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

Question

In an article in the Journal of Advertising, Weinberger and Spotts compare the use of humor in television ads in the United States and in the United Kingdom. Suppose that independent random samples of television ads are taken in the two countries. A random sample of 400 television ads in the United Kingdom reveals that 143 use humor, while a random sample of 500 television ads in the United States reveals that 128 use humor.

Ho: P1-P2 0 0 versus Ha P1-p2 (b) Test the hypotheses you set up in part a by using critical values and by setting equal to .10, .05, .01 and.001. How much evidence is there that the proportions of U.K. and U.S. ads using humor are different? (Round the proportion values to 3 decimal places. Round your answer to 2 decimal places.) 3.2792* Reject $ Ho at each value of a; extremely strong evidence (c) Set up the hypotheses needed to attempt to establish that the difference between the proportions of U.K. and U.S. ads using humor is more than .05 (five percentage points). Test these hypotheses by using a p-value and by setting a equal to .10, .05,.01, and.001. How much evidence is there that the difference between the proportions exceeds.05? (Round the proportion values to 3 decimal places Round your z value to 2 decimal places and p-value to 4 decimal places.) 3.2792 ) 3.2792 ) p-value Reject evidence $ ? Ho at each value of-.10 and = .05; strong (d) Calculate a 95 percent confidence interval for the difference between the proportion of U.K. ads using humor and the proportion of U.S. ads using humor. Interpret this interval. Can we be 95 percent confident that the proportion of U.K. ads using humor is greater than the proportion of U.S. ads using humor? (Round the proportion values to 3 decimal places. Round your answers to 4 decimal places.) 95% of Confidence Interval [ | .101 ) , | .0308 ]

Explanation / Answer

Suppose that independent random samples of television ads are taken in the two countries. A random sample of 400 television ads in the United Kingdom reveals that 143 use humor, while a random sample of 500 television ads in the United States reveals that 128 use humor.

here proportion of ads use humour in UK p^1 = 143/400 = 0.3575

proportion of ads use humor in US p^2 = 128/500 = 0.256

(b) Here pooled estimate p = (143 + 128)/ (400 + 500) = 0.3011

Standard error of proportion = sqrt [p * (1-p) * (1/n1 + 1/n2)] = sqrt [ 0.3011 * 0.6989 * (1/400 + 1/500)]

se0 = 0.0308

Here test statistic

Z = (0.3575 -  0.256)/ 0.03077 = 3.2983

p- value = 2 * Pr(Z > 3.2983) = 0.001

(c) Here

H0 : p1 - p2 <=0.05

Ha : p1 - p2 > 0.05

Z = [(p^1 - p^2) - 0.05]/ se0 = (0.3575 - 0.256 - 0.05)/ 0.030773 = 1.6735

p- value = Pr(Z > 1.6735) = 0.047

We reject H0 at alpha = 0.10 and 0.05 but not at 0.01 and 0.001

(d) 95% confidence interval = (p^1 - p^2) +- Z95% se0

= (0.3575 - 0.256) +- 1.96 * 0.030773

= 0.1015 + 0.06032

= (0.0412, 0.1618)

yes the proportion of UK ads using humor is greater than the proportion of US adults.

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