Concordia-Comm215 Mock Final Solutions

Concordia-Comm215 Mock Final Solutions - COMM 215 MOCK...

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Unformatted text preview: COMM 215 MOCK FINAL EXAM SOLUTIONS exam solutions page 1 Question 1 Using these results: achievement test 39 43 21 64 57 47 75 34 52 89 final calculus mark 65 78 52 82 92 a) Calculate the coefficient of correlation for this data. 28 73 98 56 75 ¨ x = 460, ¨ x 2 = 23634 ¨ y = 760 ¨ y 2 = 59816 ¨ xy = 36854 n ¨ xy − ¨ x ¨ y r= n(¨ x 2 ) − (¨ x ) 2 n(¨ y 2 ) − (¨ y ) 2 Solution: data: r= 10(36854) − (460)(760) 10(23634 ) − (460 ) 2 10(59816 ) − (760 ) 2 = 0.8398 l 0.84 b) Find the equation of the line of best fit for this data. Solution: b0 = 40.78 b1 = 0.766 y = 40.78 + 0.766x c) Predict the final calculus grade for a student who scored 50 on the Achievement Test. Solution: y = 40.78 + 0.766 (50) = 79.08 or 79%. Question 2 a) For a random sample of 16 summer days in Montreal, the mean level of CO was found to be 4.9 ppm. With a standard deviation of 1.2. Based on this, construct a 95% confidence interval estimate of the true mean level of CO in summer in Montreal. Solution: Since n = 16, this a small sample t-distribution situation. We know that x = 4.9 , s = 1.2, and degrees of freedom = 15, so s x = 1.2 = 0.3 16 ta/2, 15 = 2.131, so our confidence limits are 4.9 ± 2.131(0.3) = 4.9 ± 0.639 the confidence interval for the mean level of CO is [ 4.26, 5.54 ] parts per million. b) A preliminary study indicated that a new drug was effective in 75% of its trials. Accordingly, how large a sample would be required to estimate the true effective rate of the drug to within 3% with 95% confidence. Solution: We’re finding sample size given a proportion, the size of the error and the level of confidence. The formula for sample size under these conditions is: n = p(1 − p) z / 2 E 2 95% confidence means z = 1.96, we know p = 0.75 and E = 0.03 n = 0.75(0.25) 1.96 0.03 2 = 800.3333 , so sample size is 801 people. Solutions will be posted at www.prep101.com/solutions COMM 215 MOCK FINAL EXAM SOLUTIONS exam solutions page 2 Question 3 A report said the average height of CEGEP basketball players is at least 180 cm. At a recent tournament a random sample of 25 players yielded the following data height in cm of 25 Cégep basketball players 176 181 200 178 180 177 179 180 197 181 183 189 180 183 178 185 180 184 190 187 177 176 182 195 196 Do these results support the report at the 0.05 Level of Significance? Solution: Since n = 25, we’ll use the t-distribution for this upper tail test statistic. ¨ x = 4594, x = 185.76, s = 6.98, t 0.05, 24 = 1.711 Small Sample Hypothesis Test on a Mean Value of interest: , mean height of players Ha: % > 180 Hypotheses: Ho: % [ 180 Criteria: if t > 1.711, heave Ho Level of significance:  = 5 %, df = 24 Test Statistic: t = x − % 185.76 − 180 = = 4.12 s/ n 6.98/ 25 Decision: 4.12 is WAY BIGGER than 1.711, we must heave Ho. Conclusion: the test supports the claim that the average height of CEGEP basketball players is at least 180 cm.. Question 4 In a study of air-bag effectiveness, it was found that in 821 crashes of midsize cars equipped with air-bags, 46 of the crashes resulted in the hospitalization of the drivers. Use a 0.01 level of significance to test the claim that the hospitalization rate for crashes of midsize cars equipped with air-bags is less than 7.8%. Solution: lower tail hypothesis test on a proportion. First we find the sample statistics: p = 46 = 0.056 1 = p 821 p(1 − p) = 0.0094 n Large Sample Hypothesis Test on a Proportion Value of interest: p , proportion of hospitalizations from crashes with air bags. Ha: p < 0.078 Hypotheses: Ho: p m 0.078 Criteria: if z < – 2.33, heave Ho Level of significance:  = 1 %, Test Statistic: z = p−p = 0.056 − 0.078 = −2.34 0.0094 p(1 − p) n Decision: since – 2.34 is just a teeny bit less than – 2.33, we might not heave Ho. We might run another test. Conclusion: the test barely supports the claim that the hospitalization rate for crashes of midsize cars equipped with air-bags is less than 7.8%. Solutions will be posted at www.prep101.com/solutions COMM 215 MOCK FINAL EXAM SOLUTIONS exam solutions page 3 Question 5 Suppose that we want to investigate whether there is a relationship between the test scores of persons who have gone through a certain job training program and their subsequent performance on the job. A random sample of 400 cases yields these results. job performance test score Poor Fair Good Total Below average 67 64 25 156 Average 42 76 56 174 Above average 10 23 37 70 Total 119 163 118 400 Test at the 0.01 level of significance whether on the job performance of workers who took the training program is independent of their test score. Solution: this is an r × c table and a chi-square test for independence. First, we find the expected values for the 9 data cells. Degrees of freedom = (r – 1)(c – 1) = 4 e= (row total) % (column total) , so, e 11 = 156 % 119 = 46.41 etc. 400 grand total observed and expected values for test scores vs. job performance o11 = 67 o12 = 64 o13 = 25 e11 = 46.41 e12 = 63.57 e13 = 46.02 o21 = 42 e21 = 51.77 o22 = 76 o23 = 56 o31 = 10 o32 = 23 o33 = 37 e22 = 70.91 e23 = 51.33 e31 = 20.83 e32 = 28.53 e33 = 20.65 Ho: the test scores are independent of workers’ job performance. Ha: the test scores are not independent of workers’ job performance. alpha = 1% 72 =¨ dof = 4 test criteria: if 7 2 > 13.277, heave Ho. (o − e ) 2 (67 − 46.41 ) 2 (64 − 63.57 ) 2 (37 − 20.65 ) 2 = + + ..... + = 41.02 e 46.41 63.57 20.65 Decision: since 41.02 > 13.277, we must reject Ho Conclusion: the test indicates that after taking a training program the test scores and job performance are not independent of each other. Solutions will be posted at www.prep101.com/solutions COMM 215 MOCK FINAL EXAM SOLUTIONS exam solutions page 4 Question 6 These are the average weekly losses of man-hours due to accidents in 10 industrial plants before and after a certain safety program was put into operation. Use the 0.05 level of significance to test whether the safety program is effective. Solution: We do a small sample, 2-tail t-distribution test on whether the average difference = 0. before 45 57 73 83 46 34 124 26 33 17 after 36 51 60 77 44 29 119 24 35 11 difference 9 6 7 6 2 5 5 2 –2 6 statistics ¨d 2 ¨d % d = n i = 4.6, ¨ x 2 = 300, (¨ x ) = 2116 % d = n i = 4.6 sd = n ¨ x 2 − (¨ x) 2 n(n − 1) Ho: % d = 0 Level of significance:  = 5 % df = 9 Statistic: t = sd = 10(300) − 2116 = 3.13 90 Ha: % d ! 0 Criteria: if | t | > 2.262, reject Ho %d 4.6 = = 4.65 1d/ n 3.13/ 10 Decision: 4.65 > 2.262 therefore, we must reject the null hypothesis Ho. Conclusion: The result indicates that the change in the number of man hours lost due to accidents is significant. Question 7 A random sample of 120 passengers arriving at Dorval airport took an average of 24.15 minutes with s = 3.29 minutes to retrieve their luggage and get through customs. a) construct a 95% confidence interval for the average time it takes for passengers at Dorval to retrieve their luggage and get through customs. Solution: 95% confidence interval makes z = ± 1.96 x = 24.15 ! 1.96 3.29 120 the confidence interval about = 24.15 ! 0.59 is: 23.56 < < 24.74 b) By how much is the maximum error increased if we use a 99% level of confidence? Solution: 99% confidence interval makes z = ± 2.575 E = ! 