In an article in theJournal 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 142 use humor, while a random sample of 500 television ads in the United States reveals that 123 use humor.

(a)Set up the null and alternative hypotheses needed to determine whether the proportion of ads using humor in the United Kingdom differs from the proportion of ads using humor in the United States.

(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.)

Z = ______

Reject H0 at each value of a; ________ 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 ap-value and by setting 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 valueto 2 decimal places andp-value to 4 decimal places.)

Z = ____________

P-value = ____________

(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 = [___________ , _____________]