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 145 use humor, while a random sample of 500 television ads in the United States reveals that 124 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 = 145, p̂1 = x1/n1 = 0.363

For US:

n2 = 500, x2 = 124, p̂2 = x2/n2 = 0.248

a) Null and Alternative hypothesis:

Ho : p1 - p2 = 0

H1 : p1 - p2 ≠ 0

b) Critical value :

z(0.10) = ABS(NORM.S.INV((0.10/2) = 1.645

z(0.05) = ABS(NORM.S.INV((0.05/2) = 1.96

z(0.01) = ABS(NORM.S.INV((0.01/2) = 2.576

z(0.001) = ABS(NORM.S.INV((0.001/2) = 3.291

Pooled proportion:

p̄ = (x1+x2)/(n1+n2) = (145+124)/(400+500) = 0.2989

Test statistic:

z = (p̂1 - p̂2)/√ [p̄*(1-p̄)*(1/n1+1/n2)] = (0.363 - 0.248)/√[0.2989*0.7011*(1/400+1/500)] = 3.750

Reject Ho. at each value of α; very strong evidence.

(c) Null and Alternative hypothesis:

Ho : p1 - p2 = 0.05

H1 : p1 - p2 > 0.05

Test statistic:

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

= (0.363 - 0.248-0.05)/√[0.2989*0.7011*(1/400+1/500)] = 2.12

p-value = 2*(1-NORM.S.DIST(ABS(2.1167), 1)) = 0.0343

Reject H0 at each value of α = .10 and α = .05; evidence.

(d) 95% confidence interval for the difference between the proportion of U.K. ads using humor and the proportion of U.S. ads using humor:

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) ]

= (0.363 - 0.248) - 1.96*√[(0.363*0.637/400) + (0.248*0.752/500)] = 0.0546

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

= (0.363 - 0.248) + 1.96*√[(0.363*0.637/400) + (0.248*0.752/500)] = 0.1754

0.0546 < p1 -p2 < 0.1754