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. x1 = 958, x2 = 157, s1 = 77, s2 = 88. The sample size is 478 for both samples. Find the 85% confidence interval for μ1 - μ2. 781 < μ1 - μ2 < 821 800 < μ1 - μ2 < 802 794 < μ1 - μ2 < 808 793.2946 < μ1 - μ2 < 808.7054

Answer 1

Answer: $793.2946 < \mu_1 - \mu_2 < 808.7054$

Step-by-step explanation:

The confidence interval for the difference of two population mean is given by :-

$(\overline{x_1}-\overline{x}_2)\pm z_{\alpha/2}\sqrt{\dfrac{s_1^2}{n_1}+\dfrac{s^2_2}{n_2}}$

Given : $\overline{x}_1=958,\ \overline{x}_2=157$

$s_1=77,\ s_2=88$

Significance level : $\alpha=1-0.85=0.15$

Critical value : $z_{\alpha/2}=z_{0.075}=\pm1.44$

We assume that the two samples are independent simple random samples selected from normally distributed populations.

Now, the confidence interval for the difference of two population mean is given by :-

$(958-157)\pm 1.44\sqrt{\dfrac{(77)^2}{478}+\dfrac{(88)^2}{478}}\\\\\approx801\pm7.70=(801-7.7016,801+7.7016)\\\\(793.2984,808.7016)\subset(793.2946,\ 808.7054)$

Hence, the 85% confidence interval for $\mu_1-\mu_2$ is given by :-

$793.2946 < \mu_1 - \mu_2 < 808.7054$