Question

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

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)s²/σ²₀ where

n = sample size

s = sample standard deviation  

Calculations

χ² = (8 x 116.2)/136.2

   = 6.83 ANSWER 2

χ² ~ χ²_{n – 1}

So, degrees of freedom = 8 ANSWER 3

Part (f)

100(1 - α) % Confidence Interval for σ² is: [{(n - 1)s²/χ²_{n – 1, α/2}}, {(n - 1)s²/χ²_{n – 1, 1 - α/2}}] where

χ²_{n – 1, α/2} and χ²_{n – 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,

χ²_{n – 1, α/2} = χ²_8,,0.05 = 15.5073

and χ²_{n – 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