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Solutions to statistics homework problems involving descriptive and inferential statistics. The problems include calculating mean, median, mode, variance, standard deviation, and confidence intervals for different sets of data. The data sets cover various topics such as marriage counseling, sample data, and educational levels of voters. The homework also includes tests to determine if there is a statistically significant difference between means of different populations.

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Download Statistics Homework: Descriptive and Inferential Analysis - Prof. Mihaiela Gugiu and more Study notes Political Science in PDF only on Docsity! 1 PSC 280 November 23, 2010 Homework # 5_Answers Complete the following exercises by hand and make sure to include ALL steps in the homework. You can choose to either write (legible) or type (by using the Equation option in Word). If you do NOT present every single step of these analyses, you will NOT earn all the points! 1. Newsweek reported statistics on the number a visits a couple makes to a therapist while undergoing marriage counseling. With data based on this article, assume that the following show the number of visits for a sample of nine couples: X = {12, 8, 3, 13, 18, 20, 10, 9, 18}. Compute the following descriptive statistics for these data: (6 points) a) Mean b) Median X={3, 8, 9, 10, 12, 13, 18, 18, 20} c) Mode = 18 d) Variance 75.30 )33.12(9)18()9()10()20()18()13()3()8()12( 19 1 )()( 1 1 2222222222 2 1 22 Xnx n s n i i e) Standard deviation f) Standard error 33.12189102018133812 9 11 1 n i ix n x 1252/)19(]2/)1[( XXXMedian n 55.575.30)( 1 1 1 2 n i i xx n s 85.19/55.5/ nsSE 2 2. The following data have been collected from a sample of 8 items: X = {10, 8, 12, 15, 13, 11, 6, 5}. Compute the following: a) Mean (1 points) b) Standard deviation (1 points) 97.11)10(8)5()6()11()13()15()12()8()10( 18 1 )()( 1 1 222222222 2 1 22 Xnx n s n i i c) 95% confidence interval (5 points) D.f.: n-1=8-1=7 T critical value for α = .05 is 2.365 LB: 10 - (2.365*1.223) = 7.11 UB: 10 + (2.365*1.223) = 12.89 95% CI: [7.11, 12.89] 3. A simple random sample of 20 items resulted in a sample mean of 17.25 and a sample standard deviation of 3.3. Compute the following: Note: you may get slightly different numbers depending on how you do the rounding. However, the difference should be very small! a) 90% confidence interval (5 points) t.05= 1.729 90% CI: [15.97, 18.53] 105611131512810 8 11 1 n i ix n x 46.397.11)( 1 1 1 2 n i i xx n s SE 2/tX 223.18/46.3/ nsSE 738.20/3.3/ nsSE 5 Gender Party Affiliation Total Democratic Republican Independent Women 19 (18.8) 15 (17.7) 24 (21.5) 58 Men 16 (16.2) 18 (15.3) 16 (18.5) 50 Total 35 33 40 108 *Expected counts are in parentheses. a) Determine whether the relationship between party affiliation and gender is statistically significant at α = .05. (8 points) χ 2 = 1.521 D.f. = (3-1)*(2-1) = 2 For α = .05, critical value of χ 2 = 5.99147, which is larger than 1.521. Hence, the relationship is NOT statistically significant. 7. Consider the following hypothesis test: H0 : µ = 20 Ha : µ ≠ 20 Data from a sample of six items are as follows: X = {18, 20, 16, 19, 17, 18}. a) Compute the mean (1 points) b) Compute the standard deviation (1 points) 99.1)()( 1 1 2 1 22 Xnx n s n i i c) Compute the value of the test statistic t. (5 points) 18 1 1 n i ix n x 41.199.1)( 1 1 1 2 n i i xx n s 46.3 6/41.1 2018 / ns x t 6 *Note that the denominator is the standard error. d) What is your conclusion? (1 point) If we assume that α = .05, the critical t value for (n-1) = 19 degrees of freedom is 2.093. We divide alpha by 2 because the alternative hypothesis does not indicate a direction of the relationship. Since, 3.46 is larger than 2.093, we can reject the null hypothesis and conclude that the sample mean does not equal 20. 