2.575 3.29 120 = 0.77 minutes 0.77 - 0.59 = 0.18 minute increase in the maximum error. c) A random sample of 40 cans of pineapple slices has an estimated mean weight of 15.85 ounces and a standard deviation of 0.23 ounces. With what confidence can we assert that this estimate is "off" by at most 0.06 ounces? Solution: z= E = 0.06 = 1.65 . s 0.23 n 40 Confidence level is 2(.4505) = 90.1% confidence. Solutions will be posted at www.prep101.com/solutions COMM 215 MOCK FINAL EXAM SOLUTIONS exam solutions page 5 Question 8 Thirteen percent (13%) of the manuscripts submitted to The New England Journal of Medicine are accepted for publication. Harvard Medical School students and profs have submitted 50 manuscripts and they claim they have a higher than average acceptance rate. a) What are Ho and Ha, the null and alternative hypotheses to test their claim. Solution: Ho: p = 0.13 Ha: p > 0.13 b) State the decision rule for a 5% significance test of their claim. Solution: It’s an upper tail test with alpha = 5% If z > 1.65, reject Ho. c) Test their claim at a 5% level of significance if 10 of their 50 papers are accepted. Solution: The sample proportion p = 10 = 0.2 50 p−p z = 1 p = 0.20 − 0.13 = 1.47 0.048 the standard deviation 1 p = .13(.87) = 0.048 50 since 1.47 < 1.65, we cannot reject Ho. d) Find the p-value and use it to state at what level of significance we can reject their claim. Solution: z = 1.47, therefore the p-value = 0.5000 – 0.4292 = 0.0708, which is P(z > 1.47). We could reject their claim at 7.1% of significance. Question 9 The table lists data on 6 different destinations of all international travellers from Canada in percent and the number who travelled to each destination in the last month from a sample of 200 customers of a local travel agency. Run a Chi-square goodness of fit test to see if the local travellers’ choice of destination differs from the national average. Use the 5% level of significance. destination Europe Far East South Am. Middle East S.E. Asia others national % 42 % 20 % 16 % 6% 12 % 4% o = 80 o = 44 o = 34 o = 16 o = 20 e = 84 e = 40 e = 32 e = 12 e = 24 Ho: The local and national proportions are the same for the 6 destinations. Ha: The local and national proportions are NOT the same for the 6 destinations. o=6 e=8 # from 200 dof = 5 test criteria: if 7 2 > 11.070, heave Ho. alpha = 5% 72 =¨ (o − e ) 2 (80 − 84 ) 2 (44 − 40 ) 2 (6 − 8 ) 2 = + + ..... + = 3.215 e 84 40 8 Decision: since 3.215 < 11.070, we cannot reject Ho Conclusion: the local travellers’ choice of destination is not different from the national average. Solutions will be posted at www.prep101.com/solutions COMM 215 MOCK FINAL EXAM SOLUTIONS exam solutions page 6 Question 10 A job-training program offers training in advanced computer programming. We suspect that the probability that an individual who completed the program will be able to find a programming job within six months, depends on previous related work experience. The data for a random sample of 80 individuals trained in the program, are in the table: Years of Experience 0 1 more than 1 # in Program 52 16 12 # who Found a Job 20 8 10 a) At a 5% level of significance, is there enough evidence to support the hypothesis that the probability of finding a job (within six months) depends on prior related working experience? Solution: Ho: Finding a job is independent of work experience. Ha: Finding a job is NOT independent of work experience. alpha = 5% dof = 2 test criteria: if 7 2 > 5.99, heave Ho. The pooled proportion of the # who found a job = (20 + 8 + 10) ÷ 80 = 0.475 The expected value in each cell is 0.475 × the # in the program. 