To test whether the mean math score is higher for male students than for female

Question

To test whether the mean math score is higher for male students than for female students, a math test was given to a group of 8th graders. The results of the test are summarized in the following table. Assume that the distribution of math scores for both groups is normally distributed with equal standard deviations. Set up the null and alternative hypotheses to test the mean math score of 8th graders is higher for male students than for female students. a)H_0: u_1 > u_2 vs H_a:u_1 -1.645; Decision: Do not reject H_0

Explanation / Answer

Given that,
mean(x)=55.9
standard deviation , s.d1=12.96
number(n1)=14
y(mean)=48.4
standard deviation, s.d2 =12.65
number(n2)=25
null, Ho: u1 < u2
alternate,mean score is higher for male students than for female students H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.687
since our test is right-tailed
reject Ho, if to > 1.687
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (13*167.9616 + 24*160.0225) / (39- 2 )
s^2 = 162.8119
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=55.9-48.4/sqrt((162.8119( 1 /14+ 1/25 ))
to=7.5/4.2593
to=1.7608
| to | =1.7608
critical value
the value of |t | with (n1+n2-2) i.e 37 d.f is 1.687
we got |to| = 1.7608 & | t | = 1.687
make decision
hence value of | to | > | t | and here we reject Ho
p-value: right tail -ha : ( p > 1.7608 ) = 0.04326
hence value of p0.05 > 0.04326,here we reject Ho
ANSWERS
---------------
a.
null, Ho: u1 = u2
alternate, H1: u1 > u2
b.
test statistic: 1.7608
c.
critical value: 1.687

decision: t<1.771, reject Ho
p-value: 0.04326

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