Question

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

Answer 1

Answer:

a

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

  confidence interval  is  $0.3081 < p<0.4019$

b

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

   confidence interval   $0.2064 < p<0.2816$

c

 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  

      $\r p - E < p < \r p + E$

=>   $0.355 - 0.0469 < 0.355 + 0.0469$

=>    $0.3081 < p<0.4019$

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  

      $\r p - E < p < \r p + E$

=>   $0.244 - 0.0376 < 0.244 + 0.0376$

=>    $0.2064 < p<0.2816$

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