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 = 109.2. 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?


State the null and alternate hypotheses.

H_o: σ² = 136.2; H₁: σ² < 136.2 H_o: σ² < 136.2; H₁: σ² = 136.2     H_o: σ² = 136.2; H₁: σ² > 136.2 H_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 uniform population distribution. We assume a binomial population distribution.     We assume a exponential population distribution. We assume a normal 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 above this interval.     We are 90% confident that σ² lies below this interval. We are 90% confident that σ² lies within this interval.

Answer 1

a) The level of significance = 0.01

H₀: = 136.2

H₁: < 136.2

b) The test statistic = (n - 1)s²/

                                     = 10 * 109.2/136.2

                                     = 8.02

DF = 11 - 1 = 10

We assume a normal population distribution.

c) P-value = P( < 8.02)

                 = 1 - P( > 8.02)

                 = 1 - 0.6269

                 = 0.3731

P-value > 0.100

d) Since P-value > , 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) At 90% confidence interval the critical values are = = 18.307

                                                                                  = = 3.94

The 90% confidence interval for population variance is

(n - 1)s^2/ < < (n - 1)s^2/

= 10 * 109.2/18.307 < < 10 * 109.2/3.94

= 59.65 < < 277.16

The lower limit = 59.65

The upper limit = 277.16

We are 90% confident that lies within this interval.