Using the t-distribution, as we have the standard deviation for the samples, it is found that since the absolute value of the test statistic is greater than the critical value, we reject the claim that Mrs.Elite and Mrs. Bright are equally effective teachers.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, we test if they are equally effective teachers, that is, the subtraction of their means is 0, hence:

$H_0: \mu_1 - \mu_2 = 0$

At the alternative hypothesis, we test if they are not equally effective teachers, that is, the subtraction of their means is not 0, hence:

$H_1: \mu_1 - \mu_2 \neq 0$

<h3>What is the distribution of the differences?</h3>

For Mrs. Elite, we have that:

$\mu_1 = 78, \sigma_1 = 10, n_1 = 30, s_1 = \frac{10}{\sqrt{30}} = 1.82574$

For Mrs. Bright, we have that:

$\mu_2 = 85, \sigma_2 = 15, n_2 = 25, s_2 = \frac{15}{\sqrt{25}} = 3$

For the distribution of differences, we have that:

$\overline{x} = \mu_1 - \mu_2 = 78 - 85 = -7$

$s = \sqrt{s_1^2 + s_2^2} = \sqrt{1.82574^2 + 3^2} = 3.5119$

<h3>What is the test statistic?</h3>

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{s}$

In which $\mu = 0$ is the value tested at the null hypothesis.

Hence:

$t = \frac{\overline{x} - \mu}{s}$

$t = \frac{-7 - 0}{3.5119}$

$t = -1.99$

Considering a<em> two-tailed test</em>, as we are testing if the mean is different of a value, with 30 + 25 - 2 = <em>53 df and a significance level of 0.1</em>, the critical value is of $|t^{\ast}| = 1.6741$ .

Since the absolute value of the test statistic is greater than the critical value, we reject the claim that Mrs.Elite and Mrs. Bright are equally effective teachers.

To learn more about the t-distribution, you can take a look at brainly.com/question/13873630

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3 years ago