Math / Data & Statistics

QuestionIQ scores: Scores on an IQ test are normally distributed. A sample of 8 IQ scores had standard deviation $s=6$ . (a) Construct a $95 \%$ confidence interval for the population standard deviation $\sigma$ . Round the answers to at least two decimal places. (b) The developer of the test claims that the population standard deviation is $\sigma=7$ . Does this confidence interval contradict this claim? Explain.
Part: $0 / 2$
Part 1 of 2
A $95 \%$ confidence interval for the population standard deviation is $\square$ $<\sigma<$ $\square$ .

Studdy Solution

STEP 1

1. The sample size is $n = 8$ .
2. The sample standard deviation is $s = 6$ .
3. The distribution of IQ scores is normal.
4. We are constructing a $95\%$ confidence interval for the population standard deviation $\sigma$ .

STEP 2

1. Determine the chi-square critical values.
2. Calculate the confidence interval for the population variance.
3. Calculate the confidence interval for the population standard deviation.

STEP 3

Determine the chi-square critical values for a $95\%$ confidence interval with $n-1 = 7$ degrees of freedom.
For a $95\%$ confidence interval, the significance level $\alpha = 0.05$ . The critical values are found using a chi-square distribution table:
$\chi^2_{\alpha/2, n-1} = \chi^2_{0.025, 7} \quad \text{and} \quad \chi^2_{1-\alpha/2, n-1} = \chi^2_{0.975, 7}$
Using a chi-square table, we find:
$\chi^2_{0.025, 7} \approx 2.167 \quad \text{and} \quad \chi^2_{0.975, 7} \approx 16.013$

STEP 4

Calculate the confidence interval for the population variance $\sigma^2$ .
The formula for the confidence interval for the variance is:
$\left( \frac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}}, \frac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}} \right)$
Substitute the values:
$\left( \frac{7 \times 6^2}{16.013}, \frac{7 \times 6^2}{2.167} \right)$
Calculate the values:
$\left( \frac{252}{16.013}, \frac{252}{2.167} \right) \approx (15.74, 116.29)$

STEP 5

Calculate the confidence interval for the population standard deviation $\sigma$ .
Take the square root of the variance interval:
$\left( \sqrt{15.74}, \sqrt{116.29} \right) \approx (3.97, 10.79)$
The $95\%$ confidence interval for the population standard deviation $\sigma$ is:
$\boxed{3.97 < \sigma < 10.79}$

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