Question

Northwood and Eastwood are rival schools that wish to compare how their students did in a math competition. 72 randomly chosen Northwood students participated and received a mean score of 28 with a standard deviation of 8, 98 randomly chosen Eastwood students participated and received a mean score of 25 with a standard deviation of 10.

a) The Eastwood math teacher wants to show through statistics that her students receive a higher mean score in the contest. State her hypotheses.

b) Compute the test statistic and p-value for this test.

c) Reach your conclusion at the 0.05 significance level. Be sure to interpret in the context of the problem.

d) In a separate test, the Eastwood teacher finds evidence that Eastwood students performed better than Southwood students. Would this prove that the Eastwood teacher was more effective than the Southwood teacher? Explain briefly

Answer 1

Given that,
mean(x)=28
standard deviation , s.d1=8
number(n1)=72
y(mean)=25
standard deviation, s.d2 =10
number(n2)=98
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.667
since our test is right-tailed
reject Ho, if to > 1.667
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =28-25/sqrt((64/72)+(100/98))
to =2.1711
| to | =2.1711
critical value
the value of |t α| with min (n1-1, n2-1) i.e 71 d.f is 1.667
we got |to| = 2.17112 & | t α | = 1.667
make decision
hence value of | to | > | t α| and here we reject Ho
p-value:right tail - Ha : ( p > 2.1711 ) = 0.01663
hence value of p0.05 > 0.01663,here we reject Ho
ANSWERS
---------------
a.
null, Ho: u1 = u2
alternate, H1: u1 > u2
b.
test statistic: 2.1711
critical value: 1.667
c.
decision: reject Ho
p-value: 0.01663
d.
we have enough evidence to support the claim that The Eastwood math teacher wants to show through statistics that her students receive a higher mean score in the contest.