Use technology to construct the confidence intervals for the population variance sigmaσsquared2 and the population standard deviation sigmaσ. Assume the sample is taken from a normally distributed population.
cequals=0.990.99​, ssquared2equals=12.2512.25​, nequals=2525
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Part 1
The confidence interval for the population variance is

To construct the confidence interval for the population variance \(\sigma^2\) and standard deviation \(\sigma\) based on the provided sample variance \(s^2 = 12.25\), sample size \(n = 25\), and confidence level \(c = 0.99\), we will use the Chi-squared distribution.

### Steps to Calculate the Confidence Interval for Variance

1. **Determine the Degrees of Freedom:**
   \[
   df = n - 1 = 25 - 1 = 24
   \]

2. **Find the Chi-squared critical values:**
   - For a 99% confidence level, the significance level \(\alpha = 1 - 0.99 = 0.01\).
   - The critical values are found using the Chi-squared distribution:
     - \(\chi^2_{\alpha/2}\) for \(df = 24\)
     - \(\chi^2_{1 - \alpha/2}\) for \(df = 24\)

   Using a Chi-squared table or calculator:
   - \(\chi^2_{0.005, 24} \approx 42.98\)
   - \(\chi^2_{0.995, 24} \approx 9.59\)

3. **Calculate the Confidence Interval for Variance:**
   \[
   \left( \frac{(n-1)s^2}{\chi^2_{1 - \alpha/2}}, \frac{(n-1)s^2}{\chi^2_{\alpha/2}} \right)
   \]

   Plugging in the values:
   \[
   \left( \frac{24 \times 12.25}{42.98}, \frac{24 \times 12.25}{9.59} \right)
   \]

   Calculating the limits:
   - Lower limit:
   \[
   \frac{24 \times 12.25}{42.98} \approx 6.87
   \]
   - Upper limit:
   \[
   \frac{24 \times 12.25}{9.59} \approx 30.66
   \]

### Final Confidence Interval for Variance
\[
\text{Confidence interval for } \sigma^2: (6.87, 30.66)
\]

### Confidence Interval for Standard Deviation
To find the confidence interval for the population standard deviation \(\sigma\), simply take the square roots of the variance limits:
\[
\left( \sqrt{6.87}, \sqrt{30.66} \right) \approx (2.62, 5.53)
\]

### Final Confidence Interval for Standard Deviation
\[
\text{Confidence interval for } \sigma: (2.62, 5.53)
\]

Would you like more details or have any questions about this process? Here are some related questions for further exploration:

1. How does the choice of confidence level affect the width of the confidence interval?
2. What assumptions must be met for using the Chi-squared distribution in this context?
3. How can we interpret the confidence intervals for variance and standard deviation?
4. What are the implications of a larger sample size on the confidence intervals?
5. How does the method for calculating confidence intervals differ for non-normal distributions? 

**Tip:** Always check if the underlying assumptions of your statistical methods are met before interpreting results.

Use technology to construct the confidence intervals for the population variance σ² and the population standard deviation σ. Assume the sample is taken from a normally distributed population. c = 0.99, s² = 12.25, n = 25.

Learn how to construct 99% confidence intervals for the population variance and standard deviation using a sample variance of 12.25 and sample size of 25. This problem explores key concepts in statistics, including the Chi-squared distribution and confidence interval calculations. Suitable for undergraduate students.

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