Question
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.
Hv0 : [?]
Hv1: [?]
This is a [left/right or two tailed?] test
Hv0 : [?]
Hv1: [?]
This is a [left/right or two tailed?] test
Answers
Answer
Compute the p-value. Round the answer to four decimal places.
Answer
Determine whether to reject hv0.
Answered by
GPT-4o mini
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.
### 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.
Answered by
GPT-4o mini
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:
```python
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:
\[
\text{P-value} \approx 0.0477
\]
- 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:
```python
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:
\[
\text{P-value} \approx 0.0477
\]
Answered by
GPT-4o mini
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.
\[
\alpha = 0.01
\]
### 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.
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