Calculating Confidence Interval for Population Standard Deviation Using Chi-square Distribution
The image provided appears to show a statistics problem about determining the confidence interval for a population standard deviation, given a sample standard deviation and the size of the sample. Unfortunately, the image is not clear enough to read all the text, but I can provide you with a general approach to solve a problem like this based on what I can infer from the image.
To find the confidence interval for the population variance (and hence the standard deviation), you need to use the Chi-square distribution because the sample is drawn from a normally distributed population.
The formula to compute a \( (1-\alpha) \times 100\% \) confidence interval for a population variance \( \sigma^2 \) based on a sample variance \( s^2 \) and a sample size of \( n \) is:
\[
\left( \frac{(n-1) s^2}{\chi^2_{\alpha/2, n-1}}, \frac{(n-1) s^2}{\chi^2_{1-\alpha/2, n-1}} \right)
\]
Where:
- \( \chi^2_{\alpha/2, n-1} \) is the critical value for the Chi-square distribution with \( n-1 \) degrees of freedom that cuts off an area of \( \alpha/2 \) to the right-hand side.
- \( \chi^2_{1-\alpha/2, n-1} \) is the critical value for the Chi-square distribution with \( n-1 \) degrees of freedom that cuts off an area of \( \alpha/2 \) to the left-hand side.
Once you have the confidence interval for the variance, take the square root of each endpoint to get the confidence interval for the standard deviation.
Since the actual numbers and confidence level aren't readable in the image, please input the given sample size, sample standard deviation, and confidence level to perform the appropriate calculations.
Can you provide clearer information or the values so that I can help you calculate the confidence interval for the population standard deviation?
