Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Independent samples from two different populations yield the following data. $\bar { x } _ { 1 } = 958 , \bar { x } _ { 2 } = 157 , s _ { 1 } = 77 , s _ { 2 } = 88$. The sample size is 478 for both samples. Find the $85 \%$ confidence interval for $\mu _ { 1 } - \mu _ { 2 }$.
A) $781

Question

Construct the indicated confidence interval for the difference between the two population means. Assume that the two samples are independent simple random samples selected from normally distributed populations. Do not assume that the population standard deviations are equal. Independent samples from two different populations yield the following data. $\bar { x } _ { 1 } = 958 , \bar { x } _ { 2 } = 157 , s _ { 1 } = 77 , s _ { 2 } = 88$. The sample size is 478 for both samples. Find the $85 \%$ confidence interval for $\mu _ { 1 } - \mu _ { 2 }$.
A) $781 < \mu _ { 1 } - \mu _ { 2 } < 821$
B) $800 < \mu _ { 1 } - \mu _ { 2 } < 802$
C) $794 < \mu _ { 1 } - \mu _ { 2 } < 808$
D) $791 < \mu _ { 1 } - \mu _ { 2 } < 811$

Quizplus · Accepted Answer

The correct answer is D because the confidence interval is calculated as:$(\bar{x}_1 - \bar{x}_2) \pm t^* \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$where $t^*$ is the critical value from the t-distribution with $df = \frac{(s_1^2/n_1 + s_2^2/n_2)^2}{(s_1^2/n_1)^2/(n_1-1) + (s_2^2/n_2)^2/(n_2-1)}$ degrees of freedom.Plugging in the values, we get:$t^* = 1.96$$(\bar{x}_1 - \bar{x}_2) \pm t^* \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$$= (958 - 157) \pm 1.96 \sqrt{\frac{77^2}{478} + \frac{88^2}{478}}$$= 800 \pm 10$$= (790, 810)$

Construct the Indicated Confidence Interval for the Difference Between the Two