IQ scores: Scores on an IQ test are normally distributed. A sample of 

13

 IQ scores had standard deviation =

s

12

.



(a) Construct a 95

%

 confidence interval for the population standard deviation σ

. Round the answers to at least two decimal places.



(b) The developer of the test claims that the population standard deviation is =

σ

11

. Does this confidence interval contradict this claim? Explain.

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A 95

%

 confidence interval for the population standard deviation is 

<

<

σ

 

.





To solve this problem, we'll first use the formula for constructing a confidence interval for the population standard deviation (\(\sigma\)) based on the sample standard deviation (\(s\)) and the Chi-square (\(\chi^2\)) distribution. Let's go step by step.

### Part (a) – Constructing a 95% Confidence Interval for \(\sigma\)

The confidence interval for the population standard deviation \(\sigma\) is given by the following formula:

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

Where:
- \( n \) = sample size = 13
- \( s \) = sample standard deviation = 12
- \( \chi^2_{\text{right}} \) and \( \chi^2_{\text{left}} \) are the critical values from the Chi-square distribution for the desired confidence level and degrees of freedom \( df = n - 1 \) = 12.

#### Steps:
1. Find the Chi-square critical values for a 95% confidence interval with \( df = 12 \).
   - For the left tail: \(\chi^2_{\text{left}} = \chi^2_{0.025, 12}\)
   - For the right tail: \(\chi^2_{\text{right}} = \chi^2_{0.975, 12}\)

2. Plug these values into the formula to get the confidence interval for \(\sigma\).

Let's compute these values:

- For a 95% confidence interval with \(df = 12\), the Chi-square critical values are:
  - \(\chi^2_{0.025, 12} \approx 4.404\)
  - \(\chi^2_{0.975, 12} \approx 23.337\)

Now, the confidence interval is calculated as:

\[
\left( \frac{(13-1) \times 12^2}{23.337}, \frac{(13-1) \times 12^2}{4.404} \right)
\]

Simplifying:

\[
\left( \frac{12 \times 144}{23.337}, \frac{12 \times 144}{4.404} \right) = \left( \frac{1728}{23.337}, \frac{1728}{4.404} \right)
\]

Finally, the interval is approximately:

\[
(74.07, 392.46)
\]

Thus, the confidence interval for \(\sigma^2\) is \((74.07, 392.46)\).

Taking the square root of these values gives the confidence interval for \(\sigma\):

\[
\sigma \in (8.60, 19.81)
\]

### Part (b) – Does this Confidence Interval Contradict the Claim?

The developer claims that the population standard deviation \(\sigma\) is 11.

Since the 95% confidence interval for the population standard deviation is \((8.60, 19.81)\), the value of 11 falls within this interval. Therefore, **the confidence interval does not contradict the developer's claim** because the claimed value of \(\sigma = 11\) is a plausible value given the data.

---

Would you like further details or clarifications?

Here are 5 relative questions for practice:
1. How is a confidence interval for population variance different from that of population mean?
2. What is the relationship between confidence level and the width of a confidence interval?
3. How can increasing the sample size affect the confidence interval for standard deviation?
4. Why do we use the Chi-square distribution for confidence intervals of variance and standard deviation?
5. What are the assumptions behind using Chi-square distribution in this context?

**Tip:** Always ensure the assumptions for a Chi-square test hold, particularly that the data is normally distributed, when constructing confidence intervals for variance or standard deviation.

IQ scores: Scores on an IQ test are normally distributed. A sample of 13 IQ scores had standard deviation s = 12. (a) Construct a 95% confidence interval for the population standard deviation σ. Round the answers to at least two decimal places. (b) The developer of the test claims that the population standard deviation is σ = 11. Does this confidence interval contradict this claim? Explain.

Learn how to construct a 95% confidence interval for the population standard deviation in IQ scores using a sample of 13 scores with standard deviation s = 12. Explore statistical concepts such as the Chi-Square distribution and confidence intervals. Find out if this interval contradicts the claim that the population standard deviation is 11.

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