Question

how do we do this on a calculator?

Test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students at the 0.10 significance level. The null and alternative hypothesis would be: H:Ho Mc HoHo = Mc Ho: Po > Pc Ho: Po = Pc H:HO MC Ho: Po PC H : Ho < 1c H: Ho + c Hipo < pc Hipo #pc Hio > PC H:po > PC The test is: right-tailed left-tailed two-tailed The sample consisted of 30 Orange Coast students, with a sample mean GPA of 2.64 and a standard deviation of 0.06, and 30 Coastline students, with a sample mean GPA of 2.66 and a standard deviation of 0.08. The test statistic is: (to 2 decimals) The p-value is: (to 2 decimals) Based on this we: Reject the null hypothesis Fail to reject the null hypothesis Check Answer

Answer 1

Given that,
mean(x)=2.64
standard deviation , s.d1=0.06
number(n1)=30
y(mean)=2.66
standard deviation, s.d2 =0.08
number(n2)=30
null, Ho: u1 = u2
alternate, H1: u1 < u2
level of significance, α = 0.1
from standard normal table,left tailed t α/2 =1.311
since our test is left-tailed
reject Ho, if to < -1.311
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =2.64-2.66/sqrt((0.0036/30)+(0.0064/30))
to =-1.095
| to | =1.095
critical value
the value of |t α| with min (n1-1, n2-1) i.e 29 d.f is 1.311
we got |to| = 1.09545 & | t α | = 1.311
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value:left tail - Ha : ( p < -1.0954 ) = 0.14117
hence value of p0.1 < 0.14117,here we do not reject Ho
ANSWERS
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null, Ho: u1 = u2
alternate, H1: u1 < u2
test statistic: -1.095 =-1.10
critical value: -1.311
decision: do not reject Ho
p-value: 0.14
wed donot have enough evidence to support the claim that mean GPA of orange coast students is smaller than mean GPA of coastline students