Using the following data, perform an independent samples t-test. Test a nondirectional hypothesis using \(\alpha=.05\). Report the 95% confidence interval around \((\bar{x}_1 - \bar{x}_2)\). Also compute and interpret the Cohen’s \(d\) effect size. Write up the results in APA format.
\[\begin{equation} \begin{bmatrix} \textbf{Sample1} \\ 51 \\ 59 \\ 33 \\ 48 \\ 48 \\ 55 \end{bmatrix} \begin{bmatrix} \textbf{Sample2} \\ 59 \\ 76 \\ 57 \\ 71 \\ 71 \\ 66 \end{bmatrix} \end{equation}\]Preliminaries
\(n_1\) = 6, \(df_1\) = 5.
\(n_2\) = 6, \(df_2\) = 5.
\(df_{total} = 5 + 5 = 10\)
\(\bar{x}_1\) = 49
\(\bar{x}_2\) = 66.67
\(Var_1\):
Data Mean Dev Sq dev
51 49 2 4
59 49 10 100
33 49 -16 256
48 49 -1 1
48 49 -1 1
55 49 6 36\(SS_1 = 398\)
\(Var_1 = \frac{398}{6-1} = 79.6\)
\(Var_2\):
Data Mean Dev Sq dev
59 66.67 -7.67 58.83
76 66.67 9.33 87.05
57 66.67 -9.67 93.51
71 66.67 4.33 18.75
71 66.67 4.33 18.75
66 66.67 -0.67 0.45\(SS_2 = 277.34\)
\(Var_2 = \frac{277.34} {6-1} = 55.47\)
Hypothesis Test
\(Var_{pooled} = \left(\frac{5}{5 + 5}\right) 79.6+ \left(\frac{5}{5 + 5}\right) 55.47= 67.53\)
\(Var_{sampling} = \frac{67.53}{6}+\frac{67.53}{6}=22.52\)
\(StdErr = \sqrt{22.52} = 4.75\)
\(t = \frac{\left(49 - 66.67\right)}{4.75} = -3.72\)
\(t_{critical} = \pm 2.228\)
\(\fbox{ Decision: reject the null }\)
Estimated effect size
Raw effect size \((\bar{x}_1-\bar{x}_2)\) = (49 - 66.67) = -17.67
Cohen’s d = \(\frac{\left(49 - 66.67\right)}{\sqrt{67.53}} = -2.15\)
Confidence interval
CI: \(-17.67 \pm (4.75)(2.228) = \left[ -28.25, -7.09\right]\)
APA writeup
t(10)=-3.72, p <0.05, d=-2.15.
Using the following data, perform an independent samples t-test. Test a directional hypothesis (grp1 > grp2) using \(\alpha=.05\). Report the 95% confidence interval around \((\bar{x}_1 - \bar{x}_2)\). Also compute and interpret the Cohen’s \(d\) effect size. Write up the results in APA format.
\[\begin{equation} \begin{bmatrix} \textbf{Sample1} \\ 95 \\ 101 \\ 122 \\ 108 \\ 107 \end{bmatrix} \begin{bmatrix} \textbf{Sample2} \\ 86 \\ 90 \\ 73 \\ 81 \\ 106 \end{bmatrix} \end{equation}\]Preliminaries
\(n_1\) = 5, \(df_1\) = 4.
\(n_2\) = 5, \(df_2\) = 4.
