Low-fat or low-carb? Are low-fat diets or low-carb diets more effective for weight loss? A sample of 41 subjects went on a low-carbohydrate diet for six months. At the end of that time, the sample mean welght loss was 2.9 kilograms with a sample standard deviation of 4.46 kilograms. A second sample of 46 subjects went on a low-fat diet. Their sample mean weight loss was 1.5 kilograms with a standard deviation of 3.44 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 $\mu_{1}$ denote the mean weight loss on the low-carb diet and $\mu_{2}$ denote the mean weight loss on the low-fat diet. Use the $\alpha=0.1$ level and the critical value method.
Part 1 of 6
State the appropriate null and alternate hypotheses.
\[
\begin{array}{l}
H_{0}: \mu_{1}=\mu_{2} \\
H_{1}: \mu_{1}> \mu_{2}
\end{array}
\]
This is a right-tailed $\quad \mathbf{v}$ test.
Part 2 of 6
How many degrees of freedom are there, using the simple method?
The degrees of freedom using the simple method is $\square$.
Answer
The degrees of freedom using the simple method is \(\boxed{85}\).
Step 1 ：State the appropriate null and alternate hypotheses. The null hypothesis \(H_{0}: \mu_{1}=\mu_{2}\) and the alternate hypothesis \(H_{1}: \mu_{1}>\mu_{2}\). This is a right-tailed test.
Step 2 ：Calculate the degrees of freedom using the simple method. The formula for degrees of freedom is \(df = n1 + n2 - 2\).
Step 3 ：Substitute the given values into the formula. Here, \(n1 = 41\) and \(n2 = 46\). So, \(df = 41 + 46 - 2\).
Step 4 ：Simplify the expression to find the degrees of freedom. So, \(df = 85\).
Step 5 ：The degrees of freedom using the simple method is \(\boxed{85}\).
