Question

The data in TECHPRO.sav, obtained from Business Week’s (June 22, 2006) technology section, represents typical salaries of technology professionals in 13 metropolitan areas for 2003 and 2005. Suppose you want to determine if the mean salary of technology professionals at all US. Metropolitan areas have increased between 2003 and 2005.

(a) Set up the null and alternative hypothesis for the test.   

(f) Conduct the appropriate test and provide your conclusion. More specifically, I want you to examine   whether the null hypothesis should be rejected by analyzing the data with SPSS. alpha=0.05

DATA SET:

AREA	SAL 2003	Sal 2005
Silicon Valley	87.7	85.9
New York	78.6	80.3
Washington, D.C.	71.4	77.4
Los Angeles	70.8	77.1
Denver	73.0	77.1
Boston	76.3	80.1
Atlanta	73.6	73.2
Chicago	71.1	73.0
Philadelphia	69.5	69.8
San Diego	69.0	77.1
Seattle	71.0	66.9
Dallas-Ft. Worth	73.0	71.0
Detroit	62.3	64.1

Answer 1

Given that,
mean(x)=72.8692
standard deviation , s.d1=5.8895
number(n1)=13
y(mean)=74.8462
standard deviation, s.d2 =5.9602
number(n2)=13
null, Ho: u1 = u2
alternate, H1: u1 < u2
level of significance, α = 0.05
from standard normal table,left tailed t α/2 =1.782
since our test is left-tailed
reject Ho, if to < -1.782
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =72.8692-74.8462/sqrt((34.68621/13)+(35.52398/13))
to =-0.851
| to | =0.851
critical value
the value of |t α| with min (n1-1, n2-1) i.e 12 d.f is 1.782
we got |to| = 0.8507 & | t α | = 1.782
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value:left tail - Ha : ( p < -0.8507 ) = 0.20579
hence value of p0.05 < 0.20579,here we do not reject Ho
ANSWERS
---------------
a.
null, Ho: u1 = u2
alternate, H1: u1 < u2
b.
test statistic: -0.851
c.
critical value: -1.782
d.
decision: do not reject Ho
e.
p-value: 0.20579
f.
we do not have enough evidence to support the claim that if the mean salary of technology professionals at all US. Metropolitan areas have increased between 2003 and 2005.