sample_size = random.randint(5,15)
RandomData(groups = 2, n = sample_size).dependent_samples_t_test()6 Dependent-Samples t-tests
Sample Problems
6.1 About
the dependent-samples t-Test problems use the RamdomData class which requires:
- the groups variable set to 2:
groups = 2 - the sample size per group - all groups will have the same sample size
- a call to the
dependent_samples_t_test()method
In these sample problems, the per-group sample size is randomly set between 5 and 15. An example funciton call is included below
6.2 Problem 1
Given the following within-subjects data, is \(M_D\) significantly different from \({0}\)? Use a \({2}\) tailed-test with \(\alpha = {0.05}\)
| ID | A | B |
|---|---|---|
| 1 | 14 | 13 |
| 2 | 13 | 11 |
| 3 | 19 | 18 |
| 4 | 20 | 17 |
| 5 | 16 | 13 |
| 6 | 15 | 14 |
| 7 | 17 | 17 |
| 8 | 19 | 14 |
| 9 | 19 | 20 |
Summary statistics for these data:
\[M_A = {16.89}, M_B = {15.22}\] \[n = {9}\]
State the Hypotheses
\[H_0: \mu_D = 0\] \[H_1: \mu_D \ne 0\]
The decision criteria:
\(t_{crit} = \pm{2.31}, \alpha_{two-tailed} = {0.05}, df = {8}\)
Calculating the difference scores \(D = X_B - X_A\)
| ID | A | B | D |
|---|---|---|---|
| 1 | 14 | 13 | -1 |
| 2 | 13 | 11 | -2 |
| 3 | 19 | 18 | -1 |
| 4 | 20 | 17 | -3 |
| 5 | 16 | 13 | -3 |
| 6 | 15 | 14 | -1 |
| 7 | 17 | 17 | 0 |
| 8 | 19 | 14 | -5 |
| 9 | 19 | 20 | 1 |
Calculate the Mean of the Difference Scores \[M_D = \frac{\Sigma D}{n}\]
\[M_D = \frac{-15}{9}\]
\[M_D = {-1.67}\]
Create a column for the squared difference scores \(D^2\)
| ID | A | B | D | D^2 |
|---|---|---|---|---|
| 1 | 14 | 13 | -1 | 1 |
| 2 | 13 | 11 | -2 | 4 |
| 3 | 19 | 18 | -1 | 1 |
| 4 | 20 | 17 | -3 | 9 |
| 5 | 16 | 13 | -3 | 9 |
| 6 | 15 | 14 | -1 | 1 |
| 7 | 17 | 17 | 0 | 0 |
| 8 | 19 | 14 | -5 | 25 |
| 9 | 19 | 20 | 1 | 1 |
New Summary statistics for the difference scores:
\[M_D = {-1.67}, \quad \Sigma{D} = {-15}, \quad \Sigma{D^2} = {51}\]
Calculate SS of the difference scores
\[SS_D = \Sigma D^2 - \frac{(\Sigma D)^2}{n}\]
\[SS_D = {51} - \frac{225}{9}\]
\[SS_D = {51} - {25.0}\]
\[SS_D = {26.0}\]
Calculate the variance of the difference scores
\[s_D^2 = \frac{SS_D}{df}\]
\[s_D^2 = \frac{26.0}{8}\]
\[s_D^2 = {3.25}\]
Calculate the estimated standard error of the difference scores
\[s_{M_D} = \sqrt{\frac{s^2}{n}}\]
\[s_{M_D} = \sqrt{\frac{3.25}{9}}\]
\[s_{M_D} = \sqrt{0.36}\]
\[s_{M_D} = {0.6}\]
Calculate \(t_{obt}\)
\[t_{obt} = {\frac{M_D - \mu_D}{s_{M_D}}}\]
\[t_{obt} = \frac{-1.67 - 0}{0.6}\]
