evafaith
17.03.2020 •
Mathematics
A random sample of scores is obtained from a population with a mean of μ . After a treatment is administered to the individuals in the sample, the sample mean is found to be . Assuming that the sample standard deviation is , compute and the estimated Cohen’s d to measure the size of the treatment effect. Assuming that the sample standard deviation is , compute and the estimated Cohen’s d to measure the size of the treatment effect. Comparing your answers from parts a and b, how does the variability of the scores in the sample influence the measures of effect size?
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Ответ:
The question is incomplete and values are missing, from online we have the following informations,
Firstly: Null Hypothesis is H0:μ=45H0:μ=45
and Alternate Hypothesis is H1:μ≠45
Secondly: For a sample of n=16,
the t-statistics will have
(n−1) degrees of freedom, i.e. df=15. For a two-tailed test with α=0.05 and df=15, the critical value (CV) is obtained from the t-table as t=±2.1315
Now The t-statistics can be calculated using formula:
t=M−μ/sm
Calculate sm using the formula:
sm=√s^2/n
=√(8)^2/16
= √4
=2
Now, substitute 49.2 for MM, 45 for μμ and 2 for smsm in the formula
t=49.2−45/2
=4.22
=2.1
Now the Rejection rule: Reject when t−statistics<−2.1315 or t−statistics>2
Ответ:
or .80
The zero doesnt really matter.
I hope this helps!