Hypothesis Testing: Two-Sample T-Test for Means, Exams of Statistics

Download Hypothesis Testing: Two-Sample T-Test for Means and more Exams Statistics in PDF only on Docsity! Pratice Test 4 01/01/2007 1 Abe Mirza Practice Test # 4 Statistic Hypothesis Testing Topics Page Problems 1 Solutions 4-6 1- A consulting agency was asked by a large insurance company to investigate if business majors were better salespersons that those with other majors. A sample of 40 salespersons with a business degree showed that they sold an average 0f 11 insurance policies per week with a st. dev. of 1.8 policies. Another sample of 45 salespersons with a degree other than business showed that they sold an average of 9 insurance policies per week with a st. dev. of 1.35 policies. At 2.5% significance level, can it be concluded that persons with business degrees are better salespersons than those who have a degree in another area? 2-Among drivers who have had a car crash in the last year, 88 are randomly selected and categorized by age, with the results listed below. If all ages have the same crash rate, we would expected (because of the age distribution of licensed drivers) the given categories to have 16%, 44%, 27% and 13% of the subjects, respectively. At 10% significance level, test the claim that the given distribution of crashes conforms to the distribution of ages. 3- Many students suffer from math anxiety. A professor who teaches statistics offered her students a 2 – hour lecture and ways to overcome on math anxiety. The following table gives the test scores statistics of seven students before and after they attendant this lecture. Before 56 69 48 74 65 71 60 After 62 73 44 85 71 70 73 d=difference Σd = dd s= = Test at 2.5% whether attending this lecture increases the average score in statistics. 4- A real estate agent claims that the average price of a 1-week condominium time-share on Hilton Head Island, SC is at most $19,200. The population standard deviation is $2,100. A sample of 34 such units had an average selling price of $20,145. Does the evidence support the claim at a α = 0.05? 5- Experts claim that women commit 10% of murders. Is there enough evidence to reject this claim if in a sample of 67 murders, women committed 10? Use a significance level of 0.01 6- It is claimed that the average GPA of ARC students is at least 2.8. A sample of 10 ARC students showed an average GPA of 2.78 with a standard deviation of .62. Use α = 0.01 to test this claim. 7- It is claimed that the average GPA of Sierra students is more than 2.8. A sample of 56 Sierra students showed an average GPA of 2.88 with a standard deviation of 1.5 Use α = .025 to test this claim. Age Under 25 25-44 45-64 Over 64 Total O(observed)= # of Drivers 36 21 12 19 88 Pratice Test 4 01/01/2007 2 8- It is claimed that the average GPA of ARC students is 2.8. A sample of 10 students showed the following; 2.6, 3.2, 2.5, 2.8, 2.7, 3.4, 3.1, 2.2, 3.5, 2.9. Use α = of 0.01 to test this claim. 9- The following table shows the number of persons in a random sample of 210 listed according to the day of the week they prefer to do their grocery shopping. Day of the Week Mon Tue Wed Thu Fri Sat Sun O(observed) = Shoppers 9 17 12 26 36 69 41 At α = 2.5% , test the null hypothesis that the proportion of persons who prefer to do their grocery shopping on a particular day is the same for all days of the week. 10- A real estate agent claims that the average price of a 1-week condominium time-share on Pinole Island, S C is at least $23,500. The population standard deviation is $2,900. A sample of 14 such units had an average selling price of $21,545. Does the evidence support the claim at α = 0.01? 11- Experts claim that women commit 15% of homicides. Is there enough evidence to reject this claim if in a sample of 105 homicides, women committed 18? Use a significance level of 0.10. 12- A consulting agency was to asked by a large insurance company to investigate if business majors were better salespersons that those with other majors. A sample of 60 salespersons with a business degree showed that they sold an average 0f 13 insurance policies per week with a st. dev. of 2.3 policies. Another sample of 75 salespersons with a degree other than business showed that they sold an average of 16 insurance policies per week with a st. dev. of 2.58 policies. At α = 5%, can it be concluded that persons with business degrees are as good salespersons as those who have a degree in another area? 13- There was a research on the weights at birth of the children of urban and rural women. The researcher suspects there is a significant difference between the mean weights at birth of children of urban and rural women. To test this hypothesis, he selects independent random samples of weights at birth of children of mothers from each group, with the following mean weights and standard deviations. Test the researcher's belief, at α = .01. Urban mothers (µ 1 ) n1 = 75 1 3x = s1 = .11 Rural mothers (µ 2 ) n2 = 64 2 2.95x = s2 = .09 14- Abe Claims that generally 22% of his students are getting grade of “A”, 26% “B”, 24% “C”, 12% “D” and 16 they % “F”. The following table lists the grade distribution for a sample of 100 students for stat class Grade A B C D F Total O(observed)=Students 20 24 26 12 18 100 At 10% significance level, test Abe’s claim. 