Question

A set of solar batteries is used in a research satellite. The satellite can run on only one battery, but it runs best if more than one battery is used. The variance σ² of lifetimes of these batteries affects the useful lifetime of the satellite before it goes dead. If the variance is too small, all the batteries will tend to die at once. Why? If the variance is too large, the batteries are simply not dependable. Why? Engineers have determined that a variance of σ² = 23 months (squared) is most desirable for these batteries. A random sample of 26 batteries gave a sample variance of 15.2 months (squared). Using a 0.05 level of significance, test the claim that σ² = 23 against the claim that σ² is different from 23.

(a) What is the level of significance?


State the null and alternate hypotheses.

H_o: σ² = 23; H₁: σ² > 23H_o: σ² = 23; H₁: σ² ≠ 23    H_o: σ² > 23; H₁: σ² = 23H_o: σ² = 23; H₁: σ² < 23

(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 binomial population distribution.    We assume a normal population distribution.We assume a uniform 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 5% level of significance, there is insufficient evidence to conclude that the variance of battery life is different from 23.At the 5% level of significance, there is sufficient evidence to conclude that the variance of battery life is different from 23.    

(f) Find a 90% confidence interval for the population variance. (Round your answers to two decimal places.)

lower limit
upper limit

(g) Find a 90% confidence interval for the population standard deviation. (Round your answers to two decimal places.)

lower limit	months
upper limit	months

Answer 1

Given that,
population variance (σ^2) =23
sample size (n) = 26
sample variance (s^2)=15.2
null, Ho: σ^2 =23
alternate, H1 : σ^2 !=23
level of significance, α = 0.05
from standard normal table, two tailed ᴪ^2 α/2 =37.652
since our test is two-tailed
reject Ho, if ᴪ^2 o < - OR if ᴪ^2 o > 37.652
we use test statistic chisquare ᴪ^2 =(n-1)*s^2/o^2
ᴪ^2 cal=(26 - 1 ) * 15.2 / 23 = 25*15.2/23 = 16.522
| ᴪ^2 cal | =16.522
critical value
the value of |ᴪ^2 α| at los 0.05 with d.f (n-1)=25 is 37.652
we got | ᴪ^2| =16.522 & | ᴪ^2 α | =37.652
make decision
hence value of | ᴪ^2 cal | < | ᴪ^2 α | and here we do not reject Ho
ᴪ^2 p_value =0.8984
ANSWERS
---------------
a.
level of significance =0.05
null, Ho: σ^2 =23
alternate, H1 : σ^2 !=23
b.
test statistic: 16.522
d.f (n-1)=25 is 37.652
We assume a normal population distribution.
critical value: -37.652 , 37.652
c.
p-value:0.8984
P-value > 0.100
d.
decision: do not reject Ho
e.
we do not have enough evidence to support the claim that σ2 = 23 against the claim that σ2 is different from 23.
f.
CONFIDENCE INTERVAL FOR VARIANCE
ci = (n-1) s^2 / ᴪ^2 right < σ^2 < (n-1) s^2 / ᴪ^2 left
where,
s^2 = variance
ᴪ^2 right = (1 - confidence level)/2
ᴪ^2 left = 1 - ᴪ^2 right
n = sample size
since aplha =0.1
ᴪ^2 right = (1 - confidence level)/2 = (1 - 0.9)/2 = 0.1/2 = 0.05
ᴪ^2 left = 1 - ᴪ^2 right = 1 - 0.05 = 0.95
the two critical values ᴪ^2 left, ᴪ^2 right at 25 df are 37.6525 , 14.611
variance( s^2 )=15.2
sample size(n)=26
confidence interval = [ 25 * 15.2/37.6525 < σ^2 < 25 * 15.2/14.611 ]
= [ 380/37.6525 < σ^2 < 380/14.6114 ]
[ 10.09, 26.01 ]
g.
CONFIDENCE INTERVAL FOR STANDARD DEVIATION
ci = (n-1) s^2 / ᴪ^2 right < σ^2 < (n-1) s^2 / ᴪ^2 left
where,
s = standard deviation
ᴪ^2 right = (1 - confidence level)/2
ᴪ^2 left = 1 - ᴪ^2 right
n = sample size
since alpha =0.1
ᴪ^2 right = (1 - confidence level)/2 = (1 - 0.9)/2 = 0.1/2 = 0.05
ᴪ^2 left = 1 - ᴪ^2 right = 1 - 0.05 = 0.95
the two critical values ᴪ^2 left, ᴪ^2 right at 25 df are 37.6525 , 14.611
s.d( s )=3.898
sample size(n)=26
confidence interval for σ^2= [ 25 * 15.1944/37.6525 < σ^2 < 25 * 15.1944/14.611 ]
= [ 379.8601/37.6525 < σ^2 < 379.8601/14.6114 ]
[ 10.0886 < σ^2 < 25.9975 ]
and confidence interval for σ = sqrt(lower) < σ < sqrt(upper)
= [ sqrt (10.0886) < σ < sqrt(25.9975), ]
= [ 3.1763 < σ < 5.0988 ]