Need to learn to work this out to take a test on it An elect

Need to learn to work this out to take a test on it:

An electrical contractor has concluded that the average home uses at least 500 yards of electrical wiring. In a sample of 45 homes, the mean amount of wiring is 545.3 yards with a standard deviation of 166.4 yards. a) At the 5% level of significance, do you agree with the contractor?  

Solution

Formulating the null and alternative hypotheses,              
              
Ho:   u   <=   500  
Ha:    u   >   500  
              
As we can see, this is a    right   tailed test.      
              
Thus, getting the critical z, as alpha =    0.05   ,      
alpha =    0.05          
zcrit =    +   1.644853627      
              
Getting the test statistic, as              
              
X = sample mean =    545.3          
uo = hypothesized mean =    500          
n = sample size =    45          
s = standard deviation =    166.4          
              
Thus, z = (X - uo) * sqrt(n) / s =    1.826211768          
              
Also, the p value is              
              
p =    0.033909185          
              
Comparing z and zcrit (or, p and significance level), we   REJECT THE NULL HYPOTHESIS.          

I agree with the contractor. There is significant evidence that the average home uses at least 500 yards of electrical wiring. [conclusion]