15- 50 smokers were questioned about the number of hours they sleep each day. We want to test the hypothesis that the smokers need less sleep than the general public which needs an average of 7.7 hours of sleep. We follow the steps below. If the sample mean is 7.5 and the standard deviation is .5, what can you conclude? 16- Suppose that you interview 1000 exiting voters about who they voted for governor. Of the 1000 voters, 450 reported that they did not vote for the democratic candidate. Is there sufficient evidence to suggest that the democratic candidate will win the election at the .01 level? 17-The liquid chlorine added to swimming pools to combat algae has a relatively short shelf live. Records indicate the mean shelf life is 2160 hours. Another chemical was added to see if it would increase the life. A sample of nine jugs with the additive had a sample mean of 2172.44 hours with a sample standard deviation of 9.3823 hours. Test at the 0.025 level whether the chemical additive increases the shelf life. Pratice Test 4 01/01/2007 5 H0: Equal Proportions of shoppers every day of the week. H1: Unequal Proportions of shoppers every day of the week. 9 Day of the week M T W Th F Sa Su Observed 19 17 12 26 36 69 41 210 Expected 30 30 30 30 30 30 30 210 2( )O E− (19-30) 2 (17-30)2 (12-30)2 (26-30)2 (36-30)2 (69-30)2 (41-30)2 121 169 324 16 36 1251 121 2( ) /O E E− 4.03 + 5.63 + 10.8 + .53 + 1.2 + 50.7 + 4.03 = TS = 76.92 ∑ 2( )O E E − =76.92 2 14.449cv χ= = TS = 76.92 Conclusion: Reject H0 Comment: Accept H1 10 SC: 23, 500µ ≥ OC: 23, 500µ < Ho: 23, 500µ ≥ H1: 23, 500µ < (LTT) 2.650t = − 14 (21545 23500) 2900 2.52t − = −= Accept Ho Accept SC 11 SC: 0.15P = OC: 0.15P ≠ Ho: 0.15P = H1: 0.15P ≠ (TTT) 1.645z = ± 0.171 .150 0.603 .15(1 .15) 105 Z − = = − Accept Ho Accept SC 12 SC: 1 2µ µ= OC: 1 2µ µ≠ H0: 1 2 0µ µ− = H1: 1 2 0µ µ ≠− (TTT) 1.96z = ± 2 2 (13 16) 0 2.3 2.58 60 75 Z = − − = + -7.14 Reject Ho Reject SC 13 SC: 1 2µ µ≠ OC: 1 2µ µ= Ho: 1 2µ µ= H1: 1 2µ µ≠ (TTT) 2.58z = ± 2 2 (3 2.95) 0 .11 .09 75 64 2.947Z − − + = = Reject Ho Accept SC 14 H0:Stated proportions are correct. H1: Stated proportions are in correct 2χ = 7.779 2χ = ∑ 2( )O E E − = 0.75 Accept Ho Accept SC 15 SC: 7.7µ < OC: 7.7µ ≥ Ho: 7.7µ ≥ H1: 7.7µ < (LTT) 1.645z = − 50 (7.5 7.7) 0.5 2.83z − = −= Reject Ho Accept SC 16 SC: 0.5P > OC: 0.5P ≤ Ho: 0.5P ≤ H1: 0.5P > (RTT) 2.326z = .55 .5 .5(1 .5) 1000 3.16Z − = − = Reject Ho Accept SC 17 SC: 2160µ > OC: 2160µ ≤ Ho: 2160µ ≤ H1: 2160µ > (RTT) 2.306t = 9 (2172.44 2160) 3.978 9.3823 t − == Reject Ho Accept SC 19- a SC: 5µ < OC: 5µ ≥ Ho: 5µ ≥ H1: 5µ < (LTT) 1.345t = − 15(4.287 5) 1.6867 1.638 t − = −= Accept Ho Reject SC Pratice Test 4 01/01/2007 6 19- b To perform the t-test on the population mean you must assume that the data are approximately normally distributed test 19- c Use the same method as project 2. t test for normality. 19- d SC: 5µ ≤ OC: 5µ > H0: 5µ ≤ H1: 5µ > (RTT) 1.345t = 15(4.287 5) 1.6867 1.638 t − = −= Accept Ho Reject SC 20 SC: s lµ µ> OC: s lµ µ≤ H0: s lµ µ≤ H1: s lµ µ> (RTT) 1.645z = 2 2 (28.2 22.6) 0 3.0 5.1 40 40 5.6 5.98 0.936 Z − − + = = = Reject Ho Accept SC 21 H0: Die is fair. H1: Die is not fair Numbers 1 2 3 4 5 6 7 8 Observed 24 36 15 17 9 22 18 19 160 Expected 20 20 20 20 20 20 20 20 160 2( )O E− (24-20) 2 (36-20)2 (15-20)2 (17-20)2 (9-20)2 (22-20)2 (18-20)2 (19-20)2 36 256 25 9 121 64 4 1 2( ) /O E E− 1.8 + 12.8 + 1.25 + .45 + 6.05 + 3.2 + .2 + .05 = TS = 25.8 2 12.017cv χ= = TS = 25.8 Conclusion: Reject H0 Comment: Accept H1 22 SC: 190µ > OC: 190µ ≤ H0: 190µ ≤ H1: 190µ > (RTT) 1.645z = 100 (198 190) 15 5.33z − == Reject Ho Accept SC 23 SC: 0.1P = OC: 0.1P ≠ H0: 0.1P = H1: 0.1P ≠ (TTT) 1.96z = ± 0.133 .10 .10(1 .10) 150 1.361Z −= − = Accept Ho Accept SC 24 H0: Same number of returns each day of the week. H1: Different number of returns each day of the week Returns M T W R F S S Observed 14 10 12 16 26 33 29 140 Expected 20 20 20 20 20 20 20 140 2( )O E− (14-20) 2 (10-20)2 (12-20)2 (16-20)2 (26-20)2 (33-20)2 (29-20)2 36 100 64 16 36 169 81 2( ) /O E E− 36/20 100/20 64/20 16/20 36/20 9/20 81/20 1.8 + 5 + 3.2 + .8 + 1.8 + 8.45 + 4.05 = 25.01 ∑ 2( ) / EO E− = 25.01 2cv χ= = 10.645 TS = 25.01 Conclusion: Reject H0 Comment: Accept H1 SC: 2.2µ < OC: 2.2µ ≥ H0: 2.2µ ≥ H1: 2.2µ <