The 90% confidence interval estimate of a population standard deviation when a sample variance of 50 is obtained from a sample of 15 items is

A) 26.8 to 124.356.
B) 5.177 to 11.152.
C) 5.436 to 10.321.
D) 29.555 to 106.529.

Question

Kevin Galvan · Accepted Answer

Final Answer : C, Explanation :  The formula to calculate the confidence interval of the population standard deviation is:(√[ (n-1) s² / χ²_(α/2)(n-1)] , √[ (n-1) s² / χ²_(1-α/2)(n-1)])wheren = sample size,s² = sample variance,χ²_(α/2)(n-1) and χ²_(1-α/2)(n-1) are the chi-square values for the given level of confidence and degree of freedom (n-1).Substituting the values in the formula, we get(√[(15-1) 50 / χ²_(0.05/2)(14)], √[(15-1) 50 / χ²_(1-0.05/2)(14)])= (√[14*50 / 6.572], √[14*50 / 26.119])= (5.436, 10.321)Therefore, the 90% confidence interval estimate of a population standard deviation is 5.436 to 10.321. Option C is the correct answer.

The 90% confidence interval estimate of a population standard deviation when a sample variance of 50 is obtained from a sample of 15 items is

A) 26.8 to 124.356.
B) 5.177 to 11.152.
C) 5.436 to 10.321.
D) 29.555 to 106.529.

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Related questions

The 90% confidence interval estimate of a population standard deviation when a sample variance of 50 is obtained from a sample of 15 items isA) 26.8 to 124.356.B) 5.177 to 11.152.C) 5.436 to 10.321.D) 29.555 to 106.529.

Definitions

Learning Objectives

Learning Objectives

Related questions

The 90% confidence interval estimate of a population standard deviation when a sample variance of 50 is obtained from a sample of 15 items is

A) 26.8 to 124.356.
B) 5.177 to 11.152.
C) 5.436 to 10.321.
D) 29.555 to 106.529.