Question

In order to compare the performance of students in large enrollment and small enrollment sections, 35 students from large sections and 35 students from small sections of a freshman mathematics course were randomly selected. The mean and sample standard deviation of grades on the common final exam for the students from large sections were 72.8 and 7.4; for the students from small sections, the mean and standard deviation were 75.3 and 6.8. A 90% confidence interval for the difference in the mean common final exam scores between students in the two types of sections is about:

Answer 1

Answer:

The 90% confidence interval for the difference in the mean common final exam scores between students in the two types of sections is about (-5.3, 0.3).

Step-by-step explanation:

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

Central Limit Theorem

The Central Limit Theorem estabilishes 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.

35 students from large sections. The mean and sample standard deviation of grades on the common final exam for the students from large sections were 72.8 and 7.4;

This means that:

$\mu_L = 72.8, s_L = \frac{7.4}{\sqrt{35}} = 1.25$

35 students from small sections. For the students from small sections, the mean and standard deviation were 75.3 and 6.8.

This means that:

$\mu_S = 75.3, s_S = \frac{6.8}{\sqrt{35}} = 1.15$

Distribution of the difference in the mean common final exam scores between students in the two types of sections:

Has mean and standard error given by:

$\mu = \mu_L - \mu_S = 72.8 - 75.3 = -2.5$

$s = \sqrt{s_L^2+s_S^2} = \sqrt{1.25^2 + 1.15^2} = 1.7$

Confidence interval:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$.

That is z with a pvalue of $1 - 0.05 = 0.95$, so Z = 1.645.

Now, find the margin of error M as such

$M = zs = 1.645*1.7 = 2.8$

The lower end of the interval is the sample mean subtracted by M. So it is -2.5 - 2.8 = -5.3

The upper end of the interval is the sample mean added to M. So it is -2.5 + 2.8 = 0.3

The 90% confidence interval for the difference in the mean common final exam scores between students in the two types of sections is about (-5.3, 0.3).