Reference no: EM131476850
Not sure what the answer is for the last part of c. The choices are extremely strong, very strong, strong, some, or none. I know its not extremely strong or very strong.
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 140 use humor, while a random sample of 500 television ads in the United States reveals that 120 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.
H0: p1 ? p2 = 0 versus Ha: p1 ? p2 ≠ 0.
(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=3.62
(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.)
Z= 1.97
p-value = .0244
Reject H0 at each value of ? = .10 and ? = .05; ______________ Evidence
So, the null hypothesis is rejected for 0.1 and 0.05 significance levels but not rejected for 0.01 and 0.001