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 7 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.
Ho: σ2 = 136.2; H1: σ2 > 136.2Ho: σ2 < 136.2; H1: σ2 = 136.2 Ho: σ2 = 136.2; H1: σ2 ≠ 136.2Ho: 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 normal population distribution. We assume a binomial population distribution.We assume a exponential 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 below this interval.We are 90% confident that σ2 lies above this interval. We are 90% confident that σ2 lies within this interval.We are 90% confident that σ2 lies outside this interval.
The statistical software output for this problem is:
One sample variance summary hypothesis
test:
σ2 : Variance of population
H0 : σ2 = 136.2
HA : σ2 < 136.2
Hypothesis test results:
Variance | Sample Var. | DF | Chi-square Stat | P-value |
---|---|---|---|---|
σ2 | 112 | 6 | 4.9339207 | 0.4477 |
90% confidence interval results:
Variance | Sample Var. | DF | L. Limit | U. Limit |
---|---|---|---|---|
σ2 | 112 | 6 | 53.368967 | 410.91294 |
Hence,
a) Level of significance = 0.01
Hypotheses: Ho: σ2= 136.2; H1: σ2 < 136.2; Option D is correct.
b) Chi-square test statistic = 4.93
Degrees of freedom = 6
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) Lower limit = 53.37
Upper limit = 410.91
We are 90% confident that σ2 lies within this interval.
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