Questions: Scores on an IQ test are normally distributed. A sample of 11 IQ scores had standard deviation s=8. Construct a 99% confidence interval for the population standard deviation σ. Round the answers to at least two decimal places. The developer of the test claims that the population standard deviation is σ=15. Does this confidence interval contradict this claim? Explain. 99% confidence interval for the population standard deviation is <σ<

Solution Steps

Given the sample standard deviation \( s = 8 \), the sample variance \( s^2 \) is calculated as follows:

\[ s^2 = 8^2 = 64 \]

To construct a \( 99\% \) confidence interval for the variance, we use the formula:

\[ \left( \frac{(n - 1)s^2}{\chi^2_{\alpha/2}}, \frac{(n - 1)s^2}{\chi^2_{1 - \alpha/2}} \right) \]

Substituting \( n = 11 \) and \( s^2 = 64 \):

\[ CI = \left( \frac{(11 - 1) \times 64}{\chi^2_{0.005}}, \frac{(11 - 1) \times 64}{\chi^2_{0.995}} \right) \]

Using the critical values from the chi-squared distribution, we find:

\[ CI = (25.41, 296.87) \]

To find the confidence interval for the standard deviation, we take the square root of the variance confidence interval:

\[ \text{Standard Deviation CI} = \left( \sqrt{25.41}, \sqrt{296.87} \right) \approx (5.04, 17.23) \]

The developer of the test claims that the population standard deviation is \( \sigma = 15 \). We check if this claim falls within the calculated confidence interval:

\[ (5.04, 17.23) \]

Since \( 15 \) lies within this interval, the confidence interval does not contradict the claim.

Final Answer