Solution :
Here, given :
Sample size, n = 39
Sample mean, [tex]$\bar X$[/tex] = 3.6
Standard deviation of the sample, s =2
The population mean, [tex]$\mu_0 = 3$[/tex]
The significance level, [tex]$\alpha = 0.05$[/tex]
a). Therefore the hypothesis is :
[tex]$H_0 : \mu = 3 \text{ Vs} \ H_a: \mu \neq 3$[/tex]
b). The test statics is given as :
[tex]$t = \frac{\bar X - \mu_0}{\frac{s}{\sqrt n}} \rightarrow t_{n-1}$[/tex]
[tex]$t = \frac{3.6-3}{\frac{2}{\sqrt {39}}} $[/tex]
= 1.873
c). The p- value is given by :
[tex]$P(t_{d.f}>|t_{stat}|)$[/tex]
[tex]$=P(t_{39-1}> 1.873)$[/tex]
[tex]$=0.0688$[/tex]
d). The conclusion :
In this case, the p-value is [tex]$0.688 > \alpha=0.05$[/tex]
So, we do not reject [tex]$H_0$[/tex].
Therefore, we conclude that it is not a statistically significant difference between national average time for selling a home and the mean time for selling in Greene County.