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 68.4 inches with a standard deviation of 1.64 inches. A random sample of 17 non-American students had a mean height of 64.9 inches with a standard deviation of 1.75 inches. Determine the 95% 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. Find the point estimate that should be used in constructing the confidence interval.

Answer 1

Answer:

The point estimate that should be used in constructing the confidence interval is 3.5.

The 95% confidence interval for the true mean difference between the mean height of the American students and the mean height of the non-American students, in inches, is (2.25, 4.75).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

American students:

Sample of 12, mean height of 68.4 inches with a standard deviation of 1.64 inches. This means that:

$\mu_A = 68.4$

$s_A = \frac{1.64}{\sqrt{12}} = 0.4743$

Non-American students:

Sample of 17, mean height of 64.9 inches with a standard deviation of 1.75 inches. This means that:

$\mu_N = 64.9$

$s_N = \frac{1.75}{\sqrt{17}} = 0.4244$

Distribution of the difference:

$\mu = \mu_A - \mu_N = 68.4 - 64.9 = 3.5$

$s = \sqrt{s_A^2+s_N^2} = \sqrt{0.4743^2 + 0.4244^2} = 0.6365$

The point estimate that should be used in constructing the confidence interval is 3.5.

Confidence interval:

$\mu \pm zs$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$.

95% confidence level

So $\alpha = 0.05$, z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$, so $Z = 1.96$.  

The lower bound of the interval is:

$\mu - zs = 3.5 - 1.96*0.6365 = 2.25$

The upper bound of the interval is:

$\mu + zs = 3.5 + 1.96*0.6365 = 4.75$

The 95% confidence interval for the true mean difference between the mean height of the American students and the mean height of the non-American students, in inches, is (2.25, 4.75).