
Dattaprabhakar G. answered 10/04/14
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Theresa:
a. Assuming α=.05, two-tailed, state the cutoff score for the test.
It is a two-tailed t-test that assumes that the population variances for the two groups are equal. The degrees of freedom (DF) are 50 + 50 - 2 = 98, large, so normal approximation to the t can be used. For α = 0.05 the cut-off scores are Plus or minus 1.96. Actually, reading the question, (" treatment program designed to increase self-confidence.") I would use a one-tailed test but the question specifically says two-tailed.
b. Calculate the value of the standard error of the test statistic.
The test statistic is t. Its variance is DF / (DF - 2) = 98/96 = 1.0208, so the Std Error of the test statistic is 1.0104.
If the confused question-setter means by test-statistic "the difference between the sample means", please post a comment. I will give you the Std Error of the difference between the sample means" as the test statistic .
c. What is the value of the test statistic for these data?
The calculated t value is 2.1864. Difference between sample means is 1.4 (for the benefit of the confused question setter)
e. Statistical decision and interpretation? (What happened to (d)?)
Since the calculated t-value is larger than the cut-off score, the null hypothesis that there is no difference between the population means of treated adolescents and non-treated ones is rejected, at 5% level of significance, based on the data provided. Looking at the two-tailed P-value which is 0.0312, we would have arrived at the same conclusion, even when, the level of significance were 0.032, for example. (The sig level has to be greater than 0.0312, that is all)
f. What is the value of the effect size for these data?
The un-standardized effect size is the difference between means, 1.4
g. What is the estimate of statistical power based on these data?
The answer depends upon what value for the difference between population means of treated adolescents and non-treated ones you are going to assume. For example, if this difference between population means is assumed to be 50, the estimated statistical power is 100 per cent. On the other hand, if this difference is assumed to be 0.00001, the estimated power is 0.05.
Dr. G.
Theresa L.
10/05/14