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
Multiple Choice
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 a confidence interval for a variance \(\sigma^2\) is typically expressed as an inequality, for example, \(L < \sigma^2 < U\), where \(L\) and \(U\) are the lower and upper bounds derived from the \(\chi^2\) distribution.
Understand that the variance confidence interval is not symmetric because it is based on the \(\chi^2\) distribution, which is asymmetric, so the midpoint \((L + U)/2\) is not the point estimate for \(\sigma^2\).
Note that expressing the confidence interval as \(\sigma^2 = \text{point estimate} \pm \text{margin of error}\) assumes symmetry and a normal distribution, which does not hold for variance intervals.
Recall that the point estimate for variance is usually the sample variance \(s^2\), but it is not necessarily the midpoint of the confidence interval bounds.
Therefore, the correct way to express a confidence interval for variance is by stating the interval bounds explicitly, not by using a midpoint plus or minus a margin of error.
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