## Related questions with answers

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.

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?

Solution

VerifiedThe goal of this task is to compute a confidence interval for $p_1-p_2$ with a $95\%$ confidence interval.

When we have to compute a $100(1-\alpha)\%$ confidence interval for $p_1-p_2$, the formula we have to use is

$\left[ (\^p_1-\^p_2) \pm z_{\alpha/2} \cdot \sqrt{\frac{\^p_1(1-\^p_1)}{n_1}+\frac{\^p_2(1-\^p_2)}{n_2}} \right].$

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