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. The sample size is 478 for both samples. Find the \(85 \%\) confidence interval for \(\mu_{1}-\mu_{2}\).

\(\bar{x}_{1}=958, \bar{x}_{2}=157, s_{1}=77, s_{2}=88\)

A. \(800<\mu_{1}-\mu_{2}<802\)

B. \(791<\mu_{1}-\mu_{2}<811\)

C. \(793<\mu_{1}-\mu_{2}<809\)

D. \(781<\mu_{1}-\mu_{2}<821\)

Answer 1

We have given here,

For sample 

Sample mean \(=958\) Sample standard deviation \(=77\) Sample size \(=478\)

For sample 

Sample mean \(=157\) Sample standard deviation \(=88\) Sample size \(=478\)

Level of significance \(=1-0.85=0.15\) Degree of freedom \(=477\) t critical value (by using \(t\) table) \(=1.442\)

Confidence interval formula is \(\left(\bar{x}_{1}-\bar{x}_{2}\right) \pm t \sqrt{\frac{s_{1}^{2}}{n_{1}}}+\frac{s_{2}^{2}}{n_{2}}\)

\(=>(958-157) \pm 1.442 * \sqrt{\frac{(77)^{2}}{478}+\frac{(88)^{2}}{478}}\)

\(=>(793.29,808.71)\)

Lower limit for confidence interval is \(=793.29\) Upper limit for confidence interval is \(=808.71\)

Therefore, Option \(\mathrm{C}\) ) is correct.

\(793<\mu_{1}-\mu_{2}<809\)

 Option C) is correct.