Are low-fat diets or low-carb diets more effective for weight loss? A sample of 77 subjects went on a low-carbohydrate diet for six months. At the end of that time, the sample mean weight loss was 4.7 kilograms with a sample standard deviation of 7.16 kilograms. A second sample of 79 subjects went on a low-fat diet. Their sample mean weight loss was 2.6 kilograms with a standard deviation of 5.90 kilograms. Can you conclude that the mean weight loss differs between the two diets? Let uv1 denote the mean weight lost on the low-carb diet and uv2 denote the mean weight lost on the low-fat diet. Use the a=0.01 level.

Question

Hv0 : [?]
Hv1: [?]
This is a [left/right or two tailed?] test

Answer 1

To determine whether there is a significant difference in mean weight loss between the low-carbohydrate and low-fat diets, we need to formulate our hypotheses and determine the type of test.

Step 1: Set up the hypotheses

• Null Hypothesis (\(H_0\)): There is no difference in mean weight loss between the two diets. \[ H_0: \mu_1 - \mu_2 = 0 \]

• Alternative Hypothesis (\(H_1\)): There is a difference in mean weight loss between the two diets. \[ H_1: \mu_1 - \mu_2 \neq 0 \]

Where:

• \(\mu_1\) is the mean weight lost on the low-carb diet.
• \(\mu_2\) is the mean weight lost on the low-fat diet.

Step 2: Determine the test type

This is a two-tailed test because we are looking for any difference between the two means (not specifying whether one is greater than the other).

Summary of the hypotheses

• Null Hypothesis (\(H_0\)): \(\mu_1 - \mu_2 = 0\) (no difference in mean weight loss)
• Alternative Hypothesis (\(H_1\): \(\mu_1 - \mu_2 \neq 0\) (there is a difference in mean weight loss)
• Type of test: Two-tailed test

Next steps

You would then perform a statistical test (like a t-test or z-test, depending on your sample size and if the population standard deviations are known) to determine if you can reject \(H_0\) in favor of \(H_1\) at the \(\alpha = 0.01\) significance level.

Answer 2

Compute the p-value. Round the answer to four decimal places.

Answer 3

To compute the p-value for the difference in mean weight loss between the two diets, we can use a two-sample t-test. Given the following parameters:

• Low-carbohydrate diet:

  • Sample size (\(n_1\)) = 77
  • Sample mean (\(\bar{x}_1\)) = 4.7 kg
  • Sample standard deviation (\(s_1\)) = 7.16 kg

• Low-fat diet:

  • Sample size (\(n_2\)) = 79
  • Sample mean (\(\bar{x}_2\)) = 2.6 kg
  • Sample standard deviation (\(s_2\)) = 5.90 kg

Step 1: Calculate the test statistic

The formula for the t-statistic in a two-sample t-test is:

\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]

Substituting in the numbers:

\[ t = \frac{4.7 - 2.6}{\sqrt{\frac{7.16^2}{77} + \frac{5.90^2}{79}}} \]

Calculating the variances:

\[ s_1^2 = 7.16^2 = 51.4256 \] \[ s_2^2 = 5.90^2 = 34.81 \]

Now substituting into the formula:

\[ t = \frac{4.7 - 2.6}{\sqrt{\frac{51.4256}{77} + \frac{34.81}{79}}} \]

Calculating the fractions:

\[ \frac{51.4256}{77} \approx 0.6673 \] \[ \frac{34.81}{79} \approx 0.4402 \]

Now sum these up:

\[ 0.6673 + 0.4402 \approx 1.1075 \]

Now find the square root:

\[ \sqrt{1.1075} \approx 1.051 \]

Finally, calculate the t-statistic:

\[ t = \frac{2.1}{1.051} \approx 1.996 \]

Step 2: Degrees of freedom

To find the degrees of freedom for a two-sample t-test, use:

\[ df = \left(\frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}}\right) \]

Calculating:

• Numerator: \[ \left(0.6673 + 0.4402\right)^2 \approx (1.1075)^2 \approx 1.2283 \]

• Denominator: \[ \frac{(0.6673)^2}{76} + \frac{(0.4402)^2}{78} = \frac{0.4456}{76} + \frac{0.1935}{78} \approx 0.00586 + 0.00248 = 0.00834 \]

Thus,

\[ df \approx \frac{1.2283}{0.00834} \approx 147.451 \approx 147 \]

Step 3: Find the p-value

Now we can use the t-statistic \(t \approx 1.996\) and degrees of freedom \(df \approx 147\) to find the p-value.

Using a t-distribution table, or more commonly using software or a calculator:

1. Look up \(t = 1.996\) for \(df = 147\).
2. For a two-tailed test, multiply the obtained p-value by 2.

Using Python, R, or an online calculator, we would find:

from scipy.stats import t

# Compute the p-value
p_value = 2 * (1 - t.cdf(1.996, df=147))

Calculating gives a p-value:

P-value ≈ 0.0477 (rounded to four decimal places)

Thus, the final result for the p-value is:

\[ \text{P-value} \approx 0.0477 \]

Answer 4

Determine whether to reject hv0.

Answer 5

To determine whether to reject the null hypothesis (\(H_0\)), we compare the computed p-value with the significance level \(\alpha\) that was chosen for the test. In this case, the significance level is:

\[ \alpha = 0.01 \]

Step 1: Compare the p-value to \(\alpha\)

• Computed p-value: Approximately \(0.0477\)
• Significance Level: \(0.01\)

Step 2: Decision Rule

• If the p-value is less than \(\alpha\) (p-value < 0.01), then we reject the null hypothesis.
• If the p-value is greater than or equal to \(\alpha\) (p-value ≥ 0.01), then we fail to reject the null hypothesis.

Step 3: Conclusion

Since \(0.0477 > 0.01\):

\[ \text{We fail to reject } H_0. \]

Interpretation

This means that there is not enough evidence at the \(\alpha = 0.01\) significance level to conclude that there is a significant difference in mean weight loss between the low-carbohydrate diet and the low-fat diet.