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Let's walk through the hypothesis test described in the first image.

### Step 1: Define the Hypotheses
- **Null hypothesis (H₀):** The mean weight of one-year-old boys is 25 pounds, \( \mu = 25 \).
- **Alternative hypothesis (H₁):** The mean weight differs from 25 pounds, \( \mu \neq 25 \).

This is a two-tailed test since the pediatrician claims the weight differs from 25 pounds, without specifying whether it's greater or less.

### Step 2: Given Data
- Sample size (n) = 332
- Sample mean (\( \bar{x} \)) = 25.5 pounds
- Sample standard deviation (s) = 5.3 pounds
- Significance level (\( \alpha \)) = 0.01

### Step 3: Test Statistic
We will use the t-test because the sample size is large, but the population standard deviation is unknown. The test statistic is calculated using the formula:

\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
\]

Where:
- \( \bar{x} = 25.5 \)
- \( \mu_0 = 25 \)
- \( s = 5.3 \)
- \( n = 332 \)

Substituting the values:

\[
t = \frac{25.5 - 25}{5.3 / \sqrt{332}} = \frac{0.5}{0.29064} \approx 1.72
\]

### Step 4: P-Value
Using a t-distribution with degrees of freedom \( df = n - 1 = 331 \), and a calculated test statistic \( t = 1.72 \), we can use a calculator or statistical software (such as the TI-84 Plus) to find the two-tailed p-value.

For \( t = 1.72 \) with 331 degrees of freedom, the p-value is approximately 0.086.

### Step 5: Conclusion
Compare the p-value to the significance level \( \alpha = 0.01 \). Since the p-value (0.086) is greater than \( \alpha = 0.01 \), **we fail to reject the null hypothesis**. Therefore, there is not enough evidence to support the pediatrician's claim that the mean weight differs from 25 pounds.

### Final Answer for Part (d):
- **Do not reject** the null hypothesis \( H_0 \).

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Would you like to discuss any specific details, or have any questions? Here are some related questions you might find useful:

1. How do you interpret the p-value in the context of hypothesis testing?
2. Why did we use the t-distribution instead of the z-distribution in this test?
3. What happens if the significance level \( \alpha \) is changed to 0.05?
4. What is the impact of sample size on the results of a hypothesis test?
5. How can we calculate the critical value for this two-tailed test?

**Tip**: Always ensure the sample size is large enough for the t-test assumptions, especially when the population standard deviation is unknown!

The National Health Statistics Reports described a study in which a sample of 332 one-year-old baby boys were weighed. Their mean weight was 25.5 pounds with standard deviation 5.3 pounds. A pediatrician claims that the mean weight for one-year-old boys differs from 25 pounds. Do the data provide convincing evidence that the pediatrician's claim is true? Use the alpha = 0.01 level of significance and the P-value method with the TI-84 Plus calculator.

Conduct a hypothesis test to determine if the mean weight of one-year-old boys differs from 25 pounds. The sample includes 332 baby boys, with a mean weight of 25.5 pounds and standard deviation of 5.3 pounds. We use the t-distribution and P-value method at a 0.01 significance level.

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