Question

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?

Answer 1

Answer:

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