1. **State the problem:** We want to perform an independent samples t-test to compare the means of two groups: April (in glasses) and January (in glasses). 2. **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. 3. **Calculate sample means:** April data: 7.50, 8.00, 8.75, 9.25, 9.00, 8.50, 9.75, 10.50, 10.00, 11.25 $$\bar{x}_1 = \frac{7.50 + 8.00 + 8.75 + 9.25 + 9.00 + 8.50 + 9.75 + 10.50 + 10.00 + 11.25}{10} = \frac{92.50}{10} = 9.25$$ January data: 6.25, 7.00, 7.50, 6.75, 8.00, 7.25, 8.50, 7.75, 6.50, 8.00 $$\bar{x}_2 = \frac{6.25 + 7.00 + 7.50 + 6.75 + 8.00 + 7.25 + 8.50 + 7.75 + 6.50 + 8.00}{10} = \frac{70.50}{10} = 7.05$$ 4. **Calculate sample variances:** For April: $$s_1^2 = \frac{\sum (x_i - \bar{x}_1)^2}{n_1 - 1}$$ Calculations: $(7.50 - 9.25)^2 = 3.0625$ $(8.00 - 9.25)^2 = 1.5625$ $(8.75 - 9.25)^2 = 0.25$ $(9.25 - 9.25)^2 = 0$ $(9.00 - 9.25)^2 = 0.0625$ $(8.50 - 9.25)^2 = 0.5625$ $(9.75 - 9.25)^2 = 0.25$ $(10.50 - 9.25)^2 = 1.5625$ $(10.00 - 9.25)^2 = 0.5625$ $(11.25 - 9.25)^2 = 4.0$ Sum = 12.875 $$s_1^2 = \frac{12.875}{9} = 1.4306$$ For January: $(6.25 - 7.05)^2 = 0.64$ $(7.00 - 7.05)^2 = 0.0025$ $(7.50 - 7.05)^2 = 0.2025$ $(6.75 - 7.05)^2 = 0.09$ $(8.00 - 7.05)^2 = 0.9025$ $(7.25 - 7.05)^2 = 0.04$ $(8.50 - 7.05)^2 = 2.1025$ $(7.75 - 7.05)^2 = 0.49$ $(6.50 - 7.05)^2 = 0.3025$ $(8.00 - 7.05)^2 = 0.9025$ Sum = 4.765 $$s_2^2 = \frac{4.765}{9} = 0.5294$$ 5. **Calculate t-value:** $$t = \frac{9.25 - 7.05}{\sqrt{\frac{1.4306}{10} + \frac{0.5294}{10}}} = \frac{2.20}{\sqrt{0.14306 + 0.05294}} = \frac{2.20}{\sqrt{0.196}} = \frac{2.20}{0.4427} = 4.97$$ **Final answer:** $$t \approx 4.97$$ This t-value can be used to test the hypothesis about the difference between the two group means.