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.

(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.)

(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.)

(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.)

Answer 1

(a)

Let p1 and p2 be the proportion of ads using humor in the United Kingdom and United States respectively.

Null hypothesis H0: p1 = p2

Alternate Hypothesis H1; p1 p2

(b)

Critical values of Z for α equal to .10, .05, .01, and .001 are 1.645, 1.96, 2.576 and 3.29 respectively.

Pooled proportion, p = (140 + 128) / (400 + 500) = 0.298

Standard error of difference in proportions, SE =

=

= 0.0307

p1 = 140 / 400 = 0.35

p2 = 128 / 500 = 0.256

Test statistic, Z = (0.35 - 0.256) / 0.0307

= 3.062

Since the test statistic is greater than 1.645, 1.96, 2.576, we reject the H0 at  α equal to .10, .05, .01 and conclude that there is significant evidence that proportion of ads using humor in the United Kingdom is different than in the United States.

So, proportion of ads using humor in the United Kingdom is different than in the United States at α = 0.01

(c)

Null hypothesis H0: p1 - p2 = 0.05

Alternate Hypothesis H1; p1 - p2 > 0.05

Test statistic, Z = [(0.35 - 0.256) - 0.05] / 0.0307

= 1.433

Since the test statistic is not greater than 1.645, 1.96, 2.576 and 3.29 we fail to reject the H0 at  α equal to .10, .05, .01, 0.001 and conclude that there is no significant evidence that the difference between the proportions exceeds .05.

P-value = P[z > 1.433] = 0.0759

So, difference between the proportions does not exceeds .05 at α = 0.10

(d)

Z value for 95 percent confidence interval = 1.96

Margin of error = z * SE = 1.96 * 0.0307 = 0.0602

95 percent confidence interval is,

((0.35 - 0.256) - 0.0602, (0.35 - 0.256) + 0.0602)

(0.0338, 0.1542)

We're 95% confident that the interval (0.0338, 0.1542) captured the true difference between the proportions of humor ads in UK and US .

Since 95% confidence interval does not contains the value 0, we are  95 percent confident that the proportion of U.K. ads using humor is greater than the proportion of U.S. ads using humor.