There have been complaints recently from homeowners in the n

There have been complaints recently from homeowners in the north end claiming that their homes have been assessed at values that are too high compare with other parts of town. They say that the mean increase from last year to this year has been higher in their part of town than elsewhere. To test this, the assessor\'s office staff plans to select a random sample of north end properties (group 1) and a random sample of properties from other areas within the city (group 2) and perform a hypothesis test. The following sample information is available: North End OtherSample Size 20 10Sample Mean Increase $4,010 $3,845Sample St. Deviation $1,800 $1,750Assuming that the null hypothesis will be tested using an alpha level equal to 0.05, what is the critical value?

Solution

Set Up Hypothesis
Null, Ho: u1 = u2
Alternate,mean increase from last year to this year has been higher - H1: u1 > u2
Test Statistic
X(Mean)=4010
Standard Deviation(s.d1)=1800 ; Number(n1)=20
Y(Mean)=3845
Standard Deviation(s.d2)=1750; Number(n2)=10
we use Test Statistic (t) = (X-Y)/Sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =4010-3845/Sqrt((3240000/20)+(3062500/10))
to =0.241
| to | =0.241
Critical Value
The Value of |t | with Min (n1-1, n2-1) i.e 9 d.f is 1.833
We got |to| = 0.24113 & | t | = 1.833
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value:Right Tail -Ha : ( P > 0.2411 ) = 0.40743
Hence Value of P0.05 < 0.40743,Here We Do not Reject Ho

We have evidence to support mean increase from last year to this year has been higher
in their part of town than elsewhere

[ANSWER]
Critical Value
Few Professors Use d.f to Min (n1-1, n2-1) & others to (n1+n2-2)
The Value of |t | with Min (n1-1, n2-1) i.e 9 d.f is 1.833
with (n1+n2-2) i.e 28 d.f is 1.701