Question

(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 [ , ].

Answer 1

Solution :

Given that,

(a)

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

Z_/2 = 1.96

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

= 1.96 * (((0.339 * 0.661) / 387)

= 0.048

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.339 - 0.048 < p < 0.339 + 0.048

0.291 < p < 0.386

(0.291 , 0386)

(b)

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

Z_/2 = 1.96

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

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

= 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

(0.233 , 0.311)