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.

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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.