A random sample of size 

23

 from a normal distribution has standard deviation 

=*s*77

. Test 

=:*H*0σ59

 versus 

>:*H*1σ59

. Use the 

=α0.01

 level of significance.

To test the hypothesis about the population standard deviation, we use the chi-square (\(\chi^2\)) test for variance. Here's how you can perform this test:

### Step-by-Step Process

1. **State the Hypotheses:**

   - Null Hypothesis (\(H_0\)): \(\sigma = 59\)
   - Alternative Hypothesis (\(H_1\)): \(\sigma > 59\)

2. **Given Information:**

   - Sample size (\(n\)) = 23
   - Sample standard deviation (\(s\)) = 77
   - Population standard deviation under the null hypothesis (\(\sigma_0\)) = 59
   - 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 (\(\sigma_0^2\)):

   \[
   s^2 = 77^2 = 5929
   \]
   \[
   \sigma_0^2 = 59^2 = 3481
   \]

   Now, compute the test statistic:

   \[
   \chi^2 = \frac{(23-1) \cdot 5929}{3481} = \frac{22 \cdot 5929}{3481} \approx 37.4436
   \]

4. **Determine the Critical Value:**

   For a right-tailed chi-square test at \(\alpha = 0.01\) and \(n-1 = 22\) 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:
   \[
   =\text{CHISQ.INV.RT}(0.01, 22) \approx 40.2894
   \]

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 \( \chi^2 \approx 37.4436 \) is less than the critical value \( 40.2894 \), 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 population standard deviation is greater than 59.

### Answer

The test statistic is approximately \( 37.4436 \), and the critical value at \(\alpha = 0.01\) is approximately \( 40.2894 \). 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.

Learn how to perform hypothesis testing on population standard deviation using the chi-square test for variance. This detailed explanation covers a sample size of 23, sample standard deviation of 77, and a hypothesized population standard deviation of 59 at a significance level of 0.01.

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