Question

Given two independent random samples with the following results:n1=13x‾1=186s1=33 n2=13x‾2=171s2=23Use this data to find the 90% confidence interval for the true difference between the population means. Assume that the population variances are equal and that the two populations are normally distributed.Step 1 of 3 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Answer 1

Complete Question

Given two independent random samples with the following results:

   

$n_2=13\ , \= x_2=171\ s_1=23$  

Use this data to find the 90% confidence interval for the true difference between the population means. Assume that the population variances are equal and that the two populations are normally distributed.

Step 1 of 3: Find the point estimate that should be used in constructing the confidence interval.

Step 2 of 3: Find the margin of error to be used in constructing the confidence interval. Round your answer to six decimal places.

Step 3 of 3: Construct the 90% confidence interval. Round your answers to the nearest whole number.

Answer:

Step 1 of 3:

    $\= x_p = 15$

Step 2 of 3:

    $E =7.79$

Step 3 of 3:

   $7.21 < \mu_1 - \mu_2 < 22.79$

Step-by-step explanation:

Now considering the Step 1 of 3, the  point estimate that should be used in constructing the confidence interval  is mathematically represented as  

        $\= x_p = \= x_1 - \= x_2$

=>     $\= x_p = 186 - 171$

=>     $\= x_p = 15$

Now considering the Step 2 of 3

Given that the confidence level is  90% then the level of significance is mathematically represented as

     $\alpha = (100-90)\%$

=>    $\alpha = 0.10$

Generally the degree of freedom is mathematically represented as

        $df = n_1 + n_2 - 2$

=>    $df = 13 + 13 - 2$

=>   $df = 24$

From the student t-distribution table the critical value of  $\frac{\alpha }{2}$ at a degree of freedom of  $df = 24 \ is \ \ t_{\frac{\alpha }{2} ,df} = 1.711$

Generally the pooled variance is mathematically represented as    

  $s_p^2 = \frac{ (13 -1 ) 33^2 + (13 -1 ) 23^2 }{(13 - 1 )(13 - 1)}$

         $s_p^2 = 134.83$

Generally the margin of error is mathematically represented as

  $E = t_{\frac{\alpha }{2} ,df } * \sqrt{\frac{s_p^2}{n_1} +\frac{s_p^2}{n_2} }$

=>     $E = 1.711* \sqrt{\frac{134.83}{13} +\frac{134.83}{13}}$

=>     $E =7.79$

Now considering the Step 3 of 3

Generally the 90% confidence interval is mathematically represented as

              $\= x_p -E < \mu_1 - \mu_2 < \= x_p +E$

=>  $15 -7.79 < \mu_1 - \mu_2 < 15 +7.79$

=>  $7.21 < \mu_1 - \mu_2 < 22.79$