The National Health Statistics Reports described a study in which a sample of 322 one-year-old baby boys were weighed. Their mean weight was 25.1 pounds with standard deviation 5.3 pounds. A pediatrician claims that the mean weight of one-year-old boys is greater than 25 pounds. Do the data provide convincing evidence that the pediatrician's claim is true? Use the a=0.05 level of significance and the critical value method with the Critical Values for the Student's t Distribution Table.

Question

Hv0 : [?]
Hv1 : [?]

Is this right/left or two-tailed test

Answer 1

To test the pediatrician's claim that the mean weight of one-year-old boys is greater than 25 pounds, we will set up our null and alternative hypotheses as follows:

• Null Hypothesis (H0): The mean weight of one-year-old boys is less than or equal to 25 pounds, i.e., \( \mu \leq 25 \).
• Alternative Hypothesis (H1): The mean weight of one-year-old boys is greater than 25 pounds, i.e., \( \mu > 25 \).

This is a one-tailed test because the alternative hypothesis is only testing if the mean is greater than a specified value (25 pounds), not if it is different in either direction.

Level of Significance

The level of significance, \( \alpha \), is given as 0.05.

Test Statistic

We will use the t-test for this hypothesis testing since we are dealing with a sample mean and do not know the population standard deviation.

The test statistic can be calculated using the formula for the t-statistic:

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

Where:

• \( \bar{x} \) = sample mean = 25.1 pounds
• \( \mu_0 \) = population mean under the null hypothesis = 25 pounds
• \( s \) = sample standard deviation = 5.3 pounds
• \( n \) = sample size = 322

Now plug in the values:

\[ t = \frac{25.1 - 25}{5.3 / \sqrt{322}} \]

Calculating the denominator:

\[ s / \sqrt{n} = 5.3 / \sqrt{322} \approx 5.3 / 17.94 \approx 0.295 \]

Now substituting back into the t formula:

\[ t = \frac{0.1}{0.295} \approx 0.338 \]

Critical Value

We need to find the critical t-value for a one-tailed test at \( \alpha = 0.05 \) with \( n - 1 = 322 - 1 = 321 \) degrees of freedom.

Using a t-distribution table or calculator, the critical t value for \( \alpha = 0.05 \) and 321 degrees of freedom is approximately:

\[ t_{critical} \approx 1.645 \]

Decision Rule

• Reject \( H_0 \) if \( t > t_{critical} \).
• If \( t \leq t_{critical} \), we fail to reject \( H_0 \).

Conclusion

Our calculated t value of \( 0.338 \) is less than the critical t value of \( 1.645 \). Therefore:

• We fail to reject the null hypothesis \( H_0 \).

Final Interpretation

The data do not provide convincing evidence at the \( \alpha = 0.05 \) level of significance to support the pediatrician's claim that the mean weight of one-year-old boys is greater than 25 pounds.