Question

Data about salaries showed that staff nurses in Tampa earn less than staff nurses in Dallas (The Tampa Tribune, Jan 15, 2007). You are a manager at a Tampa hospital and want to test the claim of the Tribune. You conduct a follow up survey of a random sample of 25 staff nurses in Tampa and a random sample of 28 staff nurses in Dallas. You find that the sample from Tampa showed an average salary of $52000 with a standard deviation of $8500, and the sample from Dallas showed an average salary of $56500 with a standard deviation of $4650. Is this evidence, at the 5% level of significance, that the Tampa Tribune article is correct? Conduct a complete hypothesis test. Show all your work, including formulas, then scan and upload it below.

(Assume both populations are normally distributed. Also, you have no reason to believe the variances of the two populations are equal.)

Answer 1

Given that,
mean(x)=52000
standard deviation , s.d1=8500
number(n1)=25
y(mean)=56500
standard deviation, s.d2 =4650
number(n2)=28
null, Ho: u1 = u2
alternate, H1: u1 < u2
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.711
since our test is left-tailed
reject Ho, if to < -1.711
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =52000-56500/sqrt((72250000/25)+(21622500/28))
to =-2.3515
| to | =2.3515
critical value
the value of |t α| with min (n1-1, n2-1) i.e 24 d.f is 1.711
we got |to| = 2.35147 & | t α | = 1.711
make decision
hence value of | to | > | t α| and here we reject Ho
p-value:left tail - Ha : ( p < -2.3515 ) = 0.01362
hence value of p0.05 > 0.01362,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 < u2
test statistic: -2.3515
critical value: -1.711
decision: reject Ho
p-value: 0.01362
we have enough evidence to support the claim that salaries showed that staff nurses in Tampa earn less than staff nurses in Dallas.