Low-fat or low-carb? Are low-fat diets or low-carb diets more effective for weight loss? A sample of 61 subjects went on a lowcarbohydrate diet for six months. At the end of that time, the sample mean weight loss was 3.7 kilograms with a sample standard deviation of 5.67 kilograms. A second sample of 63 subjects went on a low-fat diet. Their sample mean weight loss was 2.1 kilograms with a standard deviation of 4.71 kilograms. Can you conclude that the mean weight loss of subjects having low-carb diets is greater than the mean weight loss of subjects having low-fat diets? Let , denote the mean weight loss on the lowcarb diet and $1 \mathrm{y}$ denote the mean weight loss on the low-fat diet. Use the $a=0.01$ level and the critical value method. Compute the test statistic

Step 1 ：State the null hypothesis \( H_0: \mu_1 = \mu_2 \) and the alternative hypothesis \( H_1: \mu_1 > \mu_2 \)

Step 2 ：Use the formula for the test statistic for comparing two means with unequal variances: \( t = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \)

Step 3 ：Calculate the degrees of freedom using the Welch-Satterthwaite equation: \( df = \frac{(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1 - 1} + \frac{(\frac{s_2^2}{n_2})^2}{n_2 - 1}} \)

Step 4 ：Find the critical value for a one-tailed test at the \( \alpha = 0.01 \) level

Step 5 ：Compare the test statistic to the critical value

Step 6 ：Since the test statistic (1.7064) is less than the critical value (2.3588), fail to reject the null hypothesis

Step 7 ：\( \boxed{\text{Fail to reject the null hypothesis}} \)