Question

According to my records, the population of all past PSY 240 final percentage scores has a mean (μ) of 85 and standard deviation (σ) of 7 points. The new class of 36 students had a mean (M) final percentage score of 87 points. I conducted a hypothesis test to see if the new class would have significantly differently final percentage scores from the population of past students. I was interested in any type of “difference,” whether it’s an increase or a decrease in final percentage score. The significance level for my Z test was set at α= .05.

f. Determine the critical value for Z

7. Compare the obtained Z and critical Z and then make a decision about the result of the hypothesis test: Explicitly states “reject” or “fail to reject” the null hypothesis

8. Write 1-2 sentences to conclude the results (you can simply restate the accepted hypothesis or explain it in another way)

1. Calculate the raw effect size and the standardized effect size for this test.

10. Since the average final percentage score for the new section is numerically higher than the average final percentage score from the existing pool of past students, I could also set up a directional research hypothesis. What would the written hypotheses and the notations be?

11. Using the alpha level of α= .05, determine the critical Z value for the directionalhypothesis.

l. In this statistical test, how high does the mean final percentage score from the new class have to be, at least, to be considered “significantly” higher than the pool of past final percentage scores?

Hint: In other words, when does the calculated Z equal the critical Z? What needs to be the sample mean for that to happen?

Answer 1

Given that,
population mean(u)=85
standard deviation, σ =7
sample mean, x =87
number (n)=36
null, Ho: μ=85
alternate, H1: μ!=85
level of significance, α = 0.05
from standard normal table, two tailed z α/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 87-85/(7/sqrt(36)
zo = 1.71
| zo | = 1.71
critical value
the value of |z α| at los 5% is 1.96
we got |zo| =1.71 & | z α | = 1.96
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 1.71 ) = 0.09
hence value of p0.05 < 0.09, here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=85
alternate, H1: μ!=85
g.
test statistic: 1.71
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.09
h.
we do not have enough evidence to support the claim that significantly differently final percentage scores from the population of past students.
i.
effective size = difference of mean/standard deviation
effective size = 87-85/7
effective size =0.2857 it is small
j.
Given that,
population mean(u)=85
standard deviation, σ =7
sample mean, x =87
number (n)=36
null, Ho: μ=85
alternate, H1: μ>85
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 = 87-85/(7/sqrt(36)
zo = 1.71
| zo | = 1.71
critical value
the value of |z α| at los 5% is 1.645
we got |zo| =1.71 & | z α | = 1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value : right tail - ha : ( p > 1.71 ) = 0.04
hence value of p0.05 > 0.04, here we reject Ho
ANSWERS
---------------
null, Ho: μ=85
alternate, H1: μ>85
test statistic: 1.71
critical value: 1.645
decision: reject Ho
p-value: 0.04
we have enough evidence to support the claim that average final percentage score
for the new section is numerically higher than the average final percentage score from the existing pool of past students.
k.
Z critical is 1.71 same as previous