Question

Consider the following results for two independent random samples taken from two populations.

Answer 1

Answer:

1. Point estimate Md = 2

2. The 90% confidence interval for the difference between means is (1.01, 2.99).

3. The 95% confidence interval for the difference between means is (0.82, 3.18).

Step-by-step explanation:

a) The point estimate of the difference between the two population means is the difference between sample means:  

$M_d=M_1-M_2=13.6-11.6=2$

2. We have to calculate a 90% confidence interval for the difference between means.

The sample 1, of size n1=50 has a mean of 13.6 and a standard deviation of 2.2.

The sample 2, of size n2=35 has a mean of 11.6 and a standard deviation of 3.

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

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{2.2^2}{50}+\dfrac{3^2}{35}}\\\\\\s_{M_d}=\sqrt{0.097+0.257}=\sqrt{0.354}=0.5949$

The degrees of freedom for this confidence interval are:

$df=n_1+n_2-2=50+35-2=83$

The critical t-value for a 90% confidence interval is t=1.663.

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

$MOE=t\cdot s_{M_d}=1.663 \cdot 0.5949=0.99$

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

$LL=M_d-t \cdot s_{M_d} = 2-0.99=1.01\\\\UL=M_d+t \cdot s_{M_d} = 2+0.99=2.99$

The 90% confidence interval for the difference between means is (1.01, 2.99).

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

The critical t-value for a 95% confidence interval and 83 degrees of freedom is t=1.989.

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

$MOE=t\cdot s_{M_d}=1.989 \cdot 0.5949=1.18$

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

$LL=M_d-t \cdot s_{M_d} = 2-1.18=0.82\\\\UL=M_d+t \cdot s_{M_d} = 2+1.18=3.18$

The 95% confidence interval for the difference between means is (0.82, 3.18).