\[t_{obt} = \frac{-1.67}{0.6}\]
\[t_{obt} = {-2.78}\]
Calculating Cohen’s d
\[d = \frac{M_D}{\sqrt{s^2}}\]
\[d = \frac{-1.67}{{{\sqrt{3.25}}}}\]
\[d = \frac{-1.67}{1.8}\]
\[d = {-0.93}\]
The results:
reject the null hypothesis, results are significant,
t(8) = -2.78, p < 0.05, d = -0.93
6.3 Problem 2
Given the following within-subjects data, is \(M_D\) significantly different from \({0}\)? Use a \({2}\) tailed-test with \(\alpha = {0.05}\)
| ID | A | B |
|---|---|---|
| 1 | 15 | 27 |
| 2 | 22 | 16 |
| 3 | 18 | 27 |
| 4 | 24 | 26 |
| 5 | 10 | 20 |
| 6 | 17 | 26 |
| 7 | 19 | 7 |
| 8 | 24 | 30 |
| 9 | 22 | 27 |
| 10 | 15 | 23 |
| 11 | 15 | 23 |
| 12 | 17 | 24 |
| 13 | 16 | 37 |
Summary statistics for these data:
\[M_A = {18.0}, M_B = {24.08}\] \[n = {13}\]
State the Hypotheses
\[H_0: \mu_D = 0\] \[H_1: \mu_D \ne 0\]
The decision criteria:
\(t_{crit} = \pm{2.18}, \alpha_{two-tailed} = {0.05}, df = {12}\)
Calculating the difference scores \(D = X_B - X_A\)
| ID | A | B | D |
|---|---|---|---|
| 1 | 15 | 27 | 12 |
| 2 | 22 | 16 | -6 |
| 3 | 18 | 27 | 9 |
| 4 | 24 | 26 | 2 |
| 5 | 10 | 20 | 10 |
| 6 | 17 | 26 | 9 |
| 7 | 19 | 7 | -12 |
| 8 | 24 | 30 | 6 |
| 9 | 22 | 27 | 5 |
| 10 | 15 | 23 | 8 |
| 11 | 15 | 23 | 8 |
| 12 | 17 | 24 | 7 |
| 13 | 16 | 37 | 21 |
Calculate the Mean of the Difference Scores \[M_D = \frac{\Sigma D}{n}\]
\[M_D = \frac{79}{13}\]
\[M_D = {6.08}\]
Create a column for the squared difference scores \(D^2\)
| ID | A | B | D | D^2 |
|---|---|---|---|---|
| 1 | 15 | 27 | 12 | 144 |
| 2 | 22 | 16 | -6 | 36 |
| 3 | 18 | 27 | 9 | 81 |
| 4 | 24 | 26 | 2 | 4 |
| 5 | 10 | 20 | 10 | 100 |
| 6 | 17 | 26 | 9 | 81 |
| 7 | 19 | 7 | -12 | 144 |
| 8 | 24 | 30 | 6 | 36 |
| 9 | 22 | 27 | 5 | 25 |
| 10 | 15 | 23 | 8 | 64 |
| 11 | 15 | 23 | 8 | 64 |
| 12 | 17 | 24 | 7 | 49 |
| 13 | 16 | 37 | 21 | 441 |
New Summary statistics for the difference scores:
\[M_D = {6.08}, \quad \Sigma{D} = {79}, \quad \Sigma{D^2} = {1269}\]
Calculate SS of the difference scores
\[SS_D = \Sigma D^2 - \frac{(\Sigma D)^2}{n}\]
\[SS_D = {1269} - \frac{6241}{13}\]
\[SS_D = {1269} - {480.08}\]
\[SS_D = {788.92}\]
Calculate the variance of the difference scores
\[s_D^2 = \frac{SS_D}{df}\]
\[s_D^2 = \frac{788.92}{12}\]
\[s_D^2 = {65.74}\]
Calculate the estimated standard error of the difference scores
\[s_{M_D} = \sqrt{\frac{s^2}{n}}\]
\[s_{M_D} = \sqrt{\frac{65.74}{13}}\]
\[s_{M_D} = \sqrt{5.06}\]
\[s_{M_D} = {2.25}\]
Calculate \(t_{obt}\)
\[t_{obt} = {\frac{M_D - \mu_D}{s_{M_D}}}\]
