Question

For either independent-measures or repeated-measures designs comparing two treatments, the mean difference can be evaluated with either at test or an ANOVA. The two tests are related by the equation F=12. The following data are from a repeated-measures study: Person Difference Scores 3 I 4 2 3 7 M = 4.00 T = 16 SS = 14 Treatment II 7 11 6 10 M 8.50 T-34 SS = 17 3 3 Mo 4.50 SS = 27.00 Use a repeated-measures t test with a = .05 to determine whether the mean difference between treatments is statistically significant. (Caution: ANOVA calculations are done with the X values, but fort you use the difference scores.) (Use three decimal places for the standard error and critical values, two for the pooled variance and t statistic. Use the Distributions tool at the end of the problem to find the critical value.) Estimated t Statistic Variance Standard Error Value Critical Values Conclusion: Reject the null hypothesis. The mean difference between the treatments is not statistically significant. Fail to reject the null hypothesis. The mean difference between the treatments is not statistically significant. Fail to reject the null hypothesis. The mean difference between the treatments is statistically significant. Reject the null hypothesis. The mean difference between the treatments is statistically significant Use a repeated-measures ANOVA with a = .05 to determine whether the mean difference between treatments is Statistically significant. (Use two decimal places.) Degrees of Freedom Sum of Squares Mean Square Source of Variation Between treatments Within treatments Between subjects Error Total Critical value for the F-ratio = (Use three decimal places.) Conclusion: Fail to reject the null hypothesis. The mean difference between the treatments is statistically significant. Reject the null hypothesis. The mean difference between the treatments is not statistically significant. Reject the null hypothesis. The mean difference between the treatments is statistically significant. Fail to reject the null hypothesis. The mean difference between the treatments is not statistically significant

Answer 1

First we apply two sample T-test on the given Treatment I and II,

The provided sample means are shown below: X1 = 4 X2 = 8.5 Also, the provided sample standard deviations are: Si = 2.16024 82 = 2.38047 and the sample sizes are ni 4 and n2 = 4. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = 12 Ha: Mi 7 u2 This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the degrees of freedom are df = 6. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal:


Hence, it is found that the critical value for this two-tailed test is tc = 2.447, for a = 0.05 and df = 6. The rejection region for this two-tailed test is R = {t: 1t| > 2.447}. (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: X1 - X2 (nı - 1) si +(n2-1)s ( + m ni+n2-2 4 - 8.5 (4-1)2.160242+(4-1)2.380472 (1 + +3) 4+4-2 -2.7997

therefore estimated standard error is given by:

2 (ni 1)si + (n2 1)s S7,-X2 = nu + n2 - 2

which come to be =
(4-112.160242+(41)2.380472 4442
= 2.273
and estimated variance = 2.273² = 5.16663
(please round up according to the requirement in the question)

thus t = -2.799
and critical value is $\pm$ 2.447

No we'll perform the F-test for the ANOVA and show how F-stest and t-test are equivalent as F= t²


Observation A | B | C | D 1 4 | 2 | 3 7 2 7|11| 6 10 Solution: B 4 7 2 11 3 6 7 10 ΣΑ = 16 ΣΒ = 34 B2 16 49 4 121 9 36 49 100 ΣΑ? = 78 ΣΒ' = 306


Data table Group B Total N n = 4 n2 = 4 n = 8 T1 = {xı = 16 2x = 78 x1 = 4 Sy = 2.1602 T2 = Ex2 = 34 Exż = 306 x2 = 8.5 S2 = 2.3805 Ex = 50 2x2 = 384 Overall x = 6.25 Mean Xi Std Dev S Let k = the number of different samples = 2 n=n, + n2 = 4 + 4 = 8 50 Overall x = 6.25 8 Ex= 17 + 12 = 16 + 34 = 50 → (1) (2x)2 502 = 312.5 → (2) n 8 T? 162342 + - = 353 → (3) ni 4 4 Ex = {x} + {xž = 78+306 = 384 — (4)


ANOVA: Step-1 : sum of squares between samples T SSB = (Σ = (3) - (2) ni (Ex)2 n = 353 - 312.5 = 40.5 Or SSB = En; · (; -7) = 4 x (4 - 6.25)2 + 4 x (8.5 - 6.25)2 = 40.5 Step-2: sum of squares within samples SSW = Σχ? - (Σ = (4)-(3) = 384-353 = 31 Step-3 : Total sum of squares SST = SSB + SSW = 40.5 + 31 = 71.5 Step-4 : variance between samples SSB MSB k-1 = 40.5 1


Step-5: variance within samples SSW MSW = n-k 31 8-2 31 6 = 5.1667 Step-6 : test statistic F for one way ANOVA test MSB F= MSW 40.5 5.1667 = 7.8387 the degree of freedom between samples k-1 = 1 Now, degree of freedom within samples n-k= 8 - 2 = 6 ANOVA table Source of Variation Between samples Within samples Total Sums of Squares Degrees of freedom Mean Squares F SS p-value DF MS SSB = 40.5 k- 1 = 1 MSB = 40.5 7.8387 0.0312 SSW = 31 n-k= 6 MSW = 5.1667 SST = 71.5 n-1 = 7


Ho : There is no significant differentiating between samples H1 : There is significant differentiating between samples F(1,6) at 0.05 level of significance = 5.9874 As calculated F = 7.8387 > 5.9874 So, H, is rejected, Hence there is significant differentiating between samples

F = 7.838 = t² = 2.7997²