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 the United Kingdom. They found that a substantially greater percentage of U.K. ads use humor. (a) Suppose that a random sample of 380 television ads in the United Kingdom reveals that 140 of these ads use humor. Find a point estimate of and a 95 percent confidence interval for the proportion of all U.K. television ads that use humor. (Round your answers to 3 decimal places.) pˆ = The 95 percent confidence interval is [ , ]. (b) Suppose a random sample of 470 television ads in the United States reveals that 112 of these ads use humor. Find a point estimate of and a 95 percent confidence interval for the proportion of all U.S. television ads that use humor. (Round your answers to 3 decimal places.) pˆ = The 95 percent confidence interval is [ , ]. (c) Do the confidence intervals you computed in parts a and b suggest that a greater percentage of U.K. ads use humor? , the U.K. 95 percent confidence interval is the maximum value in the confidence interval for the U.S.

Answer 1

a) $\widehat p$ = 140/380 = 0.368

At 95% confidence level, the critical value is z_0.025 = 1.96

The 95% confidence interval is

$\widehat p$ +/- z_0.025 * sqrt($\widehat p$(1 - $\widehat p$ )/n)

= 0.368 +/- 1.96 * sqrt(0.368 * (1 - 0.368)/380)

= 0.368 +/- 0.048

= 0.320, 0.416

b) $\widehat p$ = 112/470 = 0.238

The 95% confidence interval is

$\widehat p$ +/- z_0.025 * sqrt($\widehat p$(1 - $\widehat p$ )/n)

= 0.238 +/- 1.96 * sqrt(0.238 * (1 - 0.238)/470)

= 0.238 +/- 0.039

= 0.199, 0.277

c) Since the interval width in part (a) is wider than the interval width in part(b) , so the intervals suggests that a greater percentage of U.K. ads use humor.