Question

Let x represent the number of mountain climbers killed each year. The long-term variance of x...

Let x represent the number of mountain climbers killed each year. The long-term variance of x is approximately σ2 = 136.2. Suppose that for the past 9 years, the variance has been s2 = 116.2. Use a 1% level of significance to test the claim that the recent variance for number of mountain-climber deaths is less than 136.2. Find a 90% confidence interval for the population variance.

(a) What is the level of significance?

(b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)


What are the degrees of freedom?
(f) Find the requested confidence interval for the population variance. (Round your answers to two decimal places.)

lower limit
upper limit    

Homework Answers

Answer #1

Part (a)

The level of significance = 1% or 0.01 ANSWER 1

Part (b)

Chi-square statistic for the sample = χ2 = (n - 1)s220 where

n = sample size

s = sample standard deviation  

Calculations

χ2 = (8 x 116.2)/136.2

   = 6.83 ANSWER 2

χ2 ~ χ2n – 1

So, degrees of freedom = 8 ANSWER 3

Part (f)

100(1 - α) % Confidence Interval for σ2 is: [{(n - 1)s22n – 1, α/2}, {(n - 1)s22n – 1, 1 - α/2}] where

χ2n – 1, α/2 and χ2n – 1, 1 - α/2 are respectively upper and lower (α /2)% point of Chi-square distribution with (n - 1) degrees of freedom, s = sample standard deviation and n = sample size.

Now, given 90% confidence interval, α = 0.1. So,

χ2n – 1, α/2 = χ28,,0.05 = 15.5073

and χ2n – 1, 1 - α/2 = 2.7326

[Both are are obtained using Excel Function, Statistical, CHIINV(Probability, Deg_freedom) which gives t for which P(chisquae df > t) = given probability]

So, 90% confidence interval for the population variance is:

[{(8 x 116.2)/15.5073}, {(8 x 116.2)/2.7326}

= [59.95, 340.19]

i.e., Lower Limit: 59.95 and Upper Limit: 340.19 ANSWER 4

DONE

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