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 30 batteries gave a sample variance of 14.6 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.
Ho: σ2 = 23; H1: σ2 > 23
Ho: σ2 = 23; H1: σ2 ≠ 23
Ho: σ2 > 23; H1: σ2 = 23
Ho: 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 uniform population distribution.
We assume a exponential population distribution.
We assume a normal population distribution.
We assume a binomial 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 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

(a) The level of significance is 5% ie,

We have to test the claim that σ2 = 23 against the claim that σ2 is different from 23.

Therefore, the null and alternate hypothesis is

Ho: σ2 = 23; H1: σ2 ≠ 23

(b) The value of the chi-square statistic for the sample:

The test statistic is

and df = n-1

Given,

n=30

sample variance s2 = 14.6

σ2 = 23

c) p value

We have got the calculated value of chi square test statistic as 19.04

From the table of Chi-square test, we can see that for degrees of freedom = 29, the values 19.04 lies between 17.708 and 19.768. Thus we can write 0.90 < p value < 0.95

And thus using R software we get the p value as 0.920258

ie p value > 0.100

d)

Since the p-value of 0.9203 is greater than the level of significance of 0.05 (5%), we fail to reject the null hypothesis.

ie, Since the P-value > α, we fail to reject the null hypothesis.

e)

At the 5% level of significance, there is insufficient evidence to conclude that the variance of battery life is different from 23.

f)

90% confidence interval for the population variance

Sample size n = 30, Thus degrees of freedom = n-1 = 29

The desired confidence level = 90%

Now,

Hence the required 90% CI for population variance is calculated as

Thus the 90% Confidence limits for the population variance are

lower limit : 9.95 months2

upper limit : 23.91 months2

g)

90% confidence interval for the population standard deviation

Thus the 90% Confidence limits for the population variance are

lower limit : 3.15 months

upper limit : 4.89 months