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PSYCH/625 Version 2 Time to Practice � Week Four University of Phoenix Material

A set of practice problems designed to reinforce psychological concepts from Week Four of PSYCH/625.

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    Time to Practice – Week Four PSYCH/625 Version 2 1 Copyright © 2014 by University of Phoenix . All rights reserved. PSYCH/625 Version 2 Time to Practice – Week Four University of Phoenix Material Time to Practice – Week Four Complete Parts A, B, and C below. Part A Some questions in Part A require that you access data from Statistics for People Who (Think T hey) Hate Statistics. This data is available on the student website under the Student Text Resources link . 1. Using the data in the file named Ch . 11 Data Set 2, test the research hypothesis at the .05 level of significance that boys raise their hands in class more often than girls. Do this practice probl em by hand using a calculator. What is your conclusion regarding the research hypothesis? Remember to first decide whether this is a one - or two - tailed test. This is a one - tailed test because you want to determine if the mean of one group (boys) – μ 1 - is higher than the mean of the second group (girls) – μ 2. Null hypothesis: μ 1 = μ 2 Alternative hypothesis: μ 1 > μ 2 Boys X - Mean (X - Mean) 2 9 1.071429 1.147959 8 0.071429 0.005102 4 - 3.92857 15.43367 9 1.071429 1.147959 3 - 4.92857 24.29082 8 0.071429 0.005102 10 2.071429 4.290816 8 0.071429 0.005102 9 1.071429 1.147959 8 0.071429 0.005102 10 2.071429 4.290816 7 - 0.92857 0.862245 6 - 1.92857 3.719388 12 4.071429 16.57653 Sum = 111 Sum = 72.93 Mean = 7.93 S 2 = 72.93 / 13 = 5.61 Girls X - Mean (X - Mean) 2 3 - 2.3125 5.347656 5 - 0.3125 0.097656 1 - 4.3125 18.59766 2 - 3.3125 10.97266 6 0.6875 0.472656 4 - 1.3125 1.722656

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    Time to Practice – Week Four PSYCH/625 Version 2 2 Copyright © 2014 by University of Phoenix . All rights reserved. 3 - 2.3125 5.347656 6 0.6875 0.472656 7 1.6875 2.847656 9 3.6875 13.59766 7 1.6875 2.847656 3 - 2.3125 5.347656 7 1.6875 2.847656 6 0.6875 0.472656 8 2.6875 7.222656 8 2.6875 7.222656 Sum = 85 Sum = 85.44 Mean = 5.31 S 2 = 85.44 / 15 = 5.69 Degrees of Freedom Total = 16 + 14 – 2 = 28 S 2 pooled = 5.61 (13 / 28) + 5.69 (15 / 28) = 5.658 Variance for Boys: S 2 = 5.658 / n = 5.658 / 14 = 0.404 Variance for Girls: S 2 = 5.658 / n = 5.658 / 16 = 0.354 S 2 pop = 0.354 + 0.404 = 0.756 S pop = sqrt 0.756 = 0.869 T = (7.93 – 5.31) / 0.869 = 3.01 Looking at the critical values tables, the t - value at 28 degrees of freedom is 1.701 at p = 0.05 for a two - tailed test. The calculated t - value is larger so the test is significant. An independent samples t - test showed that there was a significant difference between the mean number of times boys raised their hands in class versus girls, t(28) = 3.01, 0.005 > p > 0.0025 for a two - tailed test. The p - value would be 0.0025 > p > 0.00125 for a one - tailed test. The boys would raise their hands more in class (M = 7.93, SD = 2.37) than girls (M = 5.31, SD = 2.39). We should reject the null. 2. Practice the following problems by hand just to see if you can get the numbers right. Using the following information, calculate the t test statistic. a. 𝑋 1 = 62 𝑋 2 = 60 𝑛 1 = 10 𝑛 2 = 10 𝑠 1 2 = 2 . 45 𝑠 2 2 = 3 . 16 Find the pooled S 2 = 2.45 (9 / 18) + 3.16 (9 / 18) = 2.45 Variance for 1 = 2.45 / 10 = 0.245 Variance for 2 = 0.245 Pop Variance = 0.245 + 0.245 = 0.49

