Question

The following information was obtained from independent random samples taken of two populations. Assume normally distributed populations with equal variances.
Sample 1 Sample 2
Sample Mean 45 42
Sample Variance 85 90
Sample Size 10 12
1. The 95% confidence interval for the difference between the two population means is:________.
a. 0b. 2c. 3d. 15
2. The standard error of (x-bar1)-(x-bar2) isa. 3.0b. 4.0c. 8.372d. 19.48

Answer 1

Answer:

1. The 95% confidence interval for the difference between means is (-5.34, 11.34).

2. The standard error of (x-bar1)-(x-bar2) is 4.

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{9.2195^2}{10}+\dfrac{9.4868^2}{12}}\\\\\\s_{M_d}=\sqrt{8.5+7.5}=\sqrt{16}=4$

Step-by-step explanation:

We have to calculate a 95% confidence interval for the difference between means.

The sample 1, of size n1=10 has a mean of 45 and a standard deviation of √85=9.2195.

The sample 2, of size n2=12 has a mean of 42 and a standard deviation of √90=9.4868.

The difference between sample means is Md=3.

$M_d=M_1-M_2=45-42=3$

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

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{9.2195^2}{10}+\dfrac{9.4868^2}{12}}\\\\\\s_{M_d}=\sqrt{8.5+7.5}=\sqrt{16}=4$

The critical t-value for a 95% confidence interval is t=2.086.

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

$MOE=t\cdot s_{M_d}=2.086 \cdot 4=8.34$

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

$LL=M_d-t \cdot s_{M_d} = 3-8.34=-5.34\\\\UL=M_d+t \cdot s_{M_d} = 3+8.34=11.34$

The 95% confidence interval for the difference between means is (-5.34, 11.34).