Question

A consumer agency is investigating the blowout pressures of Soap Stone tires. A Soap Stone tire is said to blow out when it separates from the wheel rim due to impact forces usually caused by hitting a rock or a pothole in the road. A random sample of 29 Soap Stone tires were inflated to the recommended pressure, and then forces measured in foot-pounds were applied to each tire (1 foot-pound is the force of 1 pound dropped from a height of 1 foot). The customer complaint is that some Soap Stone tires blow out under small-impact forces, while other tires seem to be well made and don't have this fault. For the 29 test tires, the sample standard deviation of blowout forces was 1358 foot-pounds.
(a) Soap Stone claims its tires will blow out at an average pressure of 26,000 foot-pounds, with a standard deviation of 1020 foot-pounds. The average blowout force is not in question, but the variability of blowout forces is in question. Using a 0.1 level of significance, test the claim that the variance of blowout pressures is more than Soap Stone claims it is.
Classify the problem as being a Chi-square test of independence or homogeneity, Chi-square goodness-of-fit, Chi-square for testing or estimating σ2 or σ, F test for two variances, One-way ANOVA, or Two-way ANOVA, then perform the following.
(ii) Find the sample test statistic. (Use two decimal places.)
(iii) Find the P-value of the sample test statistic. (Use four decimal places.)
(b) Find a 99% confidence interval for the variance of blowout pressures, using the information from the random sample. (Use one decimal place.)

Answer 1

Answer:

Step-by-step explanation:

Hello!

The variable of interest is:

X: Impact force needed for Sap Stone tires to blow out. (foot-pounds)

n= 29, S= 1358 foot-pound

a)

Soap Stone claims that "The tires will blow out at an average pressure of μ= 26000 foot-pounds with a standard deviation of σ= 1020 foot-pounds.

According to the consumer's complaint, the variability of the blown out forces is greater than the value determined by the company.

I)

Then the parameter of interest is the population variance (or population standard deviation) and to test the consumer's complaint you have to conduct a Chi-Square test for σ².

σ²= (1020)²= 1040400 foot-pounds²

H₀: σ² ≤ 1040400

H₁: σ² > 1040400

α: 0.01

II)

$X^2= \frac{(n-1)S^2}{Sigma^2} ~~X^2_{n-1}$

$X^2_{H_0}= \frac{(n-1)S^2}{Sigma^2}= \frac{(29-1)*(1358)^2}{1040400} = 49.63$

III)

This test is one-tailed to the right and so is the p-value. This distribution has n-1= 29-1= 28 degrees of freedom, so you can calculate the p-value as:

P(X²₂₈≥49.63)= 1 - P(X²₂₈<49.63)= 1 - 0.99289= 0.00711

⇒ The p-value is less than the significance level so the test is significant at 1%. You can conclude that the population variance of the blowout forces is less than 1040400 foot-pounds², at the same level the population standard deviation of the blow out forces is less than 1020 foot-pounds.

b)

99% CI for the variance. Using the X² statistic you can calculate it as:

$[\frac{(n-1)S^2}{X^2_{n-1;1-\alpha /2}} ;\frac{(n-1)S^2}{X^2_{n-1;\alpha /2}} ]$

$X^2_{n-1;\alpha /2}= X^2_{28; 0.005}= 13.121$

$X^2_{n-1;1-\alpha /2}= X^2_{28; 0.995}= 49.588$

$[\frac{28*(1358)^2}{49.588} ;\frac{28*(1358)^2}{13.121} ]$

[1041312.253; 3935415.898] foot-pounds²

I hope this helps!