Question #244679

The standard deviation of the apples surveyed from a sample of measurements taken from 11 apples is found to be 4.934 grams. Find the 95% confidence intervals for the standard deviation of these weights.


1
Expert's answer
2021-09-30T23:45:58-0400

The standard deviation of the apples surveyed from a sample of measurements taken from 11 apples is found to be 4.934 grams. Find the 95% confidence intervals for the standard deviation of these weights.

Find the

Lower and Upper class boundary values of Chi-square

a.)Lower = 3.25; Upper = 20.48

b.)Lower = 20.48; Upper = 3.25

c.)Lower = 21.92; Upper = 3.25

Solve the Confidence Intervals.

a.)11.8869 ≤ σ ≤ 74.9057

b.)3.4477 ≤ σ ≤8.6548

c.)3.1088 ≤ σ ≤10.6163


From chi square tables for n -1 = 10 degrees of freedom, we have here:

P(χ2<20.48)=0.975P(χ2<3.25)=0.025P(χ^2 < 20.48) = 0.975 \\ P(χ^2 < 3.25) = 0.025

Therefore, the a.)Lower = 3.25; Upper = 20.48 is the correct answer here.

The confidence interval here is obtained as:

(n1)s2χU2<σ<(n1)s2χL2(10×4.934220.48<σ<10×4.93423.253.4477<σ<8.6548\sqrt{\frac{(n-1)s^2}{χ^2_U}} < \sigma < \sqrt{\frac{(n-1)s^2}{χ^2_L}} \\ \sqrt{\frac{(10 \times 4.934^2}{20.48}} < \sigma < \sqrt{\frac{10 \times 4.934^2}{3.25}} \\ 3.4477< \sigma < 8.6548

b) is the correct answer.


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