EF 507 PS-Chapter 11 FALL 2008

1. A company produces electric devices operated by a thermostatic control. The standard deviation of the temperature at which these controls actually operate should not exceed 2.0 degrees Fahrenheit. For a random sample of 25 of these controls the sample standard deviation of operating temperatures was 2.36 degrees Fahrenheit. Stating any assumptions you need to make, test at the 5% level the null hypothesis that the population standard deviation is at most 2.0 against the alternative that is bigger.

ANSWER: The hypothesis test assumes that the population values are normally distributed

H0:s� 2.0 and H 1 : s 2.0. 2 2 Reject H0 if c > c24,0.05 = 36.42 2 2 2 (n- 1) s (24)(2.36) Test statistic: c =2 = 2 = 33.418 , therefore, we fail to reject H0 . We s 0 (2) conclude that the population standard deviation is at most 2 degrees Fahrenheit.

QUESTIONS 2 THROUGH 4 ARE BASED ON THE FOLLOWING INFORMATION: A new safety device is installed in an assembly-line operation. After the installation of the device, a random sample of ten days output gave the following results for numbers of finished components produced: 577, 642, 615, 655, 600, 570, 640, 620, 595, and 575. Management is concerned about the variability of daily output and views as undesirable any variance above 525.

2. Calculate the sample variance ANSWER: s2 = 922.322

3. Test at the 10% significance level the null hypothesis that the population variance for daily output does not exceed 525. 2 2 ANSWER: H0:s� 525 and H 1 : s 525 2 2 Reject H0 if c > c9,0.10 =14.684 2 2 (n- 1) s (9)(922.322) c =2 = =15.811, therefore, we reject H0 and conclude that the s 0 525 population variance for daily output exceeds 525.

4. Of a random sample of 150 music industry management majors, 75 rated a sense of humor as very important trait to their career performance. The same view was held by 81 of an independent random sample of 180 professional gulf management majors. Test at the 5% level against a two-sided alternative the null hypothesis that the population proportions of music industry management and professional gulf management majors who rate a sense of humor as very important are the same.

ANSWER: M = Music industry management, and G = professional gulf management

H0: PM- P G = 0 and H 1 : P M - P G 0

Reject H0 if Z< - z0.025 = -1.96 or Z > z 0.025 = 1.96

1 pˆ M = 75 / 150 = 0.50 and pˆG = 81 / 180 = 0.45 An estimate of the common proportion is given by

nF pˆ F+ n Up ˆ U (150)(0.5)+ (180)(0.45) pˆ 0 = = = 0.4727 nF+ n U 150 + 180

pˆF- p ˆ u 0.5- 0.45 Test statistic: Z = = = 0.91. pˆ0(1-p ˆ 0 )(1 nF + 1/ n U ) (0.4727)(0.5273)(1/150 + 1/180)

Therefore, do not reject H0 . We conclude that the population proportions of music industry management and professional gulf management majors who rate a sense of humor as very important are the same.

5. You are comparing the precision of two brands of stamping machines. From a random sample of 12 units of output from the Brand A machine, you find that it produces with a standard deviation of 15.2. For the Brand B machine, in a sample of 20 units of output, you find a standard deviation of 10.1. Is this sufficient evidence at the 5% significance level to conclude that Brand B machines produce with lower variance? Assume that the output of both machines follows a normal distribution.

2 2 2 2 ANSWER: H0 : H0:sA= s B vs. H 1 : s A > s B

Reject H0 if Fcalc> F crit = F11,19,0.05� F 10,19,0.05 2.38

2 2 2 2 Since Fcalc= s A/ s B = (15.2) /(10.1) = 2.26 , we fail to reject H0 at a = 0.05. There is insufficient evidence to conclude that Brand B machines produce with lower variance. QUESTIONS 6 THROUGH 8 ARE BASED ON THE FOLLOWING INFORMATION: The medical school of Wisconsin conducted an investigation to determine whether women and men complete medical school in significantly different amounts of time, on the average. Two independent random samples were selected and the following summary information concerning times to completion of medical school computed: Women Men Sample Size 80 100 Sample mean 4.5 years 4.4 years Population Standard 0.5 years 0.6 years Deviation

6. State the appropriate null and alternative hypotheses

ANSWER: H0:mW- m M = 0 vs. H1:mW- m M 0

7. Perform the appropriate test of hypothesis to determine whether there is a significant difference in time to completion of medical school between women and men. Test using a = 0.05. x- x 4.5- 4.4 Z =W M = = ANSWER: s 2s 2 (0.5) 2 (0.6) 2 1.22 W + M + nW n M 80 100

Reject H0 if Z < -z.025 = -1.96 or Z > z.025 = 1.96. Since Z = 1.22, do not reject H0 . Thus one cannot conclude there is a significant difference in mean time to completion of medical school between women and men.

2 8. Find the p-value associated with the test in Question 77 and explain how to use it for testing the hypothesis in Question 76. ANSWER: p-value =2 P(Z  1.22) = 2 (0.50 – 0.3888) = 0.2224. Since p-value > a

= 0.05, we fail to reject H0 . QUESTIONS 9 THROUGH 11 ARE BASED ON THE FOLLOWING INFORMATION: Before agreeing to purchase a large order of polyethylene sheaths for a particular type of high pressure oil-filled submarine power cable, a company wants to see conclusive evidence that the true standard deviation of sheath thickness is less than .05mm. 9. What hypotheses should be tested, and why? ANSWER: Let s denote the population standard deviation. The appropriate hypotheses are H0 :s = .05 vs. H1 :s = .05 . With this formulation, the burden of proof is on the data to show that the requirement has been met (the sheaths will not be used unless H0 can be rejected in favor of H1 .) 10. In this context, what is Type I error? ANSWER: Type I error: Conclude that the standard deviation is < .05 mm when it is really equal to 0.05 mm. 11. In this context, what is Type I error? ANSWER: Type II error: Conclude that the standard deviation is .05 mm when it is really < 0.05 QUESTIONS 12 THROUGH 16 ARE BASED ON THE FOLLOWING INFORMATION: The Ford Motor Company, as part of its quality control program, began returning to the supplier all shipments of steel that had defects or faulty chemistry. When Ford began this program, the defective rate in 100 shipments was 9%. A recent survey indicated that 2.2% in 136 shipments was defective. The quality control department wants to determine at the 0.05 level of significance if this represents a significant improvement in the quality of the steel. 12. State the null and alternative hypotheses.

ANSWER: Let P1 = proportion of defects before the quality control program, and

P2 = proportion of defects after the quality control program,

Then, the null and alternative hypotheses are H0: P 1- P 2� 0 and H 1 : P 1 > P 2 0. 13. Calculate an estimate of the common proportion

ANSWER: pˆ1 = 0.09 and pˆ 2 = 0.022. Then, an estimate of the common proportion is

n1 pˆ 1+ n 2 p ˆ 2 (100)(0.09)+ (140)(0.022) given by pˆ 0 = = = 0.0503 n1+ n 2 100 + 140 14. What is the test statistics?

pˆF- p ˆ u 0.09- 0.022 ANSWER: Z = = = 2.38 pˆ0(1-p ˆ 0 )(1 nF + 1/ n U ) (0.0503)(0.9497)(1/100 + 1/140) 15. What is the p-value? ANSWER: p-value = P(Z > 2.38) = 0.50 – 0.4913 = 0.0087

16. What is the conclusion?

ANSWER: Since p-value = 0.0087 < 0.05, reject H 0 . There is enough statistical evidence to conclude that there is a significant improvement in the quality of the steel