Mathematics, 07.10.2020 23:01 ander67061

Item2 10 points eBookPrintReferences Check my work Check My Work button is now enabledItem 2Item 2 10 points 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 142 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 500 television ads in the United States reveals that 122 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? Next Visit question mapQuestion 2 of 4 Total2 of 4 Prev

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Answer from: mstrish71oteauw

point estimate is $\r p = 0.355$ i.e 35.5 %

confidence interval is 0.3081 < p

point estimate is $\r p = 0.244$ i.e 24.4 %

confidence interval 0.2064 < p

The confidence intervals you computed in parts a and b does not suggest that a greater number of U.K. ads use humor

Step-by-step explanation:

Considering the first question

The sample size is n = 400

The number of ads that use humor is k = 142

Generally the point estimate is mathematically represented as

$\r p = \frac{k}{n}$

=> $\r p = \frac{142}{400}$

=> $\r p = 0.355$

Given that the confidence level is 95% then the level of significance is

$\alpha =( 100 -95)\%$

=> $\alpha =5\%$

=> $\alpha =0.05$

From the normal distribution table the critical value for $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } =Z_{\frac{0.05 }{2} } = 1.96$

Generally the standard error is mathematically represented as

$SE = \sqrt{\frac{\r p(1 - \r p)}{n} }$

=> $SE = \sqrt{\frac{0.355(1 - 0.355)}{400} }$

=> SE = 0.0239

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * SE$

E = 1.96 * 0.0239

E = 0.0469

Generally the 95% confidence interval is

$\r p - E < p < \r p + E$

=> 0.355 - 0.0469 < 0.355 + 0.0469

=> 0.3081 < p

Considering the second question

The sample size is n = 500

The number of ads that use humor is k = 122

Generally the point estimate is mathematically represented as

$\r p = \frac{k}{n}$

=> $\r p = \frac{122}{500}$

=> $\r p = 0.244$

Generally the standard error is mathematically represented as

$SE = \sqrt{\frac{\r p(1 - \r p)}{n} }$

=> $SE = \sqrt{\frac{0.244(1 - 0.244)}{500} }$

=> SE = 0.0192

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * SE$

E = 1.96 * 0.0192

E = 0.0376

Generally the 95% confidence interval is

$\r p - E < p < \r p + E$

=> 0.244 - 0.0376 < 0.244 + 0.0376

=> 0.2064 < p

Considering third question

Looking at the confidence interval in a(first question ) and b(second question) we see that the upper limit of both interval is not up to or greater than 0.5 i.e 50% hence the confidence intervals you computed in parts a and b does not suggest that a greater percentage of U.K ads uses humor

Answer from: Quest

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Answer from: Quest

answer: its the third option.

step-by-step explanation:

i just took the test.

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