Final answer:
The construction of a 99% confidence interval for the population variance involves using the chi-square distribution and degrees of freedom, which is not accurately reflected in the multiple-choice options given.
Step-by-step explanation:
The question concerns constructing a 99% confidence interval for the population variance of the weights of mattresses given a sample variance of 1.48 from 6 randomly selected mattresses. To find the confidence interval for the variance, we use the chi-square distribution since the sample size is small.
Firstly, we need to determine the chi-square values that correspond to the 99% confidence interval for our degrees of freedom, which is n-1 in this case. For 6 mattresses, our degrees of freedom would be 5. Using a chi-square distribution table or calculator, we find the critical values for the lower and upper tails of the distribution that capture 99% of the middle area.
The formula to calculate the confidence interval for the variance is given by:
- Variance Lower Limit = (n-1) * s^2 / χ^2 upper
- Variance Upper Limit = (n-1) * s^2 / χ^2 lower
Where n is the sample size, s^2 is the sample variance, and χ^2 lower and χ^2 upper are the chi-square values for the lower and upper critical values, respectively.
After computing these, we can find the range that constitutes the confidence interval for the population variance, and it will not be simply ±0.26, ±0.60, ±1.82, or ±3.66; it needs to be computed specifically using the chi-square critical values. Therefore, none of the multiple-choice options provided (a, b, c, d) are accurate as they do not reflect the correct calculation methodology for a confidence interval of a variance.