Assume that the two samples are independent simple random sa

Assume that the two samples are independent simple random samples selected from normally distributed populations Also assume that the population standard deviations are equal (sigma1 = sigma2), 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. Construct a 99% confidence interval for mu1 - mu2, the difference between the mean drying time for paint type A and the mean drying time for paint type B.

Solution

option (D)

the standard error of the difference in means in the population is>
sigma= root over( sigma1^2/ n1 + sigma2^2/n2)

=root over( sigma1^2((1/n1) + (1/n2)) [as popln std dev.s are equal]

=root( sigma1^2 *( 1/11+ 1/9))

now sigma1^2 is estimated by=(s1^2 + s2^2)/2 =((3.4)^2 + (3.7)^2 )/2 =12.625

so,sigma = root ( 12.625*(1/11 +1/99) ) = 1.597
d.f=(n1-1)+(n2-1)= 18

t-value at 99% conf int. with (df=18) = 2.878

so,conf. int.= [ ( 70.3 -67.8) - (2.878*1.597),( 70.3 -67.8) - (2.878*1.597) ]
which nearly equals to = [ 2.07 ,7.07 ]