\[t_{obt} = \frac{6.08 - 0}{2.25}\]
\[t_{obt} = \frac{6.08}{2.25}\]
\[t_{obt} = {2.7}\]
Calculating Cohen’s d
\[d = \frac{M_D}{\sqrt{s^2}}\]
\[d = \frac{6.08}{{{\sqrt{65.74}}}}\]
\[d = \frac{6.08}{8.11}\]
\[d = {0.75}\]
The results:
reject the null hypothesis, results are significant,
t(12) = 2.7, p < 0.05, d = 0.75
6.4 Problem 3
Given the following within-subjects data, is \(M_D\) significantly different from \({0}\)? Use a \({2}\) tailed-test with \(\alpha = {0.01}\)
| ID | A | B |
|---|---|---|
| 1 | 18 | 19 |
| 2 | 14 | 22 |
| 3 | 15 | 24 |
| 4 | 16 | 17 |
| 5 | 16 | 21 |
| 6 | 16 | 27 |
| 7 | 17 | 26 |
| 8 | 15 | 21 |
| 9 | 13 | 25 |
| 10 | 13 | 23 |
| 11 | 12 | 21 |
| 12 | 13 | 21 |
| 13 | 15 | 22 |
| 14 | 10 | 21 |
Summary statistics for these data:
\[M_A = {14.5}, M_B = {22.14}\] \[n = {14}\]
State the Hypotheses
\[H_0: \mu_D = 0\] \[H_1: \mu_D \ne 0\]
The decision criteria:
\(t_{crit} = \pm{3.01}, \alpha_{two-tailed} = {0.01}, df = {13}\)
Calculating the difference scores \(D = X_B - X_A\)
| ID | A | B | D |
|---|---|---|---|
| 1 | 18 | 19 | 1 |
| 2 | 14 | 22 | 8 |
| 3 | 15 | 24 | 9 |
| 4 | 16 | 17 | 1 |
| 5 | 16 | 21 | 5 |
| 6 | 16 | 27 | 11 |
| 7 | 17 | 26 | 9 |
| 8 | 15 | 21 | 6 |
| 9 | 13 | 25 | 12 |
| 10 | 13 | 23 | 10 |
| 11 | 12 | 21 | 9 |
| 12 | 13 | 21 | 8 |
| 13 | 15 | 22 | 7 |
| 14 | 10 | 21 | 11 |
Calculate the Mean of the Difference Scores \[M_D = \frac{\Sigma D}{n}\]
\[M_D = \frac{107}{14}\]
\[M_D = {7.64}\]
Create a column for the squared difference scores \(D^2\)
| ID | A | B | D | D^2 |
|---|---|---|---|---|
| 1 | 18 | 19 | 1 | 1 |
| 2 | 14 | 22 | 8 | 64 |
| 3 | 15 | 24 | 9 | 81 |
| 4 | 16 | 17 | 1 | 1 |
| 5 | 16 | 21 | 5 | 25 |
| 6 | 16 | 27 | 11 | 121 |
| 7 | 17 | 26 | 9 | 81 |
| 8 | 15 | 21 | 6 | 36 |
| 9 | 13 | 25 | 12 | 144 |
| 10 | 13 | 23 | 10 | 100 |
| 11 | 12 | 21 | 9 | 81 |
| 12 | 13 | 21 | 8 | 64 |
| 13 | 15 | 22 | 7 | 49 |
| 14 | 10 | 21 | 11 | 121 |
New Summary statistics for the difference scores:
\[M_D = {7.64}, \quad \Sigma{D} = {107}, \quad \Sigma{D^2} = {969}\]
Calculate SS of the difference scores
\[SS_D = \Sigma D^2 - \frac{(\Sigma D)^2}{n}\]
\[SS_D = {969} - \frac{11449}{14}\]
\[SS_D = {969} - {817.79}\]
\[SS_D = {151.21}\]
Calculate the variance of the difference scores
\[s_D^2 = \frac{SS_D}{df}\]
\[s_D^2 = \frac{151.21}{13}\]
\[s_D^2 = {11.63}\]
Calculate the estimated standard error of the difference scores
\[s_{M_D} = \sqrt{\frac{s^2}{n}}\]
\[s_{M_D} = \sqrt{\frac{11.63}{14}}\]
