Module 5 Practice problem and Homework answers

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Module 5 Practice problem and Homework answers Practice problem What is the mean for the before period? Answer: 5.3 x = 74.1 14 = 5.3 What is the mean for the after period? Answer: 6.8 x = 95.3 14 = 6.8 What is the sum of squared deviations for the before group? Answer: 28.23 What is the sum of squared deviations for the after group? Answer: 31.35 x 1 (before) x 1 X 1 -x 1 (x 1 -x 1) 2 x 2 (after) x 2 x 2 -x 2 (x 2 -x 2) 2 4.2 5.3-1.1 1.21 8 6.8 1.2 1.44 7.7 5.3 2.4 5.76 6.2 6.8-0.6 0.36 3.3 5.3-2.0 4.00 4.1 6.8-2.7 7.29 4.7 5.3-0.6 0.36 7.3 6.8 0.5 0.25 4.9 5.3-0.4 0.16 5.4 6.8-1.4 1.96 5.1 5.3-0.2 0.04 6.1 6.8-0.7 0.49 6.8 5.3 1.5 2.25 7.5 6.8 0.7 0.49 4.7 5.3-0.6 0.36 7.8 6.8 1.0 1.00 3.3 5.3-2.0 4.00 9.5 6.8 2.7 7.29 7 5.3 1.7 2.89 4.5 6.8-2.3 5.29 5.1 5.3-0.2 0.04 8.2 6.8 1.4 1.96 5.9 5.3 0.6 0.36 5.9 6.8-0.9 0.81 7.5 5.3 2.2 4.84 6.4 6.8-0.4 0.16 3.9 5.3-1.4 1.96 8.4 6.8 1.6 2.56 Σ = 28.23 Σ = 31.35

Practice problem, continued What is the standard deviation of the before group? Answer: 1.42 s 1 = (x 1 x 1) 2 N = 28.23 14 = 1.42 What is the standard deviation of the after group? Answer: 1.50 s 2 = (x 2 x 2) 2 N = 31.35 14 = 1.50 What is the value of the standard error of the difference of means? (What is the result of the sigma formula?) Answer: 0.562 σ x 1 x 2 = N 1 s 1 2 + N 2 s 2 2 N 1 + N 2 2 N 1 + N 2 N 1 N 2 σ x 1 x 2 = 14 1.422 + 14 1.50 2 14 + 14 2 14 + 14 14 14 What is the value of t? Answer: -2.67 = 14 2.02+14 2.25 28 26 = 196 28.28+31.50 26 = 2.30. 14 = 1.52 * 0.37 = 0.562 t = x 1 x 2 σ x 1 x 2 0.14 t = 5.3 6.8 0.562 = 1.5 0.562 = 2.67

Practice problem, continued What is the df for this problem? Answer: 26 df = N 1 + N 2 2 = 14 + 14 2 = 26 What is the critical t value? Answer: 1.706 For a one-tailed test at.05 significance level and 26 degrees of freedom, the degrees of freedom value is 1.706. Is the difference between groups significant? Answer: yes Because the calculated value (-2.67) has a higher magnitude than the critical value (1.706), we can reject the null hypothesis and conclude that there is a significant difference between groups. Was the manager correct that the discount increased revenue, or was her boss correct that there was no significant difference in sales before and after the discount? Who was right? Answer: the manager Because the difference between groups is significant, and because the mean sales for the before period was greater than the mean for the after period, the manager was right. Homework problems page 1 Is the research hypothesis one-tailed or two-tailed? Answer: two-tailed The set-up to this problem says that the research hypothesis is that the number of emails varies significantly by project, which predicts that there is a significant difference between projects, but not which project will have the greater number of emails. Because there is no specific direction indicated in this hypothesis, it is two-tailed. Calculate the mean for Project A. Answer: 19.7 x = 197 10 = 19.7

Homework page 1, continued Calculate the mean for Project B. Answer: 24.9 x = 249 10 = 24.9 Calculate the standard deviation for Project A. Answer: 3.13 Project A mean deviation squared deviation 17 19.7-2.7 7.29 24 19.7 4.3 18.49 23 19.7 3.3 10.89 18 19.7-1.7 2.89 15 19.7-4.7 22.09 21 19.7 1.3 1.69 17 19.7-2.7 7.29 21 19.7 1.3 1.69 17 19.7-2.7 7.29 24 19.7 4.3 18.49 Σ = 98.10 s 1 = (x 1 x 1) 2 N = 98.10 10 = 3.13

