Statistical Analysis Independent Samples t-Test Template

Research HypothesesSubject individuals who see the injury film will be more positive toward wearing a seatbelt.Null HypothesesSubject individuals who see the injury film will be less positive or the same toward wearing a seatbelt.Critical value of tThis is a one-tailed test and the alpha level has been set at .

01 with degrees of freedom equal to 50.df = N – (number of parameters being tested), in this case we are testing for the mean and the variance. Thendf = N – 2df = 52 – 2 = 50tcrit = +2.404STEP #2 - Estimated Variance for Each Sample:S21 = 4.

41        df1 = 15S22 = 5.76        df2 = 35STEP #3 - Pooled VarianceS2pooled = (df1/dftotal)S12 + (df2/dftotal)S22 = (15/50)(4.41) + (35/50)(5.76)= 1.

323 + 4.032 = 5.355S2pooled = 5.355STEP #4 - Estimated Variances for Each DomS2M1 = S2pooled/n1 = 5.

355/16 = 0.3346875S2M2 = S2pooled/n2 = 5.355/36 = 0.14875S2M1 = 0.3346875                             S2M2 = 0.

14875STEP #5 - Standard Error of Differences Between MeansS2D = S2M1 + S2M2 = 0.3346875 + 0.14875 = 0.4834375S2D = 0.4834375SD = ?S2D = ?0.

4834375 = 0.6952966992586689966SD = 0.6952966992586689966STEP #6 - t Obtainedtobt = (M1 - M2)/SD = (8.9 – 7.0)/0.6952966992586689966 = 2.

73264636812715416891tobt = 2.73264636812715416891STEP #7 – DecisionSince +2.732 exceed the critical value of +2.404. we have a significant findings and can reject the null hypotheses:EnglishSubject individuals who see the injury film will be more positive toward wearing a seatbelt.Statistical Notationt(50) = +2.732, p < 0.01