Question

The federal government would like to test the hypothesis that the average age of men filing for Social Security is higher than the average age of women with the following data: Men Women Sample mean 64.5 years 63.6 years Sample size 35 39 Population standard deviation 3.0 years 3.5 years If Population 1 is defined as men and Population 2 is defined as women, the 90% confidence interval for the difference in population means is (closest to) ________. Group of answer choices

Answer 1

Answer:

The 90% confidence interval is  $-0.3433< \mu_1 - \mu_2 < 2.1433$

Step-by-step explanation:

From the question we are told that

   The sample mean for men is  $\= x_1 = 64.5 \ years$

    The sample mean for women is  $\= x_2 = 63.6 \ years$

     The sample size for men  is   $n_1 = 35$

      The sample size for women is  $n_2 = 39$

      The standard deviation for men is $s_1 = 3.0$

       The standard deviation for women is  $s_2 = 3.5$

From the question we are told the confidence level is  90% , hence the level of significance is    

      $\alpha = (100 - 90 ) \%$

=>   $\alpha = 0.10$

Generally from the normal distribution table the critical value  of  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } = 1.645$

Generally the standard error is mathematically represented as

     $SE = \sqrt{\frac{s_1^2 }{n_1} + \frac{s_2^2}{n_2} }$

=>  $SE = \sqrt{\frac{3^2 }{35} + \frac{3.5^2}{39} }$

=>  $SE = 0.7558$

Generally the margin of error is mathematically represented as

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

=>     $E = 1.645 * 0.7558$

=>     $E = 1.2433$

Generally 90% confidence interval is mathematically represented as  

      $(\= x_1 - \= x_2) -E < \mu_1 - \mu_2 < (\= x_1 - \= x_2) -E$

=>    $(64.5 - 63.6) -1.2433< \mu_1 - \mu_2$

=>    $-0.3433< \mu_1 - \mu_2 < 2.1433$