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 28 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 28 test tires, the sample standard deviation of blowout forces was 1356 foot-pounds.

(a) Soap Stone claims its tires will blow out at an average pressure of 20,000 foot-pounds, with a standard deviation of 1026 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.

One-way ANOVAF test for two variances     Chi-square test of independenceChi-square for testing or estimating σ² or σChi-square test of homogeneityChi-square goodness-of-fitTwo-way ANOVA

(i) Give the value of the level of significance.

State the null and alternate hypotheses.

H_o: σ² = 1052676; H₁: σ² > 1052676H_o: σ² = 1052676; H₁: σ² ≠ 1052676     H_o: σ² < 1052676; H₁: σ² = 1052676H_o: σ² = 1052676; H₁: σ² < 1052676

(ii) Find the sample test statistic. (Use two decimal places.)


(iii) Find the P-value of the sample test statistic. (Use four decimal places.)

(iv) Conclude the test.

Since the P-value is greater than or equal to the level of significance α = 0.10, we fail to reject the null hypothesis.Since the P-value is less than the level of significance α = 0.10, we reject the null hypothesis.     Since the P-value is less than the level of significance α = 0.10, we fail to reject the null hypothesis.Since the P-value is greater than or equal to the level of significance α = 0.10, we reject the null hypothesis.

(v) Interpret the conclusion in the context of the application.

At the 10% level of significance, there is insufficient evidence to conclude that the variance is not greater than claimed.At the 10% level of significance, there is sufficient evidence to conclude that the variance is greater than claimed.     At the 10% level of significance, there is sufficient evidence to conclude that the variance is not greater than claimed.At the 10% level of significance, there is sufficient evidence to conclude that the variance is greater than claimed.

(b) Find a 90% confidence interval for the variance of blowout pressures, using the information from the random sample. (Use one decimal place.)
< σ² <  square foot-pounds

Answer 1

Sample Standard deviation, s = 1356
Sample size, n = 28

Test used: Chi-square for testing or estimating σ² or σ

a) i) Level of significance, α = 0.1

Null and alternative hypothesis:  
Hₒ : σ² =   1052676
H₁ : σ² >   1052676

ii) Test statistic:          
χ² = (n-1)s² / σ² = (28-1)*1356² / 1026² = 47.1616    
          
Degree of freedom:          
Df = n -1 =   27      
             
iii) p-value = CHISQ.DIST.RT(47.1616, 27) = 0.0095      
          
iv) Conclusion:   

Since the P-value is less than the level of significance α = 0.10, we reject the null hypothesis.

v) Interpret:

At the 10% level of significance, there is sufficient evidence to conclude that the variance is greater than claimed.

b) 90% confidence interval:

χ²_1-α/2 = 16.1514, χ²_α/2 = 40.1133