SMAM 314 Exam 3 Name 1. Indicate whether the following statements are true (T) or false (F) (6 points) F A. A null hypothesis that is rejected at α =.05 will always be rejected at α =.01. T B. A course on industrial safety is presented in ten plants. The number of accidents during a three month period before and after the course is recorded. It is appropriate to use a paired t test on the before minus after differences to determine the effectiveness of the program. F_C. A 95% confidence interval is usually narrower than a 90% confidence interval. 2. Pick the best choice.(4 points) D A. A test of the null hypothesis H0 p=0.5 vs. the alternative hypothesis p 0.5 is performed at α =.05. The computation of the test statistic yields Z= 2.0. The p value is A. 0.05 B.0.10 C. 0.0228 D. 0.0456 E..9544. _B B. The p value for a statistical test of hypothesis is 0.025. For which of the following levels of significance does the null hypothesis fail to be rejected? A. α =.10 B. α =.01 C. α =.05 D. α =.03 E. non of A- D.
3. The amount of shaft wear in thousands of an inch after a fixed mileage was determined for a random sample 8 internal combustion engines having copper lead as a bearing material. The sample mean and the sample standard deviation were x = 3.72,s = 1.25. A. Find (1) A 90 % two sided confidence interval.(8 points) (x t α /2,n 1 s n,x + t α /2,n 1 s n ) (3.72 (1.895)(1.25),3.72 + (1.895)(1.25) ) 8 8 (3.72.84,3.72 +.84) (2.88,4.56) (2) A 95% upper confidence bound? (4 points) UCB=4.56 CI (,4.56) B. For each of the following tests of hypothesis would you reject or fail to reject H0 at the given level of significance. Explain your answer.(3 each) (1) H 0 H 0 µ=4 H 1 µ 4 at α =.10 Fail to Reject H0 4 is in the 90% two sided CI (2) H 0 µ = 5 at α =.05 H 1 µ<5 Reject H0 5 is not in the 95% one sided CI.
4. Scientists think that robots will play a crucial role in factories in the next several decades. In an experiment to determine whether the use of robots to weave computer cables is feasible a robot was used to assemble 500 cables. The cables were examined and there were 15 defectives. The proportion of defects using human assemblers was.035. A. Perform a test of hypothesis to determine whether the proportion of defectives was lower for robots than for humans. Use α =.10. (15 points) Write the complete report. H0 p=.035 H1 p<.035 Assumption Large sample Region of rejection Z< 1.28 Calculation of test statistic Z = p = 15 500 =.03 Z =.03.035.035(.965) 500 =.608 p p 0, p = X p 0 (1 p 0 ) n n Reject or fail to reject H0 Fail to reject H0 Conclusion in plain English It does not appear that the proportion of defectives is lower for robots than humans. B. What is the p value for this test?(2 points) Pvalue = P(Z<-.608)=.272
C. What sample size is necessary to obtain a 95% two sided confidence interval with error bound at most.01?(7 points) E = Z α /2 p(1 p) N Solving for N N = z 2 α /2 p(1 E p) N = 1.96.01 N = 1118 2 (.03)(.97) = 1117.9 D. Find a 90% upper confidence bound, write down the one sided confidence interval and explain whether your results are consistent with that of the hypothesis test that was performed in part A.(10 points) UCB =.03 + 1.28 (.03)(.97) 500 (,.0397) =.0397 The results are consistent because.035 belongs to the CI. 5. The deterioration of many municipal pipeline networks across the country is a growing concern. One technology proposed for pipeline rehabilitation uses a flexible liner threaded through existing pipe. The following data was reported on tensile strength (psi) of liner specimens both when a certain fusion process was used and when it was not used. The data follow. Row No Fusion Fusion 1 2748 3027 2 2700 3356 3 2655 3359 4 2822 3297 5 2511 3125 6 3149 2910 7 3257 2889 8 3213 2902 9 3220 10 2753
A test of hypothesis was performed to determine whether the average tensile strength of specimens when the no- fusion treatment is used is significantly less than that when the fusion treatment is used. The results follow. Two-Sample T-Test and CI: Fusion, No Fusion Two-sample T for Fusion vs No Fusion SE N Mean StDev Mean Fusion 10 2903 277 88 No Fusion 8 3108 206 73 Difference = mu (Fusion) - mu (No Fusion) Estimate for difference: -205 95% upper bound for difference: 1 T-Test of difference = 0 (vs <): T-Value = -1.74 P-Value = 0.050 DF = 16 Both use Pooled StDev = 248.5659 Answer the following questions based on Minitab output. A. What is the null and the alternative hypothesis? (2 points) H0 µ 1 = µ 2 H1 µ 1 < µ 2 B. What are the assumptions under which this test is performed?(2 points) Independent random samples Normal populations Equal but unknown sd C. What is the region of rejection?(2 points) At α =.10 the region of rejection is T< 1.337 D. From the Minitab output what is the value of the test statistic and what is the p value?(2 points) T= 1.74 pvalue =.05 E. Would you reject H0 at α =.10?Explain.(4 points) Reject H0. P value <.10 and - 1.74<- 1.337 F. In plain English what to you conclude about the comparison of the average tensile strength of the two pipes at α =.10? (4 points) The pipe without fusion has significantly lower tensile strength than the pipe with fusion.
G. Would you come to the same conclusion as in F for α =.01.Explain.(4 points) No because.01<.05 6. The following summary data on bending strength (lb- in/in) of joints was obtained. Type Sample Sample Sample Standard Size Mean Deviation Without side coating 10 80.95 9.59 With side coating 10 63.23 5.96 A.(1) Find a 99% two sided confidence interval for the difference between the means of the two samples?(7 points) x 1 x 2 ± s p t α /2,n1 + n 2 2 s 2 p = (n 1)s 2 2 + (n 1 1 2 1)s 2 n 1 + n 2 2 1 + 1 n 1 n 2 80.95 63.23 ± 7.98(2.879) 1 10 + 1 10 17.72 ± 10.25 (7.47,27.97) s 2 p = 9(9.59)2 + 9(5.96) 2 = 63.74 18 s p = 7.98
(2)Based on this confidence interval would you reject the null hypothesis H0 µ 1 = µ 2 in favor of the alternative hypothesis H1 µ 1 µ 2?Explain.(3 points) Reject H0 because 0 is not in the CI. (3)Is it absolutely certain that the true difference between the bending strength of the joints is contained in the confidence interval you found in A?Explain.(3 points) No. The probability that the true difference between the bending strength is contained in the CI is 0.99 so there is a 1% chance it is not in the CI. B. Consider the Minitab output below Statistics Sample N StDev Variance 1 10 9.590 91.968 2 10 5.960 35.522 Ratio of variances = 2.589 Is there a significant difference in the variances of the two samples at α =.10. Explain your answer after comparing the ratio of the variances with an appropriate table value. Give the table value in your answer. (5 points) The table value is 3.18. Thus, there is not a significant difference in the variances.