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.
Q)
A paint manufacturer wished 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 = 77.5 hrs

s1 = 4.5 hrs

n1 = 11

Type B:
x2 = 63.6 hrs
s2 = 5.1 hrs
n2 = 9

Construct a 98% confidence interval for µ1 - µ2, the difference between the mean drying time for paint of type A and the mean drying time for paint of type B.

Answer 1

Answer:

The 98% confidence interval for the mean is (8.84, 18.96).

Step-by-step explanation:

We have to calculate a 98% confidence interval for the mean.

The sample 1, of size n1=11 has a mean of 77.5 and a standard deviation of 4.5.

The sample 1, of size n1=9 has a mean of 63.6 and a standard deviation of 5.1.

The difference between sample means is Md=.

$M_d=M_1-M_2=77.5-63.6=13.9$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{4.5^2}{11}+\dfrac{5.1^2}{9}}\\\\\\s_{M_d}=\sqrt{1.841+2.89}=\sqrt{4.731}=2.175$

The t-value for a 98% confidence interval is t=2.326.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=2.326 \cdot 2.175=5.06$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 13.9-5.06=8.84\\\\UL=M+t \cdot s_M = 13.9+5.06=18.96$

The 98% confidence interval for the mean is (8.84, 18.96).