Question

A student researcher compares the heights of American students and non-American students from the student body of a certain college in order to estimate the difference in their mean heights. A random sample of 12 American students had a mean height of 69.4 inches with a standard deviation of 2.79 inches. A random sample of 17 non-American students had a mean height of 63.3 inches with a standard deviation of 3.22 inches. Determine the 90% confidence interval for the true mean difference between the mean height of the American students and the mean height of the non-American students. Assume that the population variances are equal and that the two populations are normally distributed. Step 1 of 3 : Find the point estimate that should be used in constructing the confidence interval

Answer 1

Answer:

The 90% confidence interval for the difference between means is (4.189, 8.011).

The point estimate is the difference between sample means and has a value of Md=6.1.

Step-by-step explanation:

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

The sample 1 (American students), of size n1=12 has a mean of 69.4 and a standard deviation of 2.79.

The sample 2 (non-American students), of size n2=17 has a mean of 63.3 and a standard deviation of 3.22.

The difference between sample means is Md=6.1.

$M_d=M_1-M_2=69.4-63.3=6.1$

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

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{2.79^2}{12}+\dfrac{3.22^2}{17}}\\\\\\s_{M_d}=\sqrt{0.649+0.61}=\sqrt{1.259}=1.12$

The critical t-value for a 90% confidence interval is t=1.703.

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

$MOE=t\cdot s_{M_d}=1.703 \cdot 1.12=1.911$

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

$LL=M_d-t \cdot s_{M_d} = 6.1-1.911=4.189\\\\UL=M_d+t \cdot s_{M_d} = 6.1+1.911=8.011$

The 90% confidence interval for the difference between means is (4.189, 8.011).