To address this problem, we're constructing a $90\%$ confidence interval for the population standard deviation $\sigma$ using the sample standard deviation $s = 8$ and a sample size of $n = 7$. Here are the steps to find the confidence interval for $\sigma$:

(a) Constructing the $90\%$ Confidence Interval for the Population Standard Deviation $\sigma$

1. Identify the Chi-square distribution: The formula for the confidence interval for the population variance ($\sigma^2$) using a sample is:

   $\frac{(n-1)s^2}{\chi^2_{\text{right}}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{\text{left}}}$

   Here:

   • $n = 7$ (sample size)

   • $s = 8$ (sample standard deviation)

   • Degrees of freedom $df = n - 1 = 6$

   • Confidence level is $90\%$, so the significance level $\alpha = 0.10$

   • For a $90\%$ confidence interval:

     • $\alpha/2 = 0.05$
     • Critical values for the Chi-square distribution:
       • $\chi^2_{\text{left}} = \chi^2_{0.05, 6}$
       • $\chi^2_{\text{right}} = \chi^2_{0.95, 6}$

2. Determine Chi-square critical values:

   • $\chi^2_{0.05, 6} = 12.592$ (upper-tail critical value)
   • $\chi^2_{0.95, 6} = 1.635$ (lower-tail critical value)

3. Plug values into the formula:

   Using the formula above, calculate the confidence interval for $\sigma^2$:

   $\frac{(6) \times (8^2)}{12.592} \leq \sigma^2 \leq \frac{(6) \times (8^2)}{1.635}$

   $\frac{384}{12.592} \leq \sigma^2 \leq \frac{384}{1.635}$

   $30.5 \leq \sigma^2 \leq 234.9$

4. Take the square root of each bound:

   $\sqrt{30.5} \leq \sigma \leq \sqrt{234.9}$

   $5.52 \leq \sigma \leq 15.33$

Answer for part (a):

The $90\%$ confidence interval for the population standard deviation $\sigma$ is: $5.52 \leq \sigma \leq 15.33$

(b) Does this Confidence Interval Contradict the Developer's Claim?

The developer of the test claims that the population standard deviation is $\sigma = 1$.

• From the computed confidence interval $5.52 \leq \sigma \leq 15.33$, the claimed value of $\sigma = 1$ does not lie within this interval.
• Hence, the confidence interval contradicts the developer's claim, suggesting that the population standard deviation is likely larger than $1$.

Let me know if you want any further details or explanations! Here are some related questions:

1. What does it mean when a confidence interval includes a certain value?
2. How is the Chi-square distribution used for constructing confidence intervals?
3. What are the implications if the developer's claim lies outside the confidence interval?
4. How would the confidence interval change with a larger sample size?
5. What would a different confidence level (like 95%) imply for the interval?

Tip: Confidence intervals provide a range within which we believe the true population parameter lies, given a certain level of confidence. Always check if the hypothesized parameter value falls within this range to validate claims!