1. **Problem:** A clinical psychologist compares anxiety levels after two therapies: Cognitive Behavioural Therapy (CBT) and Mindfulness-Based Stress Reduction (MBSR). 2. **Hypotheses:** - Null hypothesis ($H_0$): There is no difference in mean anxiety levels between CBT and MBSR groups, i.e., $\mu_{CBT} = \mu_{MBSR}$. - Alternate hypothesis ($H_a$): There is a difference in mean anxiety levels, i.e., $\mu_{CBT} \neq \mu_{MBSR}$. 3. **Formula for independent samples t-test:** $$ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} $$ where $\bar{x}_1, \bar{x}_2$ are sample means, $s_1^2, s_2^2$ are sample variances, and $n_1, n_2$ are sample sizes. 4. **Calculate sample means:** - CBT data: 21, 24, 20, 25, 23, 23, 24, 21, 19, 21 $$\bar{x}_{CBT} = \frac{21+24+20+25+23+23+24+21+19+21}{10} = \frac{221}{10} = 22.1$$ - MBSR data: 28, 29, 26, 28, 37, 29, 35, 28, 25, 32 $$\bar{x}_{MBSR} = \frac{28+29+26+28+37+29+35+28+25+32}{10} = \frac{297}{10} = 29.7$$ 5. **Calculate sample variances:** - CBT variance $s_1^2$: Calculate squared deviations and sum: $$\sum (x_i - \bar{x}_{CBT})^2 = (21-22.1)^2 + (24-22.1)^2 + ... + (21-22.1)^2 = 38.9$$ $$s_1^2 = \frac{38.9}{10-1} = \frac{38.9}{9} \approx 4.322$$ - MBSR variance $s_2^2$: $$\sum (x_i - \bar{x}_{MBSR})^2 = (28-29.7)^2 + (29-29.7)^2 + ... + (32-29.7)^2 = 90.1$$ $$s_2^2 = \frac{90.1}{9} = 10.011$$ 6. **Calculate t-statistic:** $$ t = \frac{22.1 - 29.7}{\sqrt{\frac{4.322}{10} + \frac{10.011}{10}}} = \frac{-7.6}{\sqrt{0.4322 + 1.0011}} = \frac{-7.6}{\sqrt{1.4333}} = \frac{-7.6}{1.197} \approx -6.35$$ 7. **Degrees of freedom (Welch's approximation):** $$ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{(s_1^2/n_1)^2}{n_1-1} + \frac{(s_2^2/n_2)^2}{n_2-1}} = \frac{(0.4322 + 1.0011)^2}{\frac{0.4322^2}{9} + \frac{1.0011^2}{9}} = \frac{(1.4333)^2}{\frac{0.187}{9} + \frac{1.002}{9}} = \frac{2.054}{0.0208 + 0.1113} = \frac{2.054}{0.1321} \approx 15.55$$ Use $df \approx 16$. 8. **Critical value and interpretation:** - For two-tailed test at $\alpha=0.05$ and $df=16$, critical t-value $\approx \pm 2.12$. - Since calculated $t = -6.35$ is less than $-2.12$, reject $H_0$. - **Interpretation:** There is a statistically significant difference in anxiety levels between CBT and MBSR groups at the 0.05 significance level. **Final answer:** $t \approx -6.35$, $df \approx 16$, reject null hypothesis.