A local supermarket spent 250000 remodeling Prior to remodel

A local supermarket spent $250,000 remodeling. Prior to remodeling, the average daily receipts were $57,500. In a sample of 18 days following the remodeling, daily receipts were $59,800 with a standard deviation of $5,460. At the 10% significance level, what can you conclude as to whether the remodeling improved daily receipts?

Solution

Formulating the null and alternative hypotheses,              
              
Ho:   u   <=   57500  
Ha:    u   >   57500  
              
As we can see, this is a    right   tailed test.      
              
Thus, getting the critical t,              
df = n - 1 =    17          
tcrit =    +   1.33337939      
              
Getting the test statistic, as              
              
X = sample mean =    59800          
uo = hypothesized mean =    57500          
n = sample size =    18          
s = standard deviation =    5460          
              
Thus, t = (X - uo) * sqrt(n) / s =    1.787192963          
              
Also, the p value is              
              
p =    0.045874291          
              
Comparing t and tcrit (or, p and significance level), we   REJECT THE NULL HYPOTHESIS.          
              
There is significant evidence that the remodeling improved daily receipts. [conclusion]