We are given summary data for two groups: pet owners (group 1) and non-owners (group 2). We want to test the claim that the average stress level is lower for pet owners than for non-owners at a significance level of 0.05. We are given $\bar{X_1} = 15.25$, $\bar{X_2} = 21.95$, $s_1 = 3.00$, $s_2 = 4.10$, $n_1 = 27$, and $n_2 = 20$.

1. Problem Description

We are given summary data for two groups: pet owners (group 1) and non-owners (group 2). We want to test the claim that the average stress level is lower for pet owners than for non-owners at a significance level of 0.

0

5. We are given $\bar{X_1} = 15.25$, $\bar{X_2} = 21.95$, $s_1 = 3.00$, $s_2 = 4.10$, $n_1 = 27$, and $n_2 = 20$.

2. Solution Steps

We will perform a one-tailed t-test.

First, we set up the null and alternative hypotheses:

$H_0: \mu_1 = \mu_2$ (The average stress level of pet owners is equal to the average stress level of non-owners)

$H_1: \mu_1 < \mu_2$ (The average stress level of pet owners is lower than the average stress level of non-owners)

Next, we calculate the t-statistic:

$t = \frac{(\bar{X_1} - \bar{X_2}) - (\mu_1 - \mu_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$

Since we are testing if $\mu_1 < \mu_2$, we assume $\mu_1 - \mu_2 = 0$ under the null hypothesis. Thus:

$t = \frac{\bar{X_1} - \bar{X_2}}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$

Plugging in the given values:

$t = \frac{15.25 - 21.95}{\sqrt{\frac{3.00^2}{27} + \frac{4.10^2}{20}}} = \frac{-6.7}{\sqrt{\frac{9}{27} + \frac{16.81}{20}}} = \frac{-6.7}{\sqrt{\frac{1}{3} + 0.8405}} = \frac{-6.7}{\sqrt{0.3333 + 0.8405}} = \frac{-6.7}{\sqrt{1.1738}} = \frac{-6.7}{1.0834} = -6.184$

Next, we calculate the degrees of freedom:

$df = \frac{(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1 - 1} + \frac{(\frac{s_2^2}{n_2})^2}{n_2 - 1}}$

$df = \frac{(\frac{3^2}{27} + \frac{4.1^2}{20})^2}{\frac{(\frac{3^2}{27})^2}{27 - 1} + \frac{(\frac{4.1^2}{20})^2}{20 - 1}} = \frac{(\frac{9}{27} + \frac{16.81}{20})^2}{\frac{(\frac{1}{3})^2}{26} + \frac{(0.8405)^2}{19}} = \frac{(0.3333 + 0.8405)^2}{\frac{0.1111}{26} + \frac{0.7065}{19}} = \frac{(1.1738)^2}{0.0043 + 0.0372} = \frac{1.3778}{0.0415} = 33.19$

We take the smaller of $n_1 - 1 = 26$ and $n_2 - 1 = 19$, so we use $df = 19$ as a more conservative estimate. The approximation of 33.19 is closer to 19 than to 26, so we will round down to $df = 19$. Using a t-table with $df = 19$ and a significance level of $\alpha = 0.05$ for a one-tailed test, we find the critical t-value to be $t_{critical} = -1.729$.

Since our calculated t-statistic, $t = -6.184$, is less than the critical t-value, $t_{critical} = -1.729$, we reject the null hypothesis.

3. Final Answer

Reject the null hypothesis. There is sufficient evidence to support the claim that the average stress level of pet owners is lower than the average stress level of non-owners at the 0.05 significance level.