Question

Given two independent random samples with the following results: n1=8x‾1=186s1=33 n2=7x‾2=171s2=23 Use this data to find the 90% confidence interval for the true difference between the population means. Assume that the population variances are equal and that the two populations are normally distributed. Step 1 of 3: Find the point estimate that should be used in constructing the confidence interval.

Answer 1

The point of estimate for the true difference would be:

`186-171= 15`

And the confidence interval is given by:

`(186-171) -1.77 \sqrt{(33^2)/(8) +(23^2)/(7)}= -10.753`

`(186-171) +1.77 \sqrt{(33^2)/(8) +(23^2)/(7)}= 40.753`

Explanation:

For this case we have the following info given:

`\bar X_1 = 186` the sample mean for the first sample

`\bar X_2 = 171` the sample mean for the second sample

`s_1 =33` the sample deviation for the first sample

`s_2 =23` the sample deviation for the second sample

`n_1 = 8` the sample size for the first group

`n_2 = 7` the sample size for the second group

The confidence interval for the true difference is given by:

`(\bar X_1 -\bar X_2) \pm t_(\alpha/2)\sqrt{(s^2_1)/(n_1) +(s^2_2)/(n_2)}`

We can find the degrees of freedom are given:

`df = n_1 +n_2 -2 =8+7-2= 13`

The confidence level is given by 90% so then the significance would be
`\alpha=1-0.9=0.1` and
`\alpha/2=0.05` we can find the critical value with the degrees of freedom given and we got:

`t_(\alpha/2)= \pm 1.77`

The point of estimate for the true difference would be:

`186-171= 15`

And replacing into the formula for the confidence interval we got:

`(186-171) -1.77 \sqrt{(33^2)/(8) +(23^2)/(7)}= -10.753`

`(186-171) +1.77 \sqrt{(33^2)/(8) +(23^2)/(7)}= 40.753`