\[s_{M_D} = \sqrt{0.83}\]
\[s_{M_D} = {0.91}\]
Calculate \(t_{obt}\)
\[t_{obt} = {\frac{M_D - \mu_D}{s_{M_D}}}\]
\[t_{obt} = \frac{7.64 - 0}{0.91}\]
\[t_{obt} = \frac{7.64}{0.91}\]
\[t_{obt} = {8.4}\]
Calculating Cohen’s d
\[d = \frac{M_D}{\sqrt{s^2}}\]
\[d = \frac{7.64}{{{\sqrt{11.63}}}}\]
\[d = \frac{7.64}{3.41}\]
\[d = {2.24}\]
The results:
reject the null hypothesis, results are significant,
t(13) = 8.4, p < 0.01, d = 2.24
6.5 Problem 4
Given the following within-subjects data, is \(M_D\) significantly different from \({0}\)? Use a \({2}\) tailed-test with \(\alpha = {0.01}\)
| ID | A | B |
|---|---|---|
| 1 | 42 | 44 |
| 2 | 38 | 37 |
| 3 | 41 | 43 |
| 4 | 41 | 39 |
| 5 | 41 | 41 |
| 6 | 41 | 28 |
| 7 | 35 | 40 |
| 8 | 43 | 44 |
| 9 | 47 | 42 |
Summary statistics for these data:
\[M_A = {41.0}, M_B = {39.78}\] \[n = {9}\]
State the Hypotheses
\[H_0: \mu_D = 0\] \[H_1: \mu_D \ne 0\]
The decision criteria:
\(t_{crit} = \pm{3.36}, \alpha_{two-tailed} = {0.01}, df = {8}\)
Calculating the difference scores \(D = X_B - X_A\)
| ID | A | B | D |
|---|---|---|---|
| 1 | 42 | 44 | 2 |
| 2 | 38 | 37 | -1 |
| 3 | 41 | 43 | 2 |
| 4 | 41 | 39 | -2 |
| 5 | 41 | 41 | 0 |
| 6 | 41 | 28 | -13 |
| 7 | 35 | 40 | 5 |
| 8 | 43 | 44 | 1 |
| 9 | 47 | 42 | -5 |
Calculate the Mean of the Difference Scores \[M_D = \frac{\Sigma D}{n}\]
\[M_D = \frac{-11}{9}\]
\[M_D = {-1.22}\]
Create a column for the squared difference scores \(D^2\)
| ID | A | B | D | D^2 |
|---|---|---|---|---|
| 1 | 42 | 44 | 2 | 4 |
| 2 | 38 | 37 | -1 | 1 |
| 3 | 41 | 43 | 2 | 4 |
| 4 | 41 | 39 | -2 | 4 |
| 5 | 41 | 41 | 0 | 0 |
| 6 | 41 | 28 | -13 | 169 |
| 7 | 35 | 40 | 5 | 25 |
| 8 | 43 | 44 | 1 | 1 |
| 9 | 47 | 42 | -5 | 25 |
New Summary statistics for the difference scores:
\[M_D = {-1.22}, \quad \Sigma{D} = {-11}, \quad \Sigma{D^2} = {233}\]
Calculate SS of the difference scores
\[SS_D = \Sigma D^2 - \frac{(\Sigma D)^2}{n}\]
\[SS_D = {233} - \frac{121}{9}\]
\[SS_D = {233} - {13.44}\]
\[SS_D = {219.56}\]
Calculate the variance of the difference scores
\[s_D^2 = \frac{SS_D}{df}\]
\[s_D^2 = \frac{219.56}{8}\]
\[s_D^2 = {27.44}\]
Calculate the estimated standard error of the difference scores
\[s_{M_D} = \sqrt{\frac{s^2}{n}}\]
\[s_{M_D} = \sqrt{\frac{27.44}{9}}\]
\[s_{M_D} = \sqrt{3.05}\]
\[s_{M_D} = {1.75}\]
Calculate \(t_{obt}\)
\[t_{obt} = {\frac{M_D - \mu_D}{s_{M_D}}}\]
\[t_{obt} = \frac{-1.22 - 0}{1.75}\]
\[t_{obt} = \frac{-1.22}{1.75}\]
\[t_{obt} = {-0.7}\]
Calculating Cohen’s d
