In: Statistics and Probability
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 9 years, the variance has been s2 = 113.4. 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 exponential population distribution.
We assume a normal population distribution.
We assume a binomial population distribution.
We assume a uniform 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.
Given
S^2= 113.4
n= 9
significance level,
Degree of freedom=9-1=8
Critical Value are given
X^2_R = 1.6464973
Test staticstics
p-value=0.426367
...........................................
Ho: σ2 = 136.2; H1: σ2 < 136.2
chi test statistics =6.66079
degree of freedom = 8
We assume a normal population distribution.
P-value > 0.100
Since the P-value > α, we fail to reject the null hypothesis.
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
............................................................
S= 10.64894361
S^2= 113.4
n= 9
df= 8
alpha= 0.1
Area to the right of X(Right)^2=0.05
Area to the Right of X(left)^2= 0.95
ꭕ^2(left)= 2.73263
ꭕ^2(right)= 15.5073
there is 90 % chance that true population variance of the number of mountain climbers killed each year is between 58.50 and 331.987
...........................
We are 90% confident that σ2 lies within this interval. (58.50,331.987)