sample_size = random.randint(5,15)
RandomData(groups = 2, n = sample_size).independent_samples_t_test()5 Independent-Samples t-tests
Sample Problems
5.1 About
the independent-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
independent_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
5.2 Problem 1
Given the following between-subjects data, is the mean of \(Group_A\) significantly different from the mean of \(Group_B\)? Use a \({2}\) tailed-test with \(\alpha = {0.05}\)
| A | B |
|---|---|
| 62 | 78 |
| 61 | 64 |
| 63 | 74 |
| 73 | 89 |
| 62 | 72 |
| 66 | 79 |
| 55 | 79 |
The necessary summary Statistis for these data:
\[M_A = {63.14}, M_B = {76.43}\] \[SS_A = {178.86}, SS_B = {353.71}\] \[n_A = {7}, n_B = {7}\]
State the Hypotheses
\[H_0: \mu_A - \mu_B = 0\] \[H_1: \mu_A - \mu_B \ne 0\]
The decision criteria:
\(t_{crit} = \pm{2.18}, \alpha_{two-tailed} = {0.05}, df = {12}\)
Calculate the pooled variance:
\[s_p^2 = {\frac{SS_A + SS_B}{df_A + df_B}}\]
\[s_p^2 = {\frac{178.86 + 353.71}{6 + 6}}\]
\[s_p^2 = \frac{532.57}{12}\]
\[s_p^2 = {44.38}\]
Calculate the estimated error of the difference between means: \(s_{(M_A - M_B)}\)
\[s_{(M_A - M_B)} = \sqrt{\frac{s_p^2}{n_1} + \frac{s_p^2}{n_1}}\]
\[s_{(M_A - M_B)} = \sqrt{\frac{44.38}{7} + \frac{44.38}{7}}\]
\[s_{(M_A - M_B)} = \sqrt{6.34 + 6.34}\]
\[s_{(M_A - M_B)} = \sqrt{12.68}\]
\[s_{(M_A - M_B)} = {3.56}\]
Calculate \(t_{obt}\)
\[t_{obt} = {\frac{(M_A - M_B) - (\mu_A - \mu_B)}{s_{(M_A - M_B)}}}\]
\[t_{obt} = \frac{(63.14 - 76.43) - {0}}{3.56}\]
\[t_{obt} = \frac{-13.29}{3.56}\]
\[t_{obt} = {-3.73}\]
Calculate Cohen’s d
\[d = \frac{M_A - M_B}{\sqrt{s_p^2}}\]
\[d = \frac{(63.14 - 76.43)}{{{\sqrt{44.38}}}}\]
\[d = \frac{(-13.29)}{6.66}\]
\[d = {-2.0}\]
The results:
reject the null hypothesis, results are significant,
t(12) = -3.73, p < 0.05, d = -2.0
5.3 Problem 2
Given the following between-subjects data, is the mean of \(Group_A\) significantly different from the mean of \(Group_B\)? Use a \({2}\) tailed-test with \(\alpha = {0.01}\)
| A | B |
|---|---|
| 14 | 15 |
| 15 | 10 |
| 15 | 15 |
| 14 | 11 |
| 14 | 19 |
| 17 | 9 |
The necessary summary Statistis for these data:
\[M_A = {14.83}, M_B = {13.17}\] \[SS_A = {6.83}, SS_B = {72.83}\] \[n_A = {6}, n_B = {6}\]
State the Hypotheses
\[H_0: \mu_A - \mu_B = 0\] \[H_1: \mu_A - \mu_B \ne 0\]
The decision criteria:
\(t_{crit} = \pm{3.17}, \alpha_{two-tailed} = {0.01}, df = {10}\)
Calculate the pooled variance:
\[s_p^2 = {\frac{SS_A + SS_B}{df_A + df_B}}\]
\[s_p^2 = {\frac{6.83 + 72.83}{5 + 5}}\]
\[s_p^2 = \frac{79.66}{10}\]
\[s_p^2 = {7.97}\]
Calculate the estimated error of the difference between means: \(s_{(M_A - M_B)}\)
\[s_{(M_A - M_B)} = \sqrt{\frac{s_p^2}{n_1} + \frac{s_p^2}{n_1}}\]
