Question

A researcher selects a sample of n=25 individuals from a population with a mean of μ=60 and standard deviation of σ=10 and administers a treatment. The researcher predicts that the treatment will increase scores. Following the treatment, the average scores for this sample is M=65.

1. Using symbols, state the hypothesis for a one-tailed test.

2. Calculate the z-score and place this on a standardized normal distribution.

3. With a one-tail α=0.05, state the conclusion of these findings.

4. Calculate Cohen’s d and state the effect size (i.e., small, medium, large).

Answer 1

Given that,
population mean(u)=60
standard deviation, σ =10
sample mean, x =65
number (n)=25
null, Ho: μ=60
alternate, H1: μ>60
level of significance, α = 0.05
from standard normal table,right tailed z α/2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 65-60/(10/sqrt(25)
zo = 2.5
| zo | = 2.5
critical value
the value of |z α| at los 5% is 1.645
we got |zo| =2.5 & | z α | = 1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value : right tail - ha : ( p > 2.5 ) = 0.01
hence value of p0.05 > 0.01, here we reject Ho
ANSWERS
---------------
1.
null, Ho: μ=60
alternate, H1: μ>60
2.
test statistic: 2.5
critical value: 1.645
decision: reject Ho
3.
p-value: 0.01
we have enough evidence to support the claim that he researcher predicts that the treatment will increase scores
4.
cohen's d size = mean difference/standard deviation
cohen's d size =(65-60)/10 =5/10 =0.5
medium effect