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    Time to Practice – Week Four PSYCH/625 Version 2 3 Copyright © 2014 by University of Phoenix . All rights reserved. Sqrt 0.49 = 0.7 T = (62 – 60) / 0.7 = 2.86 b. 𝑋 1 = 158 𝑋 2 = 157 . 4 𝑛 1 = 22 𝑛 2 = 26 𝑠 1 2 = 2 . 06 𝑠 2 2 = 2 . 59 Find the pooled S 2 = 2.06 (21 / 46) + 2.59 (25 / 4 8) = 2.35 Variance for 1 = 2.35 / 22 = 0.107 Variance for 2 = 2.35 / 26 = 0.0903 Pop Variance = 0.107 + 0.0903 = 0.197 Sqrt 0.197 = 0.444 T = (158 – 157.4 ) / 0.444 = 1.35 c. 𝑋 1 = 200 𝑋 2 = 198 𝑛 1 = 17 𝑛 2 = 17 𝑠 1 2 = 2 . 45 𝑠 2 2 = 2 . 35 Find the pooled S 2 = 2.45 ( 16 / 32 ) + 2.35 ( 16 / 32 ) = 2.40 Variance for 1 = 2. 4 5 / 17 = 0.144 Variance for 2 = 2.35 / 17 = 0.138 Pop Variance = 0.1 44 + 0.138 = 0.282 Sqrt 0. 282 = 0.531 T = ( 200 - 198 ) / 0 .531 = 3.77 3. Using the results you got from Question 2 and a level of significance at .05, what are the two - tailed critical values associated with each? Would the null hypothesis be rejected? a) Critical value at df = 18 and alpha = 0.05, t = 1.734. The null hypothesis would be rejected because the calculated t = 2.86 is greater than critical t = 1.734 b) Critical value at approximate df = 40 and alpha = 0.05, t = 1.684. The null hypothesis would not be rejected because the calculated t = 1.35 is less than critical t = 1.684 c) Critical value at approximate df = 30 and alpha = 0.05, t = 1.697. The null hypothesis would be rejected because the calculated t = 3.77 is greater than critical t = 1.697 4. Using the data in the file named Ch . 11 Data Set 3, test the null hypothesis that urban and rural residents both have the same attitude toward gun control. Use IBM ® SPSS ® software to complete the analysis for this problem. Group Statistics Group N Mean Std. Deviation Std. Error Mean Attitude Urban 16 6.5112 1.77221 .44305 Rural 14 5.3979 3.31442 .88582

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    Time to Practice – Week Four PSYCH/625 Version 2 4 Copyright © 2014 by University of Phoenix . All rights reserved. Independent Samples Test Levene's Test for Equality of Variances t - test for Equality of Means F Sig. t df Sig. (2 - tailed) Mean Difference Std. Error Difference 95% Confidence Interval of the Difference Lower Upper Attitude Equal variances assumed 4.463 .044 1.168 28 .253 1.11339 .95311 - .83897 3.06576 Equal variances not assumed 1.124 19.273 .275 1.11339 .99044 - .95763 3.18442 An independent samples t - test was performed on the sample. The mean attitude for urban residents (M = 6.51, SD = 1.77) was slightly higher than the mean attitude for rural resident s (M = 5.40, SD = 3.31). The Levene’s Test was significant , F = 4.463, p = 0.044, which suggests that the variances of the two groups are not equal. The t - test was not significant, t (19) = 1.124, p = 0.274. There is no difference in attitude on gun control among urban and rural residents. We would fail to reject the null. 5. In the following examples, indicate whether you would perform a t test of independent means or dependent means. a. Two groups were exposed to different treatment levels for ankle sprains. Which treatment was most effective? A t - test of in dependent means would be used because the two groups were mutually exclusive and had different subjects in each group. b. A researcher in nursing wanted to know if the recovery of patients was quicker when some received additional in - home care whereas when others received the standard amount. A t - test of independent means was used because the groups were independent of each other. c. A group of adolescent boys was offered interpersonal skills counseling and then tested in September and May to see if there was any impact on family harmony. A t - test of dependent means was used because the same group of boys was tested again at two different time points . d. One group of adult men was given instructions in reducing their high blood pressure whereas another was not given any instructions. A t - test of independent means was used because two different groups of men are being tested. e. One group of men was provided access to an exercise program and tested two times over a 6 - month period for heart health. A t - test of dependent means was used because the same group of men was tested twice. 6. The data set for this problem can be found through the Sage Materials in the Student Textbook

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