8. City Homes bought 101 badly deteriorated Baltimore rowhouses and turned them into housing for the poor. The mean monthly rent for these houses was $200. Nine hundred homes in another city were built and turned over to a foundation to rent. A sample of 10 of these homes showed the following monthly rental rates: $220, $190, $250, $230, $185, $210, $240, $260, $200, $ 195. a) Using α = .05, test to see if the mean monthly rental rate for the population of 900 homes exceeds the $200 mean monthly rental rate in Baltimore. (5 points) b) What is your conclusion? (1 point) H0 : µ ≤ 200 Ha : µ > 200 * Note that this is a one-sided test because the directionality of the relationship I being specified by the alternative hypothesis. n = 10 112.690)()( 1 1 2 1 22 Xnx n s n i i 218 1 1 n i ix n x 27.26112.690)( 1 1 1 2 n i i xx n s 17.2 10/27.26 200218 / ns x t 7 For 9 degrees of freedom, the critical t value is 1.833, which is smaller than 2.17. Hence, we can reject the null hypothesis and conclude that the population mean rental rate exceeds the $200 per month rate in Baltimore. 9. Starting salary data for college graduates is reported by the College placement Council. Annual salaries in thousands of dollars for a sample of accounting majors and a sample of finance majors are shown below. Accounting Finance 28.8 28.1 26.3 29.0 25.3 24.7 23.6 27.4 26.2 25.2 25.0 23.5 27.9 29.2 23.0 26.9 27.0 29.7 27.9 26.2 26.2 29.3 24.5 24.0 a) Using α = .05, test the hypothesis that there is no difference between the mean annual starting salaries for the populations of accounting majors and finance majors. (5 points) nAccounting= 12 nFinance= 12 H0 : µ1 = µ2 Ha : µ1 ≠ µ2 12.3 )3.27(123.297.292.292.257.241.282.260.279.272.263.258.28 112 1 )()( 1 1 2222222222222 2 1 22 1 Xnx n s n i i 28.4)()( 1 1 2 1 22 1 Xnx n s n i i 6.25 3.27 Finance Accounting x x 7.3 22 08.4732.34 21212 28.4)112(12.3)112( 2 )1()1( 21 2 22 2 112 nn snsn s p 10 Hence, SSW = MSW * (nT – k) = 25 * 12 = 300 SST = SSW + SSB SSB = SST –SSW = 1500 – 300 = 1200 b) Using α = .05 level of significance, test the null hypothesis that the means of the four populations are equal. What is your conclusion? (4 points) For 3 degrees of freedom numerator and 12 degrees of freedom denominator, F3,12 = 3.49, which is smaller than the value obtained above, i.e., F = 16. Hence, we can reject the null hypothesis and conclude that the means of the four populations are not equal. 13. Random samples of 25 observations were selected from each of three populations. For these data, SSB = 120 and SSW = 216. a) Set up and complete the ANOVA table. (5 points) Source of Variation Sum of Squares Degrees of Freedom Mean Square F Between 120 2 60 20 Within 216 72 3 Total 336 75 Note: there are 3 samples drawn from 3 different populations and each sample has 25 observations. D.f.(SSB) = k-1= 3-1=2 D.f. (SSW) = nT- k = 75 – 3 = 72 D.f. (Total) = 2 + 72 = 75 SST = 120 + 216 = 336 kn SSW MSW T 16 25 400* MSW MSB F 60 2 120 1 k SSB MSB 3 72 216 kn SSW MSW T 20 3 60* MSW MSB F 11 b) At the α = .05 level of significance, what is the critical F value? (2 points) F2,72 = 3.07 (since the table has values up to 60 df, we can use that value to make our decision). H c) At the α = .05 level of significance, can we reject the null hypothesis that the means of the three populations are equal? (2 point) Yes, we can reject the null hypothesis because 20 is higher than the critical value.