\[s_{(M_A - M_B)} = \sqrt{\frac{7.97}{6} + \frac{7.97}{6}}\]
\[s_{(M_A - M_B)} = \sqrt{1.33 + 1.33}\]
\[s_{(M_A - M_B)} = \sqrt{2.66}\]
\[s_{(M_A - M_B)} = {1.63}\]
Calculate \(t_{obt}\)
\[t_{obt} = {\frac{(M_A - M_B) - (\mu_A - \mu_B)}{s_{(M_A - M_B)}}}\]
\[t_{obt} = \frac{(14.83 - 13.17) - {0}}{1.63}\]
\[t_{obt} = \frac{1.66}{1.63}\]
\[t_{obt} = {1.02}\]
Calculate Cohen’s d
\[d = \frac{M_A - M_B}{\sqrt{s_p^2}}\]
\[d = \frac{(14.83 - 13.17)}{{{\sqrt{7.97}}}}\]
\[d = \frac{(1.66)}{2.82}\]
\[d = {0.59}\]
The results:
fail to reject the null hypothesis, results not significant,
t(10) = 1.02, p > 0.01, d = 0.59
5.4 Problem 3
Given the following between-subjects data, is the mean of \(Group_A\) significantly different from the mean of \(Group_B\)? Use a \({2}\) tailed-test with \(\alpha = {0.01}\)
| A | B |
|---|---|
| 24 | 23 |
| 29 | 23 |
| 17 | 19 |
| 33 | 20 |
| 34 | 17 |
| 31 | 32 |
| 30 | 15 |
| 27 | 20 |
| 33 | 19 |
| 26 | 22 |
| 26 | 23 |
| 29 | 35 |
| 34 | 26 |
The necessary summary Statistis for these data:
\[M_A = {28.69}, M_B = {22.62}\] \[SS_A = {276.77}, SS_B = {383.08}\] \[n_A = {13}, n_B = {13}\]
State the Hypotheses
\[H_0: \mu_A - \mu_B = 0\] \[H_1: \mu_A - \mu_B \ne 0\]
The decision criteria:
\(t_{crit} = \pm{2.8}, \alpha_{two-tailed} = {0.01}, df = {24}\)
Calculate the pooled variance:
\[s_p^2 = {\frac{SS_A + SS_B}{df_A + df_B}}\]
\[s_p^2 = {\frac{276.77 + 383.08}{12 + 12}}\]
\[s_p^2 = \frac{659.85}{24}\]
\[s_p^2 = {27.49}\]
Calculate the estimated error of the difference between means: \(s_{(M_A - M_B)}\)
\[s_{(M_A - M_B)} = \sqrt{\frac{s_p^2}{n_1} + \frac{s_p^2}{n_1}}\]
\[s_{(M_A - M_B)} = \sqrt{\frac{27.49}{13} + \frac{27.49}{13}}\]
\[s_{(M_A - M_B)} = \sqrt{2.11 + 2.11}\]
\[s_{(M_A - M_B)} = \sqrt{4.22}\]
\[s_{(M_A - M_B)} = {2.05}\]
Calculate \(t_{obt}\)
\[t_{obt} = {\frac{(M_A - M_B) - (\mu_A - \mu_B)}{s_{(M_A - M_B)}}}\]
\[t_{obt} = \frac{(28.69 - 22.62) - {0}}{2.05}\]
\[t_{obt} = \frac{6.07}{2.05}\]
\[t_{obt} = {2.96}\]
Calculate Cohen’s d
\[d = \frac{M_A - M_B}{\sqrt{s_p^2}}\]
\[d = \frac{(28.69 - 22.62)}{{{\sqrt{27.49}}}}\]
\[d = \frac{(6.07)}{5.24}\]
\[d = {1.16}\]
The results:
reject the null hypothesis, results are significant,
t(24) = 2.96, p < 0.01, d = 1.16
5.5 Problem 4
Given the following between-subjects data, is the mean of \(Group_A\) significantly less than the mean of \(Group_B\)? Use a \({1}\) tailed-test with \(\alpha = {0.05}\)
| A | B |
|---|---|
| 41 | 65 |
| 30 | 57 |
| 33 | 77 |
| 59 | 72 |
| 69 | 68 |
| 44 | 66 |
| 37 | 80 |
| 73 | 76 |
| 53 | 43 |
| 43 | 73 |
| 41 | 80 |
| 74 | 53 |
| 64 | 80 |
| 41 | 49 |
The necessary summary Statistis for these data:
\[M_A = {50.14}, M_B = {67.07}\] \[SS_A = {2937.71}, SS_B = {1950.93}\] \[n_A = {14}, n_B = {14}\]
