Complete Question
Given two independent random samples with the following results:
[tex]n_2=13\ , \= x_2=171\ s_1=23[/tex]
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:
[tex]\= x_p = 15[/tex]
Step 2 of 3:
[tex]E =7.79[/tex]
Step 3 of 3:
[tex] 7.21 < \mu_1 - \mu_2 < 22.79[/tex]
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
[tex]\= x_p = \= x_1 - \= x_2[/tex]
=> [tex]\= x_p = 186 - 171[/tex]
=> [tex]\= x_p = 15[/tex]
Now considering the Step 2 of 3
Given that the confidence level is 90% then the level of significance is mathematically represented as
[tex]\alpha = (100-90)\%[/tex]
=> [tex]\alpha = 0.10[/tex]
Generally the degree of freedom is mathematically represented as
[tex]df = n_1 + n_2 - 2[/tex]
=> [tex]df = 13 + 13 - 2[/tex]
=> [tex]df = 24[/tex]
From the student t-distribution table the critical value of [tex]\frac{\alpha }{2}[/tex] at a degree of freedom of [tex]df = 24 \ is \ \ t_{\frac{\alpha }{2} ,df} = 1.711[/tex]
Generally the pooled variance is mathematically represented as
[tex]s_p^2 = \frac{ (13 -1 ) 33^2 + (13 -1 ) 23^2 }{(13 - 1 )(13 - 1)}[/tex]
[tex]s_p^2 = 134.83 [/tex]
Generally the margin of error is mathematically represented as
[tex]E = t_{\frac{\alpha }{2} ,df } * \sqrt{\frac{s_p^2}{n_1} +\frac{s_p^2}{n_2} }[/tex]
=> [tex]E = 1.711* \sqrt{\frac{134.83}{13} +\frac{134.83}{13}}[/tex]
=> [tex]E =7.79[/tex]
Now considering the Step 3 of 3
Generally the 90% confidence interval is mathematically represented as
[tex]\= x_p -E < \mu_1 - \mu_2 < \= x_p +E[/tex]
=> [tex] 15 -7.79 < \mu_1 - \mu_2 < 15 +7.79[/tex]
=> [tex] 7.21 < \mu_1 - \mu_2 < 22.79[/tex]