0.475 × 52 = 24.7 72 =¨ 0.475 × 16 = 7.6 and 0.475 × 12 = 5.7. (o − e ) 2 (20 − 24.7 ) 2 (8 − 7.6 ) 2 (10 − 5.7 ) 2 = + + = 4.159 e 5.7 24.7 7.6 Decision: since 4.159 < 5.99, we cannot reject Ho Conclusion: The test indicates that finding a job is independent of work experience. b) At a 10% level of significance, is there any evidence against the hypothesis that the percentage of the 80 individuals in the program with “zero”, “one year” and “more than one year” of experience are respectively 70%, 15%, and 15%? Solution: Ho: p 1 = .70; p 2 = .15; p 3 = .15. Ha: at least one of the proportions is different from the stated values. alpha = 10% dof = 2 test criteria: if 7 2 > 4.60517, heave Ho. the expected values are: (.70)(80) = 56; 72 =¨ (.15)(80) = 12 ; (.15)(80) = 12 (o − e ) 2 (52 − 56 ) 2 (16 − 12 ) 2 (12 − 12 ) 2 = + + = 1.6193 e 56 12 12 Decision: since 1.6193 < 4.60517, we cannot reject Ho Conclusion: there is not enough evidence to support the claim that the proportions are different from those suggested. Solutions will be posted at www.prep101.com/solutions COMM 215 MOCK FINAL EXAM SOLUTIONS exam solutions page 7 Question 11 An insurance company is analyzing the profiles of its individual policyholders as they relate to the amount of life insurance coverage carried. Three independent variables are bein g considered: policyholders annual income (X 1), policyholders number of children under 21 (X 2) and policyholders age (X 3). The equation for the line of regression is ^ y = 2.43 + 0.32X 1 + 8.89X 2 + 0.14X 3 where the variables are measured in thousands of dollars. a) How much life insurance would a 40 year old woman be expected to carry if she has an annual income of $50,000 and 3 children? Solution: We set X1 = 50 and X2 = 3 and X3 = 40 to get ^ y = 2.43 + 0.32(50) + 8.89(3) + 0.14(40) = 50.7 She should carry $50 700 coverage. b) How much does the amount of coverage increase with each child? Solution: since the coefficient of X 2 = 8.89, each child increases the coverage by $8 890. c) If two 40-year-old policyholders each have five children, but one makes $40,000 per year and the other makes $50,000 per year, how much difference would you expect there to be in their respective insurance coverages? Solution: Since the only thing different is the annual income, the difference will be 50(0.32) – 40 (0.32) = 10(0.32) = 3.2 which is $3 200. Solutions will be posted at www.prep101.com/solutions COMM 215 MOCK FINAL EXAM SOLUTIONS exam solutions page 8 Question 12: This is the Excel Regression output for data on the annual income of 5 executives, related to the executive’s age and the number of years of college he or she has completed. SUMMARY OUTPUT Regression Statistics Multiple R R Square Adjusted R Square Standard Error Observations 0.9998 0.9995 0.9990 237.1248 5 ANOVA df SS 233,239,544 112,456 233,352,000 MS 116,619,772 56,228 Coefficients Standard Error -16740 1629.26 1961 35.29 5976 93.88 t Stat -10.27 55.56 63.65 Regression Residual Total 2 2 4 Intercept age (years) years of college F 2074 Significance F 0.0005 P-value 0.0093 0.0003 0.0002 a) What is the equation for the line of multiple regression? (use x 1 for age, x 2 for college years) Solution: From the column marked Coeffiecients we find b0 the constant, b 1 the coefficient of the “age” variable, and b 2, the coefficient of the “years of college completed” variable. ^ y = −16740 + 1961 x 1 + 5976 x 2 b) Use the equation from part (a) to estimate the annual income of a 39-year-old executive who has completed 4 years of college. Solution: Set x 1 = 39 and x 2 = 4 to get ^ y = −16740 + 1961 (39) + 5976(4) = 83, 643 his annual income would be $83,600 (rounded to the nearest hundred). c) By how much will a 36-year-old executive’s income increase once he completes one more year of college? Solution: Since b 2 = 5976, his income will increase by $5976 or about $6000. d) Find the coefficient of determination and interpret its meaning. Solution: According to the Excel sheet R² = .9995 which means that 99.95% of the variations in an executive’s annual income are due to variations in age and number of years of college completed. Solutions will be posted at www.prep101.com/solutions ...
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