Question

An experiment is conducted to determine whether intensive tutoring (covering a great deal of material in a fixed amount of time) is more effective than paced tutoring (covering less material in the same amount of time). Two randomly chosen groups are tutored separately and then administered proficiency tests. The data collected and summarized below: Group1 (the intensive group) consisted of 15 students and resulted in an average grade of 49.55 with a standard deviation of 10. group 2 (the paced group) consisted of 17 students and resulted in an average grade of 42.29 with a standard deviation of 6.52. Assume that the population variances of both groups are different. Find the test statistic to test if intensive tutoring is more effective than paced tutoring at the significance level of 0.01.

Answer 1

Answer:

The test statistic is t = 2.33.

Step-by-step explanation:

Before testing the hypothesis, we need to understand the central limit theorem and subtraction of 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.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

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.

Group1 (the intensive group) consisted of 15 students and resulted in an average grade of 49.55 with a standard deviation of 10

This means that:

$\mu_1 = 49.55, s_1 = \frac{10}{\sqrt{15}} = 2.582$

Group 2 (the paced group) consisted of 17 students and resulted in an average grade of 42.29 with a standard deviation of 6.52.

This means that $\mu_2 = 42.49, s_2 = \frac{6.52}{\sqrt{17}} = 1.581$

Test if intensive tutoring is more effective than paced tutoring:

At the null hypothesis, we test that it is the same, that is, the means are the same and the subtraction is 0. So

$H_0: \mu_1 - \mu_2 = 0$

At the alternate hypothesis, we test that it is more effective, that is, the subtraction of the means is larger than 0. So

$H_a: \mu_1 - \mu_2 > 0$

The test statistic is:

$t = \frac{X - \mu}{s}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis and s is the standard error.

0 is tested at the null hypothesis:

This means that $\mu = 0$

From the two samples:

$X = \mu_1 - \mu_2 = 49.55 - 42.49 = 7.06$

$s = \sqrt{s_1^2+s_2^2} = \sqrt{2.582^2+1.581^2} = 3.03$

Test statistic:

$t = \frac{X - \mu}{s}$

$t = \frac{7.06 - 0}{3.03}$

$t = 2.33$

The test statistic is t = 2.33.