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 140 use humor, while a random sample of 500 television ads in the United States reveals that 128 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 α 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 H0 at each value of α; 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 α 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 p-value H0 at each value of α = .10 and α = .05; evidence. (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 [ , ] the entire interval is above zero.

Answer 1

(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.

The hypothesis being tested is:

H₀: p₁ = p₂

H_a: p₁ ≠ p₂

(b) Test the hypotheses you set up in part a by using critical values and by setting α 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.)

p1	p2	pc
0.35	0.256	0.2978	p (as decimal)
140/400	128/500	268/900	p (as fraction)
140.	128.	268.	X
400	500	900	n
	0.094	difference
	0.	hypothesized difference
	0.0307	std. error
	3.06	z
	.0022	p-value (two-tailed)

α	z-critical	Decision when z = 3.06
0.001	3.29	Fail to reject Ho
0.01	2.575	Reject Ho
0.05	1.96	Reject Ho
0.1	1.645	Reject Ho

(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 α 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.)

The hypothesis being tested is:

H₀: p₁ - p₂ 0.05

H_a: p₁ - p₂ < 0.05

p1	p2
0.35	0.256	p (as decimal)
140/400	128/500	p (as fraction)
140.	128.	X
400	500	n
	0.094	difference
	0.05	hypothesized difference
	0.0308	std. error
	1.43	z
	.0767	p-value (one-tailed, upper)

α	z-critical	Decision when z = 1.43
0.001	3.29	Fail to reject Ho
0.01	2.575	Fail to reject Ho
0.05	1.96	Fail to reject Ho
0.1	1.645	Reject Ho

(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.)

0.0336	confidence interval 95.% lower
0.1544	confidence interval 95.% upper
0.0604	margin of error