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 the United Kingdom. They found that a substantially greater percentage of U.K. ads use humor. (a) Suppose that a random sample of 400 television ads in the United Kingdom reveals that 147 of these ads use humor. Find a point estimate of and a 95 percent confidence interval for the proportion of all U.K. television ads that use humor. (Round your answers to 3 decimal places.) (b) Suppose a random sample of 455 television ads in the United States reveals that 119 of these ads use humor. Find a point estimate of and a 95 percent confidence interval for the proportion of all U.S. television ads that use humor. (Round your answers to 3 decimal places.) (c) Do the confidence intervals you computed in parts a and b suggest that a greater percentage of U.K. ads use humor

Answer 1

Answer:

a)

The point estimate is 0.368.

The 95 percent confidence interval for the proportion of all U.K. television ads that use humor is (0.321, 0.415).

b)

The point estimate is 0.262.

The 95 percent confidence interval for the proportion of all U.S. television ads that use humor is (0.222, 0.302).

c)

Yes, because the lower bound of the confidence interval for UK proportions is higher than the upper bound of the confidence interval for US proportions.

Step-by-step explanation:

(a) Suppose that a random sample of 400 television ads in the United Kingdom reveals that 147 of these ads use humor.

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

Sample of 400:

This means that $n = 400$

Point estimate:

$\pi = \frac{147}{400} = 0.368$

The point estimate is 0.368.

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.368 - 1.96\sqrt{\frac{0.368*0.632}{400}} = 0.321$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.368 + 1.96\sqrt{\frac{0.368*0.632}{400}} = 0.415$

The 95 percent confidence interval for the proportion of all U.K. television ads that use humor is (0.321, 0.415).

(b) Suppose a random sample of 455 television ads in the United States reveals that 119 of these ads use humor.

Sample of 455:

This means that $n = 455$

Point estimate:

$\pi = \frac{119}{455} = 0.262$

The point estimate is 0.262.

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.262 + 1.96\sqrt{\frac{0.262*0.738}{455}} = 0.222$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.262 + 1.96\sqrt{\frac{0.262*0.738}{455}} = 0.302$

The 95 percent confidence interval for the proportion of all U.S. television ads that use humor is (0.222, 0.302).

(c) Do the confidence intervals you computed in parts a and b suggest that a greater percentage of U.K. ads use humor?

Yes, because the lower bound of the confidence interval for UK proportions is higher than the upper bound of the confidence interval for US proportions.