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 387 television ads in the United Kingdom reveals that 131 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.)   

pˆp^ =

The 95 percent confidence interval is [,].

(b) Suppose a random sample of 493 television ads in the United States reveals that 134 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.)

pˆp^ =

The 95 percent confidence interval is [,].

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

(Click to select)YesNo , the U.K. 95 percent confidence interval is (Click to select)not aboveabove the maximum value
in the confidence interval for the U.S.

Answer 1

(A) solution

Given that,

n = 387

x = 131

Point estimate = sample proportion = = x / n = 131 / 387= 0.339

1 - = 1 - 0.339 = 0.661

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.96}

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 (((0.339 - 0.661) /387

E = 0.047

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.339 - 0.047 < p < 0.339 + 0.047

0.292 <P< 0.386

The 95% confidence interval for the population proportion p is :[0.292 , 0.385]

(B) SOLUTION

Given that,

n = 493

x = 134

Point estimate = sample proportion = = x / n = 134 / 493= 0.272

1 - = 1 - 0.272 = 0.728

At 95% confidence level

= 1 - 95%

=1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.96}

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.96 (((0.272 - 0.728) /493)

E = 0.039

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.272 - 0.039 < p < 0.272 + 0.039

0.233 <P< 0.311

The 95% confidence interval for the population proportion p is :[0.233 , 0.311]