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 12 years, the variance has been s2 = 111.3. 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. Ho: σ2 < 136.2; H1: σ2 = 136.2 Ho: σ2 = 136.2; H1: σ2 < 136.2 Ho: σ2 = 136.2; H1: σ2 > 136.2 Ho: 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 normal population distribution. We assume a binomial population distribution. We assume a uniform population distribution. We assume a exponential population distribution.

(c) Find or estimate the P-value of the sample test statistic. P-value > 0.100 0.050 < P-value < 0.100 0.025 < P-value < 0.050 0.010 < P-value < 0.025 0.005 < P-value < 0.010 P-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 136.2 At 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 outside this interval. We are 90% confident that σ2 lies within this interval. We are 90% confident that σ2 lies below this interval. We are 90% confident that σ2 lies above this interval.

Answer 1

a)

level of significance =0.01

Ho: σ2 = 136.2; H1: σ2 < 136.2

b)

test statistic X2 =(n-1)s2/ σ2=

8.99

degree of freedom=n-1=11

We assume a normal population distribution.

c)

P-value > 0.100

d)

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

e)

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

f)

	here n =	12
	s2=	111.300
Critical value of chi square distribution for n-1=11 df and 90 % CI
Lower critical value χ2L=		4.575
Upper critical valueχ2U=		19.675
for Confidence interval of Variance:
Lower bound =(n-1)s2/χ2U=		62.23
Upper bound =(n-1)s2/χ2L=		267.62

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