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 10 years, the variance has been s² = 109.9. 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.

(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

b)

Test statistic,
Χ^2 = (n-1)*s^2/σ^2
Χ^2 = (10 - 1)*109.9/136.2
Χ^2 = 7.26

degrees of freedom = 10 -1 = 9

c)

Here s = 10.4833 and n = 10
df = 10 - 1 = 9

α = 1 - 0.9 = 0.1
The critical values for α = 0.1 and df = 9 are Χ^2(1-α/2,n-1) = 3.325 and Χ^2(α/2,n-1) = 16.919

CI = (9*10.4833^2/16.919 , 9*10.4833^2/3.325)
CI = (58.46 , 297.47)

Lower limit = 58.46
Upper limit = 297.47

## if you take s = 10.48 upto 2 decimal answer would be change CI = (58.42 , 297.29)

Lower limit = 58.42
Upper limit = 297.29