Let x represent the number of mountain climbers killedeach year. The long-term variance of x is approximately?² = 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 variancefor number of mountain-climber deaths is less than 136.2. Find a90% confidence interval for the population variance. (a) What isthe level of significance?


State the null and alternate hypotheses.

H_o: ?² = 136.2;H₁: ?² < 136.2H_o: ?² < 136.2;H₁: ?² =136.2     H_o:?² = 136.2; H₁:?² > 136.2 H_o:?² = 136.2; H₁:?² ? 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 originaldistribution?

We assume a uniform population distribution. We assume abinomial population distribution.     We assumea exponential population distribution. We assume a normalpopulation distribution.

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

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 orfail to reject the null hypothesis?

Since the P-value > ?, we fail to rejectthe null hypothesis. Since the P-value > ?, wereject the null hypothesis.     Since theP-value ? ?, we reject the null hypothesis. Sincethe P-value ? ?, we fail to reject the nullhypothesis.

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

At the 1% level of significance, there is insufficient evidenceto conclude that the variance for number of mountain climber deathsis less than 136.2 At the 1% level of significance, there issufficient evidence to conclude that the variance for number ofmountain climber deaths is less than136.2    

(f) Find the requested confidence interval for the populationvariance. (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 ?² lies outsidethis interval. We are 90% confident that ?²lies above this interval.     We are 90%confident that ?² lies below this interval. Weare 90% confident that ?² lies within thisinterval.