Statistics Exam Practice Questions, Exams of Mathematics

Download Statistics Exam Practice Questions and more Exams Mathematics in PDF only on Docsity! 10 n 0.5 p MATH/GM 533 Final Exam 1. (TCO D) PuttingPeople2Work has a growing business placing out-of-work MBAs. They claim they can place a client in a job in their field in less than 36 weeks. You are given the following data from a sample. Sample size: 100 Population standard deviation: 5 Sample mean: 34.2 Formulate a hypothesis test to evaluate the claim. (Points : 10) Ho: µ = 36; Ha: µ ≠ 36 Ho: µ ≥ 36; Ha: µ < 36 Ho: µ ≤ 34.2; Ha: µ > 34.2 Ho: µ > 36; Ha: µ ≤ 36 Ans. b. H0 must always have equal sign, < 36 weeks 2. (TCO B) The Republican party is interested in studying the number of republicans that might vote in a particular congressional district. Assume that the number of voters is binomially distributed by party affiliation (either republican or not republican). If 10 people show up at the polls, determine the following: Binomial distribution X P(X) cumulati ve probabili ty 0 0.00098 0.00098 1 0.00977 0.01074 2 0.04395 0.05469 1 3 0.11719 0.17188 4 0.20508 0.37695 5 0.24609 0.62305 6 0.20508 0.82813 7 0.11719 0.94531 8 0.04395 0.98926 9 0.00977 0.99902 2 6 4 6 4 55 4 5 4 3 5 66 1 6 7 1 6 Ans. a. Range = maximum - minimum Maximum observation in the data = 67 Minimum observation in the data = 19 range = 67 - 19 =48 sales/ month (Millions $) b. Median Arrange the data in the ascending order That is 19, 23, 23, 34, 34, 34, 45, 45, 54, 56, 56 67 Here, sample size n = 12 (even) So median is the average of two middle observation Mid=(34+45)/2 = 39.5 sales/month Millions ($) c. From the Mega stat output we have sample standard deviation as s=15.39 Therefore, the range of the data that would contain 68% of the result That is R +- s 5 (48-15.39, 48+15.39) (32.61, 63.39) Reference: (TCO A) Company ABC had sales per month as listed below. Using the MegaStat output given, determine: 6 (A) Range (5 points) (B) Median (5 points) (C) The range of the data that would contain 68% of the results. (5 points) Raw data: sales/month (Millions of $) 19 34 23 34 56 45 35 36 46 47 19 23 count 12 mean 34.75 sample variance 146.20 sample standard deviation 12.09 minimum 19 maximum 56 range 37 Stem and Leaf plot for # 1 stem unit = 10 leaf unit = 1 count 12.00000 mean 34.75000 sample variance 146.2045 5 sample standard deviation 12.09151 minimum 19.00000 maximum 56.00000 range 37.00000 7 population standard deviation 2.083 standard error of the mean 0.538 tolerance interval 95.45% lower 95.548 tolerance interval 95.45% upper 104.15 2 margin of error 4.302 1st quartile 98.850 median 99.200 3rd quartile 100.55 0 interquartile range 1.700 mode 103.00 0 (Points : 25) Answer: To construct 95% confidence interval for the data from the Chris Cross Manufacturing, the margin of error is: E= zc (standard deviation divided by square root of n) In this case, zc= z alpha divided by 2 = z 0.05/2 = z 0.025 = 1.96 (1-confidence level)/2 (1-.95)/2 .05/2=.025 looking at z table, Table A.3 Cumulative Areas under the Standard Normal Curve, is -1.96, that is 1.96 and the standard deviation is 2.083 the sample size, n=16 Standard error of mean is, SE= standard deviation divided by square root of n = 2.083/4 =0.52075 10 E= zc(standard deviation divided by square root of n) =(1.96) (.52075) E=1.02067 95% confidence interval for the true population mean is, (x bar - z*(standard deviation divided by square root of n), x bar + z*(standard deviation divided by square root of n) Here z*0.025=1.96 11 x bar =99.850 E=1.02067 therefore, 95% confidence interval for the true population mean u = (x bar- Em x bar + E) = (99.850-1.02067, 99.850 + 1.02067) =(98.82933, 100.87067) Given, Tesla's requirement is that 95% of the axles are 100 cm +- 2 cm The confidence interval from our sample belongs to the prescribed Confidence interval. Therefore, Tesla select Chris Cross Manufacturing as a vendor. Reference: Chegg Tesla Motors needs to buy axles for their new car. They are considering using Chris Cross Manufacturing as a vendor. Tesla’s requirement is that 95% of the axles are 100 cm ± 5 cm. The following data is MegaStat output from a test run from Chris Cross Manufacturing. Descriptive statistics count: 16 mean: 99.938 sample variance: 2.313 sample standard deviation: 1.521 minimum: 97 maximum: 102.9 range: 5.9 population variance: 2.169 population standard deviation: 1.473 standard error of the mean: 0.380 tollerance interval 95.45% lower: 96.896 tolerance interval 95.45% upper: 102.979 12 5. (TCO D) A PC manufacturer claims that no more than 2% of their machines are defective. In a random sample of 100 machines, it is found that 4.5% are defective. The manufacturer claims this is a fluke of the sample. At a .02 level of significance, test the manufacturer's claim, and explain your answer. 15 Test and CI for One Proportion Test of p = 0.02 vs p > 0.02 Sam pl e X N Sample p 98% Lower Bound Z-Value P-Value 1 4 10 0 0.0400 0 0 0.000000 1.43 0.077 Ans. Set up the hypotheses: H0: p <= 0.02 Ha: p > 0.02 This is a one tailed test, since we will only reject for high proportions. Since we are using a 0.02 level of significance (it's just chance that the hypotheses happen to have the same value as this), we'll reject the null hypothesis if our P Value is less than 0.02. The computed P value from Megastat was 0.077. This is higher than the significance level. Therefore, we do not reject H0:. We can say that the proportion is still less than or equal to 2%, and this was a chance of occurrence. Decision: There is no sufficient evidence that the proportion of defectives is more than 2%. Reference: Set up the hypotheses: H0: p <= 16 0.02 Ha: p > 0.02 This is a one tailed test, since we will only reject for high proportions. Since we are using a 0.02 level of significance (it's just chance that the hypotheses happen to have the same value as this), we'll reject the null hypothesis if our P Value is less than 0.02. The computed P value from Megastat was 0.0371. This is higher than the significance level. Therefore, we do not reject H0:. 17 (B)What percentage of Americans will exercise less than 15 minutes? If 1000 Americans were evaluated, how many would you expect to have exercised less than 15 minutes? (15 points) (Points : 30) 20 if X=24 P(Xex) =P(Xe24) t-H. 24- ie oc 22 = WZ < 3.727 )irem nomial tables) = 0.9999 P(X 2x) =1-FiXe24) =] - 0.5599 = (0001 (fram normal tables) =Fi _pyll-15.8 2- | 21-158 ajP(11<X<21) =F¢ ae = P(? 182<2<2 364) =P(Z<2 364) - P(Z<2. 182) = 0.991 - 0.9854 =f) 0056 bIPCM=15) = 0.3579 } 21