Homework page 1, continued Calculate the standard deviation for Project B. Answer: 7.03 Project B mean deviation squared deviation 28 24.9 3.1 9.61 30 24.9 5.1 26.01 32 24.9 7.1 50.41 13 24.9-11.9 141.61 32 24.9 7.1 50.41 28 24.9 3.1 9.61 20 24.9-4.9 24.01 19 24.9-5.9 34.81 32 24.9 7.1 50.41 15 24.9-9.9 98.01 Σ = 494.90 s 2 = (x 2 x 2) 2 N Calculate t. = 494.90 10 Answer: -2.02 = 7.03 t = x 1 x 2 σ x 1 x 2, σ x 1 x 2 = N 1 s 1 2 +N 2 s 2 2 N 1 +N 2 2 N 1+N 2 N 1 N 2 σ x 1 x 2 = 10 3.132 + 10 7.03 2 10 + 10 2 10 + 10 10 10 = 10 9.80+10 49.42 20 = 18 100 98.0+494.2 0.2 = 32.9. 2 18 = 5.74 * 0.45 = 2.58 t = 19.7 24.9 2.58 = 5.2 2.58 = 2.02

Homework page 1, continued What is the df? Answer: 18 df = N 1 + N 2 2 = 10 + 10 2 = 18 What is the critical t value? Answer: 2.101 For a two-tailed test at.05 significance level and 18 degrees of freedom, the degrees of freedom value is 2.101. Is the difference between groups significant? Answer: no Because the calculated value (-2.02) has a lower magnitude than the critical value (2.101), we fail to reject the null hypothesis and conclude that there is not a significantly different number of emails between projects. page 2 Is the research hypothesis one-tailed or two-tailed? Answer: one-tailed Calculate the t value. Answer: 1.69 The set-up to this problem says that the research hypothesis is that one salon has had a significantly higher number of recent new clients than the other one has. Because this hypothesis predicts a specific direction, it is one-tailed. t = x 1 x 2 σ x 1 x 2, σ x 1 x 2 = N 1 s 1 2 +N 2 s 2 2 N 1 +N 2 2 N 1+N 2 N 1 N 2 σ x 1 x 2 = 26 2.862 + 26 2.95 2 26 + 26 2 26 + 26 26 26 = 26 8.18+26 8.70 52 50 = 676 212.68+226.20 50 = 2.96 * 0.28 = 0.83 0.08 = 8.78. 08

Homework page 2, continued t = 9.2 7.8 0.83 What is the df? Answer: 50 = 1.4 0.83 = 1.69 df = N 1 + N 2 2 = 26 + 26 2 = 50 What is the critical t value? Answer: 1.671 For a one-tailed test at.05 significance level and 50 degrees of freedom, the degrees of freedom value is 1.671. (Note: when the specific df value you re working with isn t in the t table, round up to the next value. In this case, we ve used df=60.) Is the difference between groups significant? Answer: yes Because the calculated value (1.69) has a higher magnitude than the critical value (1.671), we can reject the null hypothesis and conclude that the downtown salon did have a significantly greater number of recent new clients than the suburb salon did. Homework page 3 Is the research hypothesis one-tailed or two-tailed? Answer: one-tailed The research hypothesis suggests that the amount of recyclable material that is thrown in the trash will be significantly lower after the campaign than before it. Because one period is predicted to be significantly higher than the other, this is a one-tailed hypothesis.

Homework page 3, continued Calculate the t value. Answer: 4.59 t = x 1 x 2 σ x 1 x 2, σ x 1 x 2 = N 1 s 1 2 +N 2 s 2 2 N 1 +N 2 2 N 1+N 2 N 1 N 2 σ x 1 x 2 = 31 1.432 + 30 1.41 2 31 + 30 2 31 + 30 31 30 = 31 2.04+30 1.99 61 59 = 930 63.24+59.70 59 = 1.44 * 0.26 = 0.37 0.07 = 2.08. 07 t = 19.6 17.9 0.37 = 1.7 0.37 = 4.59 What is the df? Answer: 59 df = N 1 + N 2 2 = 30 + 31 2 = 59 What is the critical t value? Answer: 1.671 For a one-tailed test at.05 significance level and 59 degrees of freedom, the degrees of freedom value is 1.671. Is the difference between groups significant? Answer: yes Because the calculated value (4.59) has a higher magnitude than the critical value (1.671), we can reject the null hypothesis and conclude that there was significantly less recyclable material thrown in the trash after the campaign than before it.