Question

1)- 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 29 batteries gave a sample variance of 15.6 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.

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 normal population distribution.We assume a exponential population distribution.    We assume a uniform population distribution.We assume a binomial 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

2)- Anystate Auto Insurance Company took a random sample of 376 insurance claims paid out during a 1-year period. The average claim paid was $1550. Assume σ = $252.

Find a 0.90 confidence interval for the mean claim payment. (Round your answers to two decimal places.)

lower limit	$
upper limit	$

Find a 0.99 confidence interval for the mean claim payment. (Round your answers to two decimal places.)

lower limit	$
upper limit	$

Answer 1

Q 1) a)

State the null and alternate hypotheses.

Ho: σ² = 23; H1: σ² ≠ 23

b) Under H0, the test statistic is

Degrees of freedom = n-1= 28

What assumptions are you making about the original distribution?

We assume a normal population distribution.

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

The P-Value is .8984

P-value > 0.100

(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.

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

f) 90% CI for population variance

df = 28

The critical value of Chi square are

Therefore the CI is

Lower Limit = 10.57

Upper limit = 25.80

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

Lower Limit = 3.25

Upper limit = 5.08