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. Also assume that the population standard deviations are equal (σ1= 2 ), so that the standard error of the difference between means is obtained by pooling the sample variances. A paint manufacturer wanted to compare the drying times of two different types of paint. Independent simple random samples of 11 cans of type A and 9 cans of type B were selected and applied to similar surfaces. The drying times, in hours, were recorded. The summary statistics are as follows.Type A: X1= 71.5hr, S1=3.4 hr N1=11Type B: X2=68.5 hr, S2= 3.6 hr, N2= 9Construct a 99% confidence interval for μ1-μ2 , the difference between the mean drying time for paint type A and the mean drying time for paint type B.

Answer 1

The indicated confidence interval for the difference between the two population means is (-1.5159, 7.5159)

Explanation:

Let the drying times of type A be the first population and the drying times of type B be the second population. Then

We have small sample sizes
`n_(1) = 11` and
`n_(2) = 9`, besides
`\bar{x}_(1) = 71.5`,
`s_(1) = 3.4` ,
`\bar{x}_(2) = 68.5` and
`s_(2) = 3.6`. Therefore, the pooled

estimate is given by

`s_(p)^(2) = ((n_(1)-1)s_(1)^(2)+(n_(2)-1)s_(2)^(2))/(n_(1)+n_(2)-2) = ((11-1)(3.4)^(2)+(9-1)(3.6)^(2))/(11+9-2) = 12.1822`

The 99% confidence interval for the true mean difference between the mean drying time of type A and the mean drying time of type B is given by

`(\bar{x}_(1)-\bar{x}_(2))\pm t_(0.01/2)s_(p)\sqrt{(1)/(11)+(1)/(9)}`, i.e.,

`(71.5-68.5)\pm t_(0.005)(3.4903)\sqrt{(1)/(11)+(1)/(9)}`

where
`t_(0.005)` is the 0.5th quantile of the t distribution with (11+9-2) = 18 degrees of freedom. So

`3\pm(-2.8784)(3.4903)(0.4495)`, i.e.,

the indicated confidence interval for the difference between the two population means is (-1.5159, 7.5159)