**Babies**: A sample of 

25

 one-year-old girls had a mean weight of 

24.1

 pounds with a standard deviation of 

4.3

 pounds. Assume that the population of weights is normally distributed. A pediatrician claims that the standard deviation of the weights of one-year-old girls is greater than 

5

 pounds. Do the data provide convincing evidence that the pediatrician's claim is true? Use the 

=α0.01

 level of significance.

To test the pediatrician's claim that the standard deviation of the weights of one-year-old girls is greater than 5 pounds, we can use the chi-square test for the population variance.

### Step-by-Step Process

1. **State the Hypotheses:**

   - Null Hypothesis (\(H_0\)): \(\sigma = 5\)
   - Alternative Hypothesis (\(H_1\)): \(\sigma > 5\)

2. **Given Information:**

   - Sample size (\(n\)) = 25
   - Sample standard deviation (\(s\)) = 4.3
   - Population standard deviation under the null hypothesis (\(\sigma_0\)) = 5
   - Significance level (\(\alpha\)) = 0.01

3. **Calculate the Test Statistic:**

   The test statistic for the chi-square test for variance is given by:

   \[
   \chi^2 = \frac{(n-1) \cdot s^2}{\sigma_0^2}
   \]

   Where:
   - \(n-1\) = degrees of freedom
   - \(s^2\) = sample variance
   - \(\sigma_0^2\) = hypothesized population variance

   First, calculate the sample variance (\(s^2\)) and the hypothesized population variance (\(\sigma_0^2\)):

   \[
   s^2 = 4.3^2 = 18.49
   \]
   \[
   \sigma_0^2 = 5^2 = 25
   \]

   Now, compute the test statistic:

   \[
   \chi^2 = \frac{(25-1) \cdot 18.49}{25} = \frac{24 \cdot 18.49}{25} = \frac{443.76}{25} \approx 17.75
   \]

4. **Determine the Critical Value:**

   For a right-tailed chi-square test at \(\alpha = 0.01\) and \(n-1 = 24\) degrees of freedom, we need to find the critical value from the chi-square distribution table or using a chi-square calculator.

   Using a chi-square table or Excel, you can find the critical value:

   In Excel:
   \[
   =\text{CHISQ.INV.RT}(0.01, 24) \approx 42.98
   \]

5. **Make the Decision:**

   - If the test statistic (\(\chi^2\)) is greater than the critical value, reject the null hypothesis.
   - If the test statistic (\(\chi^2\)) is less than or equal to the critical value, do not reject the null hypothesis.

   Since \( \chi^2 \approx 17.75 \) is less than the critical value \( 42.98 \), we do not reject the null hypothesis.

### Conclusion

Based on the sample data and using the chi-square test for variance at the \(\alpha = 0.01\) significance level, we do not have enough evidence to conclude that the standard deviation of the weights of one-year-old girls is greater than 5 pounds.

### Answer

The test statistic is approximately \( 17.75 \), and the critical value at \(\alpha = 0.01\) is approximately \( 42.98 \). Since the test statistic is less than the critical value, we do not reject the null hypothesis.

Would you like more details or have any questions?

Here are some questions you might want to ask:
1. How is the chi-square test for variance different from other hypothesis tests?
2. Why is the chi-square distribution used for testing variances?
3. What are the assumptions required for the chi-square test for variance?
4. How can we improve the accuracy of our hypothesis test?
5. Can you explain how to perform this test using software tools?

**Tip:** The chi-square test for variance is sensitive to deviations from normality. Ensure that your data is approximately normally distributed before applying this test for accurate results.

Explore whether the data provide convincing evidence that the standard deviation of one-year-old girls' weights is greater than 5 pounds using a chi-square test for variance. Detailed step-by-step solution with sample size, standard deviation, and significance level provided.

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