\(df_{total} = 4 + 4 = 8\)
\(\bar{x}_1\) = 106.6
\(\bar{x}_2\) = 87.2
\(Var_1\):
Data Mean Dev Sq dev
95 106.6 -11.6 134.56
101 106.6 -5.6 31.36
122 106.6 15.4 237.16
108 106.6 1.4 1.96
107 106.6 0.4 0.16\(SS_1 = 405.2\)
\(Var_1 = \frac{405.2}{5-1} = 101.3\)
\(Var_2\):
Data Mean Dev Sq dev
86 87.2 -1.2 1.44
90 87.2 2.8 7.84
73 87.2 -14.2 201.64
81 87.2 -6.2 38.44
106 87.2 18.8 353.44\(SS_2 = 602.8\)
\(Var_2 = \frac{602.8} {5-1} = 150.7\)
Hypothesis Test
\(Var_{pooled} = \left(\frac{4}{4 + 4}\right) 101.3+ \left(\frac{4}{4 + 4}\right) 150.7= 126\)
\(Var_{sampling} = \frac{126}{5}+\frac{126}{5}=50.4\)
\(StdErr = \sqrt{50.4} = 7.1\)
\(t = \frac{\left(106.6 - 87.2\right)}{7.1} = 2.73\)
\(t_{critical} = +1.86\)
\(\fbox{ Decision: reject the null }\)
Estimated effect size
Raw effect size \((\bar{x}_1-\bar{x}_2)\) = (106.6 - 87.2) = 19.4
Cohen’s d = \(\frac{\left(106.6 - 87.2\right)}{\sqrt{126}} = 1.73\)
Confidence interval
CI: \(19.4 \pm (7.1)(1.86) = \left[ 6.19, 32.61\right]\)
APA writeup
t(8)=2.73, p <0.05, d=1.73.
Redo problem 2 using a two-tailed test with \(\alpha=.01\) and a 99% confidence interval.
All values carry over from problem 2 except for \(t_{critical}\), the statistical decision, and the confidence interval.
Hypothesis Test
\(t_{critical} = \pm 3.355\)
\(\fbox{ Decision: fail to reject the null }\)
Confidence interval
CI: \(19.4 \pm (7.1)(3.355) = \left[ -4.42, 43.22\right]\)
APA writeup
t(8)=2.73, p >=0.01, d=1.73.
Using the following data, perform an independent samples t-test. Test a nondirectional hypothesis using \(\alpha=.01\). Report the 99% confidence interval around \((\bar{x}_1 - \bar{x}_2)\). Also compute and interpret the Cohen’s \(d\) effect size. Write up the results in APA format.
\[\begin{equation} \begin{bmatrix} \textbf{Sample1} \\ -0.26 \\ -0.86 \\ -1.17 \\ 0.17 \\ -0.44 \end{bmatrix} \begin{bmatrix} \textbf{Sample2} \\ 0.82 \\ 0.06 \\ 1.05 \\ -1.17 \\ 2.49 \end{bmatrix} \end{equation}\]Preliminaries
\(n_1\) = 5, \(df_1\) = 4.
\(n_2\) = 5, \(df_2\) = 4.
\(df_{total} = 4 + 4 = 8\)
\(\bar{x}_1\) = -0.51
\(\bar{x}_2\) = 0.65
\(Var_1\):
Data Mean Dev Sq dev
-0.26 -0.51 0.25 0.06
-0.86 -0.51 -0.35 0.12
-1.17 -0.51 -0.66 0.44
0.17 -0.51 0.68 0.46
-0.44 -0.51 0.07 0.00\(SS_1 = 1.08\)
\(Var_1 = \frac{1.08}{5-1} = 0.27\)
\(Var_2\):
Data Mean Dev Sq dev
0.82 0.65 0.17 0.03
0.06 0.65 -0.59 0.35
1.05 0.65 0.40 0.16
-1.17 0.65 -1.82 3.31
2.49 0.65 1.84 3.39\(SS_2 = 7.24\)
\(Var_2 = \frac{7.24} {5-1} = 1.81\)
Hypothesis Test
\(Var_{pooled} = \left(\frac{4}{4 + 4}\right) 0.27+ \left(\frac{4}{4 + 4}\right) 1.81= 1.04\)
\(Var_{sampling} = \frac{1.04}{5}+\frac{1.04}{5}=0.42\)
\(StdErr = \sqrt{0.42} = 0.65\)
\(t = \frac{\left(-0.51 - 0.65\right)}{0.65} = -1.78\)
\(t_{critical} = \pm 3.355\)
\(\fbox{ Decision: fail to reject the null }\)
Estimated effect size
Raw effect size \((\bar{x}_1-\bar{x}_2)\) = (-0.51 - 0.65) = -1.16
Cohen’s d = \(\frac{\left(-0.51 - 0.65\right)}{\sqrt{1.04}} = -1.14\)
Confidence interval
CI: \(-1.16 \pm (0.65)(3.355) = \left[ -3.34, 1.02\right]\)
APA writeup
t(8)=-1.78, p >=0.01, d=-1.14.