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?

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

30 units^2

step-by-step explanation:

area is l × w

the length is 3

and the width is 10

3 × 10= 30 units squared