Mathematics, 23.07.2021 01:30 lizdeleon248

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.

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Answer from: TH3L0N3W0LF

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).

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