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 $H_0$.

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!