Construct 98% confidence intervals for the a) population variance and the b)population standard deviation σ The number of hours of reserve capacity of 10 randomly selected automotive batteries. 1.71, 1.87, 1.58, 1.61, 1.78, 1.98, 1.36, 1.58, 1.47, 2.05, Assume the sample is taken from a normally distributed population.

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Yusuf Keller · Accepted Answer

Use STDEV function in excel to find the standard deviation. Then, s=0.2215.And the variance iss2=(0.2215)2=0.04906Step 2At 98% level,α=1−98%=1−0.98=0.02AndThe degree of freedom =n−1=10−1=9Step 3The critical values areX{α,n−1}2=X{0.02,9}2=19.679AndX{1−α,n−1}2=X{0.98,9}2 =2.5324Step 4The 98% confidence interval for variance is[(n−1)s2x{α,n−1}2≤σ2≤(n−1)s2x{1−α,n−1}2]=9(0.04906)19.679≤σ2≤9(0.04906)2.5324=0.4415419.679≤σ2≤0.441542.5324=0.022≤σ2≤0.174The 98% confidence interval for standard deviation is[(n−1)s2X{α,n−1}2≤σ≤(n−1)s2′X{1−α,n−1}2]=0.022≤σ≤0.174=0.150≤σ≤0.417

Construct 98% confidence intervals for the a) population variance and the b)population standard deviation sigma The number of hours of reserve capacit

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