At a local college, 65 female students were randomly selected and it was found that their...

At a local college, 65 female students were randomly selected and it was found that their mean monthly income
was $609. Seventy-five male students were also randomly selected and their mean monthly income was found to
be $651. Test the claim that male students have a higher monthly income than female students. Assume the
population standard deviation for the females is $121.50 and for the males $131. Use α = 0.01

Solutions

Expert Solution

Given that,
mean(x)=651
standard deviation , σ1 =131
number(n1)=75
y(mean)=609
standard deviation, σ2 =121.5
number(n2)=65
null, Ho: u1 = u2
alternate, H1: μ1 > u2
level of significance, α = 0.01
from standard normal table,right tailed z α/2 =2.326
since our test is right-tailed
reject Ho, if zo > 2.326
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=651-609/sqrt((17161/75)+(14762.25/65))
zo =1.97
| zo | =1.97
critical value
the value of |z α| at los 0.01% is 2.326
we got |zo | =1.967 & | z α | =2.326
make decision
hence value of | zo | < | z α | and here we do not reject Ho
p-value: right tail -Ha : ( p > 1.97 ) = 0.02459
hence value of p0.01 < 0.02459,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: μ1 > u2
test statistic: 1.97
critical value: 2.326
decision: do not reject Ho
p-value: 0.02459
we do not have enough evidence to support the claim that male students have a higher monthly income than female students.

orchestra answered 10 months ago

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