1. **State the problem:** We want to test if the average stress level among pet owners (group 1) is lower than that among non-owners (group 2) at the 0.05 significance level. 2. **Set hypotheses:** - Null hypothesis $H_0$: $\mu_1 = \mu_2$ (no difference in means) - Alternative hypothesis $H_a$: $\mu_1 < \mu_2$ (mean stress level of pet owners is less) 3. **Given data:** - Group 1 (pet owners): $\bar{x}_1 = 16.25$, $s_1 = 4.00$, $n_1 = 29$ - Group 2 (non-owners): $\bar{x}_2 = 20.95$, $s_2 = 5.10$, $n_2 = 25$ 4. **Calculate the test statistic:** We use a two-sample t-test for means with unequal variances (Welch's t-test): $$ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} $$ Calculate numerator: $$ 16.25 - 20.95 = -4.7 $$ Calculate denominator: $$ \sqrt{\frac{4.00^2}{29} + \frac{5.10^2}{25}} = \sqrt{\frac{16}{29} + \frac{26.01}{25}} = \sqrt{0.5517 + 1.0404} = \sqrt{1.5921} \approx 1.2617 $$ Calculate $t$: $$ t = \frac{-4.7}{1.2617} \approx -3.726 $$ 5. **Calculate degrees of freedom (df) using Welch-Satterthwaite equation:** $$ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} $$ Calculate numerator: $$ (0.5517 + 1.0404)^2 = 1.5921^2 = 2.535 $$ Calculate denominator: $$ \frac{0.5517^2}{28} + \frac{1.0404^2}{24} = \frac{0.3044}{28} + \frac{1.0824}{24} = 0.01087 + 0.0451 = 0.05597 $$ Calculate $df$: $$ df = \frac{2.535}{0.05597} \approx 45.3 $$ Use $df \approx 45$. 6. **Find critical t-value:** For a one-tailed test at $\alpha = 0.05$ and $df=45$, critical value $t_{crit} \approx -1.68$. 7. **Decision:** Since $t = -3.726 < -1.68$, we reject the null hypothesis. 8. **Conclusion:** There is sufficient evidence at the 0.05 significance level to support the claim that pet owners have a lower average stress level than non-owners.