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 11 years, the variance has been s² = 112.0. 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?


State the null and alternate hypotheses.

H_o: σ² = 136.2; H₁: σ² ≠ 136.2H_o: σ² = 136.2; H₁: σ² < 136.2    H_o: σ² < 136.2; H₁: σ² = 136.2H_o: σ² = 136.2; H₁: σ² > 136.2

(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?


What assumptions are you making about the original distribution?

We assume a binomial population distribution.We assume a normal population distribution.    We assume a exponential population distribution.We assume a uniform population distribution.

(c) Find or estimate the P-value of the sample test statistic.

P-value > 0.1000.050 < P-value < 0.100    0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?

Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis.    Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 1% level of significance, there is insufficient evidence to conclude that the variance for number of mountain climber deaths is less than 136.2At the 1% level of significance, there is sufficient evidence to conclude that the variance for number of mountain climber deaths is less than 136.2    

(f) Find the requested confidence interval for the population variance. (Round your answers to two decimal places.)

lower limit
upper limit

Interpret the results in the context of the application.

We are 90% confident that σ² lies within this interval.We are 90% confident that σ² lies above this interval.    We are 90% confident that σ² lies outside this interval.We are 90% confident that σ² lies below this interval.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 11 years, the variance has been s² = 112.0. 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?


State the null and alternate hypotheses.

H_o: σ² = 136.2; H₁: σ² ≠ 136.2H_o: σ² = 136.2; H₁: σ² < 136.2    H_o: σ² < 136.2; H₁: σ² = 136.2H_o: σ² = 136.2; H₁: σ² > 136.2

(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?


What assumptions are you making about the original distribution?

We assume a binomial population distribution.We assume a normal population distribution.    We assume a exponential population distribution.We assume a uniform population distribution.

(c) Find or estimate the P-value of the sample test statistic.

P-value > 0.1000.050 < P-value < 0.100    0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?

Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis.    Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 1% level of significance, there is insufficient evidence to conclude that the variance for number of mountain climber deaths is less than 136.2At the 1% level of significance, there is sufficient evidence to conclude that the variance for number of mountain climber deaths is less than 136.2    

(f) Find the requested confidence interval for the population variance. (Round your answers to two decimal places.)

lower limit
upper limit

Interpret the results in the context of the application.

We are 90% confident that σ² lies within this interval.We are 90% confident that σ² lies above this interval.    We are 90% confident that σ² lies outside this interval.We are 90% confident that σ² lies below this interval.

Answer 1

Question 1

H_o: σ² = 136.2; H₁: σ² < 136.2   

α = 0.01

Test Statistic :-

χ² = ( ( 11-1 ) * 112 ) / 136.2
χ² = 8.22

degrees of freedom = n - 1 = 11 - 1 = 10

We assume a normal population distribution.  

Decision based on P value
P value = P ( χ² > 8.2232 )
P value = 0.393

P-value > 0.100

Reject null hypothesis if P value < α = 0.01
Since P value = 0.393 > 0.01, hence we fail to reject the null hypothesis
Conclusion :- We Fail to Reject H0
Since the P-value > α, we fail to reject the null hypothesis.

At the 1% level of significance, there is insufficient evidence to conclude that the variance for number of mountain climber deaths is less than 136.2

((n-1)S² / χ² (0.1/2)) < σ² < ((n-1)S² / χ² (1 - 0.1/2) )
(( 11-1 ) 112 / χ² (0.1/2) ) < σ² < ((11-1)112 / χ² (1 - 0.1/2) )
χ² (0.1/2) = 18.307
χ² (1 - 0.1/2) ) = 3.9403
Lower Limit = (( 11-1 ) 112^2 / χ² (0.1/2) ) = 61.1788
Upper Limit = (( 11-1 ) 112^2 / χ² (0.1/2) ) = 284.2423
90% Confidence interval is ( 61.1788 , 284.2423 )
( 61.18 < σ² < 284.24 )
We are 90% confident that σ² lies within this interval.