In constructing a 90% confidence interval estimate for the difference between the means of two normally distributed populations, where the unknown population variances are assumed not to be equal, summary statistics computed from two independent samples are as follows: $n _ { 1 } = 40$ , $\bar { x } _ { 1 } = 95$ , $s _ { 1 } = 12.5$ . $n _ { 2 } = 30$ , $\bar { x } _ { 2 } = 75$ , $s _ { 2 } = 35.5$ . The lower confidence limit is:
A) 30.086.
B) 8.542.
C) 0.914.
D) 31.458.

Question

Quizplus · Accepted Answer

Using the formula for the confidence interval when population variances are not assumed to be equal:

$$\bar{x}_{1}-\bar{x}_{2}\pm t_{\alpha/2, df}\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}$$

where

$$df=\frac{\left(\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}\right)^{2}}{\frac{1}{n_{1}-1}\left(\frac{s_{1}^{2}}{n_{1}}\right)^{2}+\frac{1}{n_{2}-1}\left(\frac{s_{2}^{2}}{n_{2}}\right)^{2}}-2$$

Substituting the given values, we have:

$$\bar{x}_{1}-\bar{x}_{2}\pm t_{0.05, df}\sqrt{\frac{(12.5)^{2}}{40}+\frac{(35.5)^{2}}{30}}$$

where

$$df=\frac{\left(\frac{(12.5)^{2}}{40}+\frac{(35.5)^{2}}{30}\right)^{2}}{\frac{1}{40-1}\left(\frac{(12.5)^{2}}{40}\right)^{2}+\frac{1}{30-1}\left(\frac{(35.5)^{2}}{30}\right)^{2}}-2$$

Simplifying, we get:

$$\bar{x}_{1}-\bar{x}_{2}\pm t_{0.05, 43.632}\times 8.5418$$

Using a t-table or calculator, we find that the t-value corresponding to a 90% confidence level with 43.632 degrees of freedom is approximately 1.684.

Thus, the confidence interval is given by:

$$95-75\pm1.684\times8.5418$$

Solving, we get the lower confidence limit to be approximately 8.542. Therefore the answer is (B).

In Constructing a 90% Confidence Interval Estimate for the Difference n1=40n _ { 1 } = 40n1​=40

In Constructing a 90% Confidence Interval Estimate for the Difference $n _ { 1 } = 40$