\[d = \frac{M_D}{\sqrt{s^2}}\]
\[d = \frac{-1.22}{{{\sqrt{27.44}}}}\]
\[d = \frac{-1.22}{5.24}\]
\[d = {-0.23}\]
The results:
fail to reject the null hypothesis, results not significant,
t(8) = -0.7, p > 0.01, d = -0.23
6.6 Problem 5
Given the following within-subjects data, is \(M_D\) significantly different from \({0}\)? Use a \({2}\) tailed-test with \(\alpha = {0.01}\)
| ID | A | B |
|---|---|---|
| 1 | 47 | 40 |
| 2 | 43 | 53 |
| 3 | 35 | 45 |
| 4 | 38 | 55 |
| 5 | 40 | 43 |
| 6 | 41 | 46 |
| 7 | 32 | 44 |
| 8 | 39 | 40 |
| 9 | 38 | 40 |
Summary statistics for these data:
\[M_A = {39.22}, M_B = {45.11}\] \[n = {9}\]
State the Hypotheses
\[H_0: \mu_D = 0\] \[H_1: \mu_D \ne 0\]
The decision criteria:
\(t_{crit} = \pm{3.36}, \alpha_{two-tailed} = {0.01}, df = {8}\)
Calculating the difference scores \(D = X_B - X_A\)
| ID | A | B | D |
|---|---|---|---|
| 1 | 47 | 40 | -7 |
| 2 | 43 | 53 | 10 |
| 3 | 35 | 45 | 10 |
| 4 | 38 | 55 | 17 |
| 5 | 40 | 43 | 3 |
| 6 | 41 | 46 | 5 |
| 7 | 32 | 44 | 12 |
| 8 | 39 | 40 | 1 |
| 9 | 38 | 40 | 2 |
Calculate the Mean of the Difference Scores \[M_D = \frac{\Sigma D}{n}\]
\[M_D = \frac{53}{9}\]
\[M_D = {5.89}\]
Create a column for the squared difference scores \(D^2\)
| ID | A | B | D | D^2 |
|---|---|---|---|---|
| 1 | 47 | 40 | -7 | 49 |
| 2 | 43 | 53 | 10 | 100 |
| 3 | 35 | 45 | 10 | 100 |
| 4 | 38 | 55 | 17 | 289 |
| 5 | 40 | 43 | 3 | 9 |
| 6 | 41 | 46 | 5 | 25 |
| 7 | 32 | 44 | 12 | 144 |
| 8 | 39 | 40 | 1 | 1 |
| 9 | 38 | 40 | 2 | 4 |
New Summary statistics for the difference scores:
\[M_D = {5.89}, \quad \Sigma{D} = {53}, \quad \Sigma{D^2} = {721}\]
Calculate SS of the difference scores
\[SS_D = \Sigma D^2 - \frac{(\Sigma D)^2}{n}\]
\[SS_D = {721} - \frac{2809}{9}\]
\[SS_D = {721} - {312.11}\]
\[SS_D = {408.89}\]
Calculate the variance of the difference scores
\[s_D^2 = \frac{SS_D}{df}\]
\[s_D^2 = \frac{408.89}{8}\]
\[s_D^2 = {51.11}\]
Calculate the estimated standard error of the difference scores
\[s_{M_D} = \sqrt{\frac{s^2}{n}}\]
\[s_{M_D} = \sqrt{\frac{51.11}{9}}\]
\[s_{M_D} = \sqrt{5.68}\]
\[s_{M_D} = {2.38}\]
Calculate \(t_{obt}\)
\[t_{obt} = {\frac{M_D - \mu_D}{s_{M_D}}}\]
\[t_{obt} = \frac{5.89 - 0}{2.38}\]
\[t_{obt} = \frac{5.89}{2.38}\]
\[t_{obt} = {2.47}\]
Calculating Cohen’s d
\[d = \frac{M_D}{\sqrt{s^2}}\]
\[d = \frac{5.89}{{{\sqrt{51.11}}}}\]
\[d = \frac{5.89}{7.15}\]
\[d = {0.82}\]
The results:
fail to reject the null hypothesis, results not significant,
t(8) = 2.47, p > 0.01, d = 0.82