Test the claim that the mean GPA of Orange Coast students is larger than the mea
ID: 3323010 • Letter: T
Question
Test the claim that the mean GPA of Orange Coast students is larger than the mean GPA of Coastline students at the 0.025 significance level. The null and alternative hypothesis would be: The test is: right-tailed two-tailed left-tailed The sample consisted of 25 Orange Coast students, with a sample mean GPA of 2.29 and a standard deviation of 0.03, and 25 Coastline students, with a sample mean GPA of 2.26 and a standard deviation of 0.06 The test statistic is: The p-value is: Based on this we (to 2 decimals) (to 2 decimals) Reject the null hypothesis Fail to reject the null hypothesisExplanation / Answer
Given that,
mean(x)=2.29
standard deviation , s.d1=0.03
number(n1)=25
y(mean)=2.26
standard deviation, s.d2 =0.06
number(n2)=25
null, Ho: u1 <= u2
alternate, H1: u1 > u2
level of significance, = 0.025
from standard normal table,right tailed t /2 =2.06
since our test is right-tailed
reject Ho, if to > 2.06
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =2.29-2.26/sqrt((0.0009/25)+(0.0036/25))
to =2.24
| to | =2.24
critical value
the value of |t | with min (n1-1, n2-1) i.e 24 d.f is 2.06
we got |to| = 2.23607 & | t | = 2.06
make decision
hence value of | to | > | t | and here we reject Ho
p-value:right tail - Ha : ( p > 2.2361 ) = 0.01745
hence value of p0.025 > 0.01745,here we reject Ho
ANSWERS
---------------
a.
null, Ho: u1 <= u2
alternate, H1: u1 > u2
b.
right tailed test
c.
test statistic: 2.24
critical value: 2.06
d.
p-value: 0.01745
e.
decision: reject Ho
we have enough evidence to support the claim that mean of orange coast students greater than the coastline students
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