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 142 use humor, while a random sample of 500 television ads in the United States reveals that 126 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

For UK :

n1 = 400, x1 = 142

p̂1 = x1/n1 = 0.3550

For US :

n2 = 500, x2 = 126

p̂2 = x2/n2 = 0.2520

a) Null and Alternative hypothesis:

Ho : p1 - p2 = 0

H1 : p1 - p2 ≠ 0

b)

Test statistic:

z = (p̂1 - p̂2)/√ [(p̂1*(1-p̂1)/n1)+(p̂2*(1-p̂2)/n2) ]

= (0.355 - 0.252)/√[(0.355*0.645/400) + (0.252*0.748/500)] = 3.343

Critical value :

z_0.10 = NORM.S.INV(0.1/2) = 1.645

z_0.05 = NORM.S.INV(0.1/2) = 1.960

z_0.01 = NORM.S.INV(0.1/2) = 2.576

z_0.001 = NORM.S.INV(0.1/2) = 3.291

z = 3.343, reject Ho at each value of α.

evidence: strong

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c) Null and Alternative hypothesis:

Ho : p1 - p2 = 0.05

H1 : p1 - p2 > 0.05

Test statistic:

z = [(p̂1 - p̂2) - (p1-p2)]/√ [(p̂1*(1-p̂1)/n1)+(p̂2*(1-p̂2)/n2) ]

= (0.355 - 0.252 - 0.05)/√[(0.355*0.645/400) + (0.252*0.748/500)] = 1.720

p-value :

p-value = 1- NORM.S.DIST(1.7201, 1) = 0.0427

Reject Ho at α = 0.10, 0.05.

d) 95% Confidence interval for the difference:

At α = 0.05, two tailed critical value, z_c = NORM.S.INV(0.05/2) = 1.960

Lower Bound = (p̂1 - p̂2) - z_c*√ [(p̂1*(1-p̂1)/n1)+(p̂2*(1-p̂2)/n2) ]

= (0.355 - 0.252) - 1.96*√[(0.355*0.645/400) + (0.252*0.748/500)] = 0.043

Upper Bound = (p̂1 - p̂2) + z_c*√ [(p̂1*(1-p̂1)/n1)+(p̂2*(1-p̂2)/n2) ]

= (0.355 - 0.252) + 1.96*√[(0.355*0.645/400) + (0.252*0.748/500)] = 0.163

0.043 < p1 -p2 < 0.163