Question

The Metro Real Estate Association is preparing a pamphlet that they feel might be of interest to prospective home buyers in the Southlake and Northwood areas of the city. One item if interest is the length of time the seller occupied the home. A sample of 17 homes sold recently in Southlake revealed that the mean length of ownership was 9.2 years with a standard deviation of 2.9 years. A sample of 15 homes in Northwood revealed that the mean length of ownership was 7.4 years with a standard deviation of 2.1 years. Can we conclude that the Southlake residents owned their homes for a longer period of time compare to residents in the Northwood area? What is the decision rule? a = 0.05.

Answer 1

Given data

For southlake resident

n1 = 17             ( sample size for southlake resident )

1 = 9.2          ( sample mean i.e mean length of owenership for southlake resident )

T

s1 = 2.9          ( sample standard deviation for southlake resident )

For Northwood resident

n2 = 15

2 = 7.4          ( sample mean i.e mean length of owenership for Northwood resident )

T

s2 = 2.1      ( sample standard deviation for Northwood residentt )

To Test :-

H0 : u1 $\leq$ u2       [ Southlake resident may not owned their homes for longer period than Northwood residentt ]

H1 : u1 > u2       [ Southlake resident may owned their homes for longer period than Northwood residentt ]

Test Statistics TS :-

TS = $\frac{\bar{x1}-\bar{x2}}{SE}$

Calculation -

SE =

V sd/nl+ sd/n2

where

OS
=    $\frac{(n1-1)*s1^{2}+(n2-1)*s2^{2}}{n1+n2-2}$

        =    $\frac{(17-1)*2.9^{2}+(15-1)*2.1^{2}}{17+15-2}$

= 6.543333                    { after calculation }

OS

Thus

SE =

V sd/nl+ sd/n2

      = $\sqrt{ 6.543333/17 + 6.543333/15 }$

SE = 0.906159                   { after calculation }

So we have

T
1 = 9.2     ;      
T
2 = 7.4      ;     SE = 0.906159

TS = $\frac{\bar{x1}-\bar{x2}}{SE}$

TS = $\frac{9.2-7.4}{0.906159 }$

TS = 1.986406

So calculated test statistics value is TS = 1.986406

To find t-critical value $t_{\alpha }$ :-

Since alternative hypothesis is of " > " type , so this is one-tail test , and hence its critical t-value will be given by $t_{\alpha }$ , where $t_{\alpha }$ is t-distributed with n1+n2-2 = 17+15-2 = 30 degree of freedom with $\alpha$ = 0.05

Now $t_{\alpha }$ can be obtained from statistical book or more accuratly from any software like R/Excel .

From R

> qt(1-0.05,df=30)
[1] 1.697261

Hence t-critical value $t_{\alpha }$ = 1.697261

Rejection criteria ( Decision Rule ) :-

We reject null hypothesis is calculated test statistics is greater than t-critical value $t_{\alpha }$

Now here we have TS = 1.986406 > 1.697261

i.e   TS > $t_{\alpha }$

Since , test statistics value TS greater than t-critical value , we reject null hypothesis at 0.05 level of significance .

Conclusion :-

We reject null hypothesis H0 , and hence we conclude that , Southlake resident owned their homes for longer period of time compare to resident in the Northwood area .