Let x represent the number of mountain climbers killed each year. The long-term variance of x is approximately 2 = 136. Suppose that for the past 6...

xxσ26s2116.2

(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 exponential population distribution.

We assume a binomial population distribution.

We assume a uniform population distribution.

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 below this interval.

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

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