For my PSY230, I implemented a new guided practice so that students are guided t

Question

For my PSY230, I implemented a new guided practice so that students are guided through a problem set similar to those in the assignment before completing the assignment. The goal is to provide an additional opportunity for practicing the statistical skills and to enhance the students’ performance on the weekly assignments.

In order to see if this new element was helpful to students, I compared the average assignment grade from the new section to the existing pool of past assignment grades from all my previous sections of PSY230 prior to implementing the new guided practice. In other words, the new section with the guided practice is my sample, which will be compared to the population of all past students.

According to my records, the population of all past assignment grades have a mean () of 38 points and standard deviation () of 7 points. The new class of 64 students had a mean grade (M) of 40 points. I conducted a hypothesis test to see if the new class would perform significantly differently from the population of past students. I was interested in any type of “difference,” whether it’s an improvement or a decrease in assignment grade. The significance level for my Z test was set at = .05.

In this statistical test, how high does the mean assignment grade from the new class have to be, at least, to be considered “significantly” higher than the pool of past grades?

Hint: In other words, when does the calculated Z equal the critical Z? What needs to be the sample mean for that to happen? (2 points total: 1 for answer, 1 for formula/work)

Explanation / Answer

Given that,
population mean(u)=38
standard deviation, =7
sample mean, x =40
number (n)=64
null, Ho: =38
alternate, H1: !=38
level of significance, = 0.05
from standard normal table, two tailed z /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 = 40-38/(7/sqrt(64)
zo = 2.28571
| zo | = 2.28571
critical value
the value of |z | at los 5% is 1.96
we got |zo| =2.28571 & | z | = 1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 2.28571 ) = 0.02227
hence value of p0.05 > 0.02227, here we reject Ho
ANSWERS
---------------
null, Ho: =38
alternate, H1: !=38
test statistic: 2.28571
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.02227

we have evidence to claim,new class would perform significantly differently from the population of past students

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