In an article in the Journal of Advertising , Weinberger and Spotts compare the

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 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 (Click to select)= 0 versus Ha: p1 ? p2 (Click to select)= 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           

(Click to select)Do not RejectReject H0 at each value of ?; (Click to select)very strongsomestrongextremely strongnone 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 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

p-value

(Click to select)RejectDo not Reject H0 at each value of ? = .10 and ? = .05; (Click to select)somenonevery strongstrongextremely strong 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                      [ , ]

(Click to select)NoYes the entire interval is above zero.

z

p-value

Explanation / Answer

p1 = 140/400 = 0.35

p2 = 120/500 = 0.24

1 denotes for UK and 2 for US

a. H0 : p1 = p2

H1: p1 not equal to p2

b. Pooled sample proportion p = (p1*n1 + p2*n2)/(n1+n2) = 260/900 = 0.2889

Standard Error in proportions SE = sqrt(p*(1-p)*(1/n1 + 1/n2)) = sqrt(0.2889*0.7111*(1/400+1/500)) = 0.0304

z = (p1 - p2)/SE = (0.35-0.24)/0.0304 = 3.618

p value <0.001

So, Ho is rejected at all significance levels

c. We need to ckeck if p1 - p2 > 0.05

H0 : p1 - p2 = 0.05

H1 : p1 - p2 > 0.05

z = (p1 - p2 - 0.05)/SE = (0.35-0.24-0.05)/0.0304 = 1.97

p value = 0.0242

So, the null hypothesis is rejected for 0.1 and 0.05 significance levels but not rejected for 0.01 and 0.001

d. 95% CI for difference = 0.11 - 0.0304*Z0.975, 0.11 + 0.0304*Z0.975 = 0.11 - 1.96*0.0304, 0.11 + 1.96*0.0304 = 0.0504, 0.1696

0 does not lie in this interval so we can be 95% confident that proportion of UK is more than US

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