Multiple Choice
A delivery service tracks the weights of its packages. A sample of 20 packages has a variance of 4.5 lbs2. Construct a 95% conf. int. for the population variance. Assume a normal distribution.
8. Sampling Distributions & Confidence Intervals: Proportion
Confidence Intervals for Population Variance
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What is wrong with expressing the confidence interval as ?
A
5.1 is not the midpoint between 3.8 and 6.4.
B
The values 3.8 and 6.4 are impossible because variance must be less than 3.
C
The point estimate for σ2 is not the midpoint of a confidence interval and 1.3 is not a margin of error since the distribution is asymmetric.
D
Confidence intervals can only be written for means or proportion, not for variance.
Verified step by step guidance1
Recognize that the confidence interval for the variance \(\sigma^2\) is typically derived using the \(\chi^2\) (chi-square) distribution, which is asymmetric, not symmetric like the normal distribution.
Understand that because of this asymmetry, the confidence interval for \(\sigma^2\) is expressed as an inequality: \(L < \sigma^2 < U\), where \(L\) and \(U\) are the lower and upper bounds calculated from the chi-square distribution quantiles.
Note that the point estimate for \(\sigma^2\) (usually the sample variance) is not necessarily the midpoint of the confidence interval bounds \(L\) and \(U\), so writing \(\sigma^2 = 5.1 \pm 1.3\) is incorrect if 5.1 is not the midpoint of 3.8 and 6.4.
Recognize that the margin of error concept (the \(\pm\) part) applies to symmetric intervals, such as those for means or proportions, but does not apply directly to variance confidence intervals due to the skewness of the \(\chi^2\) distribution.
Therefore, the correct way to express a confidence interval for variance is as a range between two values, not as a point estimate plus or minus a margin of error.
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