Answer :
To find the lower bound for a 98% confidence interval of the standard deviation based on the given sample, we can use the Chi-Square distribution. Here's a step-by-step explanation:
Determine the sample variance and size:
- The sample variance [tex]s^2[/tex] is given as 0.0004.
- The sample size [tex]n[/tex] is 16.
Find the degrees of freedom (df):
- The degrees of freedom are calculated as [tex]n - 1[/tex].
- For this sample, [tex]df = 16 - 1 = 15[/tex].
Determine the Chi-Square distribution critical value:
- To find the lower bound, we need to find the critical value from the Chi-Square distribution table that corresponds to the lower tail of the 98% confidence level.
- For a 98% confidence level, the lower critical value [tex]\chi^2_{1-\alpha/2}[/tex] with 15 degrees of freedom is needed. This equates to [tex]\alpha = 0.02[/tex], so we need [tex]\chi^2_{0.99}[/tex].
- Using a Chi-Square distribution table or calculator, [tex]\chi^2_{0.99} \approx 28.299[/tex]. (Note: Tables may vary slightly, so ensure to use the correct value based on the source)
Calculate the lower bound of the confidence interval for the standard deviation:
[tex]\text{Lower bound} = \sqrt{\frac{(n-1) \cdot s^2}{\chi^2_{0.99}}}[/tex]
[tex]\text{Lower bound} = \sqrt{\frac{15 \cdot 0.0004}{28.299}}[/tex]
[tex]\text{Lower bound} = \sqrt{\frac{0.006}{28.299}}[/tex]
[tex]\text{Lower bound} = \sqrt{0.0002122}[/tex]
[tex]\text{Lower bound} \approx 0.015[/tex]
Therefore, the lower bound for the 98% confidence interval of the standard deviation is approximately 0.015. This result gives you the smallest possible standard deviation accounting for the variability seen in the bottle samples with 98% confidence.
