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

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 122 use humor. Compare these two populations using the given information and by answering the following questions. Be 90% confident and use the pooled estimate of the proportion when using Minitab/Minitab Express to calculate the answer.

1) What is the sample proportions for the UK and US are?

2) What is the 90% confidence interval is (round to 3 decimals)?

3) What does the confidence interval mean?

4) The p-value supporting this analysis is (round to 3 decimals)?

5) What conclusion can you draw from the p-value?

6) What conclusion can you draw from the p-value and confidence interval?

Explanation / Answer

1) What is the sample proportions for the UK and US are?

P UK = 142/400 = 0.355

P US = 122 / 500 = 0.244

2) What is the 90% confidence interval is (round to 3 decimals)?

alpha / 2 = 0.05   Z=1.64

P = ( 142+122) / 900 = 0.293

I: ( 0.355 - 0.244) +/- 1.64 * SRQT ( 0.293*0.707* 1/400 + 1/500 )

0.111 +/- 0.05

0.061 < p < 0.161

3) What does the confidence interval mean?

we are 90% confident that the true difference of the population proportion is between 0.061 and 0.161

4) The p-value supporting this analysis is (round to 3 decimals)?

p value : 0.05

5) What conclusion can you draw from the p-value?

since p value is small , there is a difference between the proportion

6) What conclusion can you draw from the p-value and confidence interval?

Since zero is not included in the interval that means that P UK is greater tha P US

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