Question

In constructing 95% 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:

n1=50 x1_bar=756 s1=35

n2=50 x2_bar=762 s2=30



The lower confidence limit is:

A.
-77.3

B.
-18.78

C.
-8.62

D.
-6.78

Answer 1

Option B.

Explanation:

Given information:

`n_1=50, \overline{x}_1=756, s_1=35`

`n_2=50, \overline{x}_1=762, s_2=30`

We need to find 95% confidence interval estimate for the difference between the means of two normally distributed populations.

Formula for confidence interval:

`CI=(\overline{x}_1-\overline{x}_2)\pm z*\sqrt{(s_1^2)/(n_1)+(s_2^2)/(n_2)}`

From the standard normal table it is clear that the z value at 95% confidence interval is 1.96.

Substitute the given values in the above formula.

`CI=\left(756-762\right)\pm 1.96\sqrt{(35^(2))/(50)+(30^(2))/(50)}`

`CI=\left(-6\right)\pm 1.96√(42.5)`

`CI=-6\pm 12.78`

`CI=(-18.78, 6.78)`

The lower confidence limit is −18.78.

Therefore, the correct option B.