A magazine includes a report on the energy costs per year for 32-inch liquid crystal display (LCD) televisions. The article states that
1414
randomly selected 32-inch LCD televisions have a sample standard deviation of
$3.843.84.
Assume the sample is taken from a normally distributed population. Construct
9898%
confidence intervals for (a) the population variance
sigmaσsquared2
and (b) the population standard deviation
sigmaσ.
Interpret the results.
(a) The confidence interval for the population variance is
(nothing,nothing).
(Round to two decimal places as needed.)
Given that sample size n = 14
alpha level = 1- confidence level = 1-0.98 = 0.02
sample standard deviation s = 3.84
(A) confidence interval for variance
(B) Confidence interval for standard deviation is given as
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