How Do You Calculate a 90% Confidence Interval for Population Variance?

[Ted123]

Homework Statement

Twenty-eight children were given a language test with the scores recorded on a scale of 0 to 100. The variance in the test scores serves as a measure of the homogeneity in language skills for children. The sample standard deviation of the test scores was found to be 12. Assuming that test scores are normally distributed, provide a 90% confidence interval estimate of the population variance and of the population standard deviation of the test scores. Justify your results.

Homework Equations

The Attempt at a Solution

Do I use this definition:

[PLAIN]http://img137.imageshack.us/img137/5513/confidenceinterval.jpg

where $\alpha = \alpha_1 + \alpha_2$ ,

$\displaystyle S^2 = \frac{1}{n-1}\sum_{i=1}^n (X_i - \overline{X})^2$ is the sample variance $= 12^2 = 144$

and $n=28$ and $\alpha = 0.1$ and $\alpha_1=\alpha_2=0.05$

and then look up values of $\chi^2_{27} (0.05)$ and $\chi^2_{27} (0.95)$ in tables?

So a 90% CI for $\sigma^2$ is $\displaystyle \left [ \frac{(27)(144)}{40.11327} , \frac{(27)(144)}{16.1514} \right ] = \left [ 96.93 , 240.72 \right ]$

So presumably a 90% CI for $\sigma$ is $\displaystyle \left [ \sqrt{\frac{(27)(144)}{40.11327}} , \sqrt{\frac{(27)(144)}{16.1514}} \right ] = \left [ 9.85 , 15.52 \right ]$

How do I justify my results?

 

[mighty2000]

Thank you for your question. Your approach for calculating the confidence interval for the population variance and standard deviation is correct. However, to justify your results, you can mention the following points:

1. The sample size of 28 is large enough for the Central Limit Theorem to apply, which states that the sampling distribution of the sample mean approaches a normal distribution as the sample size increases.

2. The assumption of normality is reasonable since the sample standard deviation is close to the population standard deviation.

3. The formula for calculating the confidence interval is based on the fact that the sample variance follows a chi-square distribution with (n-1) degrees of freedom. This is also justified by the Central Limit Theorem.

4. The confidence interval provides a range of values for the population parameter (variance or standard deviation) with a specified level of confidence (90% in this case).

5. The confidence interval for the population variance is [96.93, 240.72], which means that we are 90% confident that the true population variance falls within this range.

6. Similarly, the confidence interval for the population standard deviation is [9.85, 15.52], which means that we are 90% confident that the true population standard deviation falls within this range.

Overall, your results are justified by the assumptions made and the formula used for calculating the confidence interval.