Question

Sample of 77 subjects did a diet for 6 months. Their sample mean weight loss was 4.7 with a simple standard deviation of 7.16.
A second sample of 79 went on a different diet. Their mean weight loss was 2.6 with standard deviation of 5.90
Can you conclude that the mean weight loss differs? Use a=0.01
State the appropriate null and alternate hypothesis

Answer 1

Computer the p value round 4 places

Answer 2

Do we reject Hv0

Answer 3

To determine whether there is a significant difference in mean weight loss between the two diets, we can set up the null and alternative hypotheses as follows:

**Null Hypothesis (\(H_0\))**: There is no difference in mean weight loss between the two diets. Mathematically, this can be expressed as:
\[ H_0: \mu_1 = \mu_2 \]
where \(\mu_1\) is the mean weight loss for Diet 1 and \(\mu_2\) is the mean weight loss for Diet 2.

**Alternative Hypothesis (\(H_a\))**: There is a difference in mean weight loss between the two diets. This can be expressed as:
\[ H_a: \mu_1 \neq \mu_2 \]

This is a two-tailed test since we are looking for any difference, not specifying a direction.

Given the significance level (\(\alpha\)) of 0.01, you would proceed with a statistical test (e.g., a t-test for independent samples) to determine whether to reject or fail to reject the null hypothesis based on the calculated p-value and the test statistic.

Answer 4

To calculate the p-value for testing the difference in means between the two diets, we can use a two-sample t-test. First, we need to use the given data to compute the test statistic.

### Given Data
- Sample 1 (Diet 1):
- \( n_1 = 77 \)
- \( \bar{x}_1 = 4.7 \)
- \( s_1 = 7.16 \)

- Sample 2 (Diet 2):
- \( n_2 = 79 \)
- \( \bar{x}_2 = 2.6 \)
- \( s_2 = 5.90 \)

### Step 1: Calculate the pooled standard deviation (\(s_p\))
The pooled standard deviation is calculated using the formula:

\[
s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}}
\]

Calculating each component:

- \(s_1^2 = 7.16^2 = 51.4256\)
- \(s_2^2 = 5.90^2 = 34.81\)

Now, substituting the values:

\[
s_p = \sqrt{\frac{(77 - 1)(51.4256) + (79 - 1)(34.81)}{(77 + 79 - 2)}}
\]
\[
= \sqrt{\frac{(76 \cdot 51.4256) + (78 \cdot 34.81)}{154}}
\]

Calculating the numerators:

\[
76 \cdot 51.4256 = 3908.7936
\]
\[
78 \cdot 34.81 = 2713.18
\]

Adding these:

\[
3908.7936 + 2713.18 = 6621.9736
\]

Now, the pooled standard deviation:

\[
s_p = \sqrt{\frac{6621.9736}{154}} \approx \sqrt{43.02068} \approx 6.63
\]

### Step 2: Calculate the t-statistic

The t-statistic is calculated using the formula:

\[
t = \frac{\bar{x}_1 - \bar{x}_2}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}
\]

Substituting the values:

\[
t = \frac{4.7 - 2.6}{6.63 \sqrt{\frac{1}{77} + \frac{1}{79}}}
\]
\[
= \frac{2.1}{6.63 \sqrt{\frac{1.01299}{77}}}
\]
\[
= \frac{2.1}{6.63 \cdot 0.1131} \approx \frac{2.1}{0.74976} \approx 2.80
\]

### Step 3: Degrees of Freedom

The degrees of freedom for the test is given by:

\[
df = n_1 + n_2 - 2 = 77 + 79 - 2 = 154
\]

### Step 4: Find the p-value

Now we would look up the calculated \(t\) value in a t-distribution table or use a calculator/statistical software to find the p-value for \(|t| = 2.80\) with \(df = 154\).

Using software or an online calculator, you can find that:

\[
p \approx 0.0055 \, (\text{two-tailed})
\]

### Final Result
Thus, the p-value is approximately **0.0055**.

Answer 5

To determine whether to reject the null hypothesis (\(H_0\)), we need to compare the p-value to the significance level (\(\alpha\)) that you have set for the test.

### Given:
- p-value: **0.0055**
- Significance level (\(\alpha\)): **0.01**

### Decision Rule:
1. If the p-value is less than or equal to \(\alpha\), we reject the null hypothesis.
2. If the p-value is greater than \(\alpha\), we do not reject the null hypothesis.

### Conclusion:
Since the p-value (0.0055) is less than the significance level (0.01), we **reject the null hypothesis (\(H_0\))**.

This indicates that there is statistically significant evidence to conclude that there is a difference in mean weight loss between the two diets.