State the Hypotheses
\[H_0: \mu_A - \mu_B \geq 0\] \[H_1: \mu_A - \mu_B \lt 0\]
The decision criteria:
\(t_{crit} = {-1.71}, \alpha_{one-tailed} = {0.05}, df = {26}\)
Calculate the pooled variance:
\[s_p^2 = {\frac{SS_A + SS_B}{df_A + df_B}}\]
\[s_p^2 = {\frac{2937.71 + 1950.93}{13 + 13}}\]
\[s_p^2 = \frac{4888.64}{26}\]
\[s_p^2 = {188.02}\]
Calculate the estimated error of the difference between means: \(s_{(M_A - M_B)}\)
\[s_{(M_A - M_B)} = \sqrt{\frac{s_p^2}{n_1} + \frac{s_p^2}{n_1}}\]
\[s_{(M_A - M_B)} = \sqrt{\frac{188.02}{14} + \frac{188.02}{14}}\]
\[s_{(M_A - M_B)} = \sqrt{13.43 + 13.43}\]
\[s_{(M_A - M_B)} = \sqrt{26.86}\]
\[s_{(M_A - M_B)} = {5.18}\]
Calculate \(t_{obt}\)
\[t_{obt} = {\frac{(M_A - M_B) - (\mu_A - \mu_B)}{s_{(M_A - M_B)}}}\]
\[t_{obt} = \frac{(50.14 - 67.07) - {0}}{5.18}\]
\[t_{obt} = \frac{-16.93}{5.18}\]
\[t_{obt} = {-3.27}\]
Calculate Cohen’s d
\[d = \frac{M_A - M_B}{\sqrt{s_p^2}}\]
\[d = \frac{(50.14 - 67.07)}{{{\sqrt{188.02}}}}\]
\[d = \frac{(-16.93)}{13.71}\]
\[d = {-1.23}\]
The results:
reject the null hypothesis, results are significant,
t(26) = -3.27, p < 0.05, d = -1.23
5.6 Problem 5
Given the following between-subjects data, is the mean of \(Group_A\) significantly different from the mean of \(Group_B\)? Use a \({2}\) tailed-test with \(\alpha = {0.05}\)
| A | B |
|---|---|
| 15 | 26 |
| 12 | 9 |
| 15 | 29 |
| 14 | 20 |
| 26 | 11 |
| 23 | 15 |
| 13 | 10 |
| 27 | 12 |
The necessary summary Statistis for these data:
\[M_A = {18.12}, M_B = {16.5}\] \[SS_A = {264.88}, SS_B = {410.0}\] \[n_A = {8}, n_B = {8}\]
State the Hypotheses
\[H_0: \mu_A - \mu_B = 0\] \[H_1: \mu_A - \mu_B \ne 0\]
The decision criteria:
\(t_{crit} = \pm{2.14}, \alpha_{two-tailed} = {0.05}, df = {14}\)
Calculate the pooled variance:
\[s_p^2 = {\frac{SS_A + SS_B}{df_A + df_B}}\]
\[s_p^2 = {\frac{264.88 + 410.0}{7 + 7}}\]
\[s_p^2 = \frac{674.88}{14}\]
\[s_p^2 = {48.21}\]
Calculate the estimated error of the difference between means: \(s_{(M_A - M_B)}\)
\[s_{(M_A - M_B)} = \sqrt{\frac{s_p^2}{n_1} + \frac{s_p^2}{n_1}}\]
\[s_{(M_A - M_B)} = \sqrt{\frac{48.21}{8} + \frac{48.21}{8}}\]
\[s_{(M_A - M_B)} = \sqrt{6.03 + 6.03}\]
\[s_{(M_A - M_B)} = \sqrt{12.06}\]
\[s_{(M_A - M_B)} = {3.47}\]
Calculate \(t_{obt}\)
\[t_{obt} = {\frac{(M_A - M_B) - (\mu_A - \mu_B)}{s_{(M_A - M_B)}}}\]
\[t_{obt} = \frac{(18.12 - 16.5) - {0}}{3.47}\]
\[t_{obt} = \frac{1.62}{3.47}\]
\[t_{obt} = {0.47}\]
Calculate Cohen’s d
\[d = \frac{M_A - M_B}{\sqrt{s_p^2}}\]
\[d = \frac{(18.12 - 16.5)}{{{\sqrt{48.21}}}}\]
\[d = \frac{(1.62)}{6.94}\]
\[d = {0.23}\]
The results:
fail to reject the null hypothesis, results not significant,
t(14) = 0.47, p > 0.05, d = 0.23