SMAM 314 Exam 3 Name. 1. Mark the following statements true or false. (6 points 2 each)

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1 SMAM 314 Exam 3 Name 1. Mark the following statements true or false. (6 points 2 each) F A. A t test on independent samples is appropriate when the results of an algebra test are being compared for the same students before and after a course in remedial algebra. F B. You have real data from a population with unknown mean. You have % confidence intervals for the true mean population mean. You can easily tell which of these confidence intervals contain the true population mean. F. C. You conduct a test of hypothesis at α =.01 and fail to reject the null hypothesis. You will always fail to reject the null hypothesis at α = Pick the best choice. (6 points 2 each) d A. A test of hypothesis µ = 10 vs. µ 10 is performed at α =.05. The standard deviation is known to be 1. Let Z be the standard normal random variable. The region of rejection is a <Z <1.96 b. Z>1.96 c. Z>1.645 and Z< d. Z>1.96 and Z < 1.96 e. Z>2.576 and Z < d B. A weak positive linear association would be present if two variables X and Y have correlation coefficient a..92 b. 1.5 c..92 d..092 e b C. The linear equation y =1.1x 3.5 has a. slope 1.1 and y intercept 3.5 b. slope 1.1 and y intercept 3.5 c. slope 1.1 and y intercept 3.5 d. slope 3.5 and y intercept 1.1 e. slope 3.5 and y intercept 1.1.

2 3. The following data are about the degree of polymerization of paper specimens Degree of Polymerization Variable N N* Mean SE Mean StDev Degree of Polymerization A. Find a 95% two sided confidence interval for the true average degree of polymerization.(8 points) s x ± t n ± (2.120)(3.67) (430.51,445.92) B.Interpret the confidence interval that was obtained in part A of the problem.(4 points) The probability that the confidence interval contains the true average degree of polymerization is.95. B. Based only on the results of the confidence interval calculation in part A of the problem what would be your conclusion for each of the following tests of hypothesis at α =.05 Write a sentence explaining your answer.(8 points 4 each) (1) µ=440 µ 440 You would fail to reject and conclude that the average degree of polymerization is not significantly different from 440 because 440 is inside the confidence interval.

3 (2) µ=450 µ 450 You would reject and conclude that the average degree of polymerization is significantly different from 450 because 450 is outside the confidence interval. 4. Mist is generated when metal removing fluids are used in machining operations to cool and lubricate the tool and workpiece. The data below gives X= fluid flow velocity for a 5% soluble oil (cm/sec) and Y the extent of mist droplets having diameters smaller than 10 mg/m 3. x y A. Using your calculator fit the regression line. (8 points) Y= x B. Is the data positively or negatively correlated? Explain. (4 points) R =.964 As x increases so does y. The data is positively correlated. C. What percentage of the variation is accounted for by the model?(2 points) The model accounts for 93% of the variation D. Predict y when x=200. (4 points) Y= (200)+.4041=.5283

4 5.In a sample of 150 households in a certain city 110 had internet access. Conduct a test of hypothesis at α =.05 to determine whether it is reasonable to assume that more than 70% of the households in this city have internet access and find the p value. Write the complete report. (20 points) p=0.7 p>0.7 Assumptions Large sample Region of rejection Z>1.645 Calculation of the test statistic p p Z = 0, p = X p 0 (1 p 0 ) n n p = =.733 Z = (.7)(.3) 150 = =.882 Reject or fail to reject Fail to reject. Conclusion There insufficient evidence to say that more than 70% of the homes have internet access. P value =P(Z>.882)=.189

5 No computations are required for this problem. Base your answers on the Minitab output. 6. A sample of ten diesel trucks were run with both hot and cold engines to determine whether this would affect fuel economy. The results were in miles per gallon were Truck Hot Cold A test of hypothesis was conducted to determine whether there is a significant difference in the fuel consumptions for the hot and cold engines. The results follow. Paired T-Test and CI: Hot, Cold Paired T for Hot - Cold N Mean StDev SE Mean Hot Cold Difference % CI for mean difference: (0.2865, ) T-Test of mean difference = 0 (vs not = 0): T-Value = 8.08 P-Value = A. What is the null and the alternative hypothesis (2 points) µ H = µ C µ H µ C B. What assumptions are needed to perform this test?(2 points) The differences are normally distributed. Paired data. C. What is the region of rejection?(4 points) T>2.262 T< 2.262

6 D. What is the value of the test statistic and what is the p value?(2 points) T=8.08 pvalue =0 E. At α =.05 would you reject or fail to reject? Explain.(3 points) You would reject at α =.05 because the calculated T>2.262 and the p value =0 <.05 F. Is there a significant difference in gas consumption for the two kinds of engines? Explain. (3 points) There is a significant difference in gas consumption because the null hypothesis was rejected at α =.05. G. Is the confidence interval result consistent with your answer to part E?(3 points) Yes because it does not contain zero. H. Would you reject at α =.01? Explain.(3 points) Yes because the p value is zero. I. Calculate a 99% confidence interval on the difference in the average gas mileage for the two kinds of engines and tell whether your results are consistent with those obtained in H?(8 points) zero. d ± t s d n.3980 ± 3.25(.0493) (.238,.558) This result is consistent with that obtained in H because it does not contain

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