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To solve the problem, we need to determine whether the "healthy" diet resulted in a significant difference in weight after three weeks. This can be done using a paired \( t \)-test since we have two related samples (before and after weights of the same subjects).

### Step 1: State the hypotheses
- **Null Hypothesis (H₀):** There is no difference in the mean weights before and after the diet. (\(\mu_d = 0\))
- **Alternative Hypothesis (H₁):** There is a difference in the mean weights before and after the diet. (\(\mu_d \neq 0\))

### Step 2: Calculate the differences
For each subject, calculate the difference between the "before" and "after" weights:

\[
\text{Difference (d)} = \text{Before} - \text{After}
\]

So, the differences for each subject are:

- A: \( 150 - 152 = -2 \)
- B: \( 195 - 190 = 5 \)
- C: \( 188 - 185 = 3 \)
- D: \( 197 - 191 = 6 \)
- E: \( 204 - 200 = 4 \)
- F: \( 175 - 170 = 5 \)
- G: \( 160 - 162 = -2 \)
- H: \( 180 - 179 = 1 \)

### Step 3: Calculate the mean and standard deviation of the differences
\[
\text{Mean of differences } (\overline{d}) = \frac{-2 + 5 + 3 + 6 + 4 + 5 - 2 + 1}{8} = \frac{20}{8} = 2.5
\]

\[
\text{Standard deviation of differences } (s_d) = \sqrt{\frac{\sum{(d_i - \overline{d})^2}}{n-1}}
\]

I'll calculate this in the next step.

### Step 4: Calculate the test statistic
\[
t = \frac{\overline{d}}{s_d / \sqrt{n}}
\]

### Step 5: Determine the critical value and make a decision
Using the \( t \)-distribution table and a 0.05 significance level for a two-tailed test with \( n-1 = 7 \) degrees of freedom, compare the calculated \( t \)-value with the critical \( t \)-value.

### Step 6: Conclusion
Depending on whether the \( t \)-value falls within the critical region, we will either reject or fail to reject the null hypothesis.

I'll proceed with the detailed calculations next.Here are the calculated values:

- **Mean of Differences (\(\overline{d}\))**: 2.5
- **Standard Deviation of Differences (\(s_d\))**: 3.1623
- **\(t\)-statistic**: 2.2361
- **Critical \(t\)-value**: 2.3646 (for a two-tailed test at the 0.05 significance level with 7 degrees of freedom)

### Step 6: Conclusion
The calculated \(t\)-statistic (2.2361) is slightly less than the critical \(t\)-value (2.3646). Since the \(t\)-statistic does not exceed the critical value, we **fail to reject the null hypothesis**.

### Final Conclusion:
At the 0.05 level of significance, there is not enough evidence to conclude that the diet resulted in a significant difference in weight.

---

Do you have any questions or would you like a more detailed explanation? Here are five related questions:

1. What would be the conclusion if the significance level were 0.10 instead of 0.05?
2. How is the paired \(t\)-test different from an independent \(t\)-test?
3. Can the same method be used if the sample size was larger, say 30 subjects?
4. What assumptions must be met to validly use a paired \(t\)-test?
5. How would the results change if one of the weight differences was an extreme outlier?

**Tip:** Always ensure that the data meets the assumptions of the test you are using; for a paired \(t\)-test, this includes the assumption that the differences are normally distributed.

Learn how to conduct a paired t-test to determine if there is a significant difference in weight before and after a diet intervention. Includes step-by-step calculations of mean differences, standard deviation, and the t-statistic. Suitable for advanced undergraduate students and researchers.

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