male z. An experiment is conducted to compare the starting salaries of male and
ID: 3358216 • Letter: M
Question
male z. An experiment is conducted to compare the starting salaries of male and te college graduates who find jobs. Pairs are formed by choosing a male and a Te with the same major and similar grade point averages. Suppose a random sa of nine pairs is recorded. The results are given below. mple 8 Pair 6 Male 28900 31500 30100 28500 33100 27800 29500 30400 28200 Female 28800 31600 29900 28500 32700 28000 2920o 30100 28400 Diff (d) 100 -100 200 400 -200 300 300200 We want to investigate if the average salary of the males exceeds the females. a. State the null and alternative hypotheses. Given = 0.05, find the critical valueExplanation / Answer
Given that,
mean(x)=29777.7778
standard deviation , s.d1=1709.3696
number(n1)=9
y(mean)=29688.8889
standard deviation, s.d2 =1576.7407
number(n2)=9
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, = 0.05
from standard normal table,right tailed t /2 =1.75
since our test is right-tailed
reject Ho, if to > 1.75
calculate pooled variance s^2= (n1-1*s1^2 + n2-1*s2^2 )/(n1+n2-2)
s^2 = (8*2921944.4294 + 8*2486111.235) / (18- 2 )
s^2 = 2704027.8322
we use test statistic (t) = (x-y)/sqrt(s^2(1/n1+1/n2))
to=29777.7778-29688.8889/sqrt((2704027.8322( 1 /9+ 1/9 ))
to=88.8889/775.1742
to=0.1147
| to | =0.1147
critical value
the value of |t | with (n1+n2-2) i.e 16 d.f is 1.75
we got |to| = 0.1147 & | t | = 1.75
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value: right tail -ha : ( p > 0.1147 ) = 0.45507
hence value of p0.05 < 0.45507,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 0.1147
critical value: 1.75
decision: do not reject Ho
p-value: 0.45507
salary of male exceeds female
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