising, Weinberger andSpotts compare the use of humor in television ads in the UnitedStates and in the United Kingdom. Suppose that independent randomsamples of television ads are taken in the two countries. A randomsample of 400 television ads in the United Kingdom reveals that 143use humor, while a random sample of 500 television ads in theUnited States reveals that 126 use humor.

(a) Set up the null and alternative hypotheses needed todetermine whether the proportion of ads using humor in the UnitedKingdom differs from the proportion of ads using humor in theUnited States.

(b) Test the hypotheses you set up in part a by using criticalvalues and by setting α equal to .10, .05, .01, and .001. How muchevidence is there that the proportions of U.K. and U.S. ads usinghumor are different? (Round the proportion values to 3 decimalplaces. Round your answer to 2 decimal places.)

(c) Set up the hypotheses needed to attempt to establish thatthe difference between the proportions of U.K. and U.S. ads usinghumor is more than .05 (five percentage points). Test thesehypotheses by using a p-value and by setting α equal to .10, .05,.01, and .001. How much evidence is there that the differencebetween the proportions exceeds .05? (Round the proportion valuesto 3 decimal places. Round your z value to 2 decimal places andp-value to 4 decimal places.) (d) Calculate a 95 percent confidenceinterval for the difference between the proportion of U.K. adsusing humor and the proportion of U.S. ads using humor. Interpretthis interval. Can we be 95 percent confident that the proportionof U.K. ads using humor is greater than the proportion of U.S. adsusing humor? (Round the proportion values to 3 decimal places.Round your answers to 4 decimal places.)

Answer & Explanation Solved by verified expert
3.6 Ratings (607 Votes)
Given that sample one x1 143 n1 400 p1 x1n10358 sample two x2 126 n2 500 p2 x2n20252 null Ho p1 p2 alternate H1 p1 p2 level of significance 01 from standard normal table two tailed z 2 1645 since our test is twotailed reject Ho if zo 1645 OR if zo 1645 we use test statistic z p1p2pq1n11n2 zo 03580252sqrt0299070114001500 zo 3436 zo 3436 critical value the value of z at los 01 is 1645 we got zo 3436 z 1645 make decision hence value of zo z and here we reject Ho pvalue two tailed double the one tail Ha p 34356 00006 hence value of p01 00006here we reject Ho ANSWERS a null Ho p1 p2 alternate H1 p1 p2 b i level of significance 010 test statistic 3436 critical value 1645 1645 decision reject Ho pvalue 00006 we have enough evidence to support the claim that whether the proportion of ads using humor in the United Kingdom differs    See Answer
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