A corporation owns several convenience stores that are open 24 hours a day. It i

Question

A corporation owns several convenience stores that are open 24 hours a day. It is interested in knowing whether there is a relationship between time of day and size of purchase. One of the stores is selected at random to be invovled in a study. Store records are collected over a period of several weeks and then 300 purchases are randomly selected. Since the register also prints the time of the purchase, this random selection procedure yields both amount of time and purchase. The information is summarized in the table.

a. Is there a relationship between time and size of purchase? Use a=.05 and report the p-value.

b. Use a 90% confidence interval to estimate the proportion of customers who spend $20 or less during the period 8 am to 4 pm.

$20 or less 20.01 - $100 Over $100 8am to 4pm 65 38 14 4pm to 12 midnight 61 49 10 12 midnight to 8am 29 27 7

Explanation / Answer

a)

Doing an Expected Value Chart,          
          
60.45   44.46   12.09  
62   45.6   12.4  
32.55   23.94   6.51  

Using chi^2 = Sum[(O - E)^2/E],          
          
chi^2 =    3.132187893      
          
With df = (a - 1)(b - 1), where a and b are the number of categories of each variable,          
          
a =    3      
b =    3      
          
df =    4      
          
Thus, the critical value is          
          
significance level =    0.05      
          
chi^2(critical) =    9.487729037      
          
Also, the p value is          
          
P =    0.53595286      
          
As P > 0.05, we   FAIL TO REJECT THE NULL HYPOTHESIS.  

Hence, there is no significant evidence that there is a relationship between time and size of purchase at 0.05 level. [CONCLUSION]

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b)

There are 65 out of 117 who spent 20 or less at 8am to 4pm.

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.555555556          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.045938791          
              
Now, for the critical z,              
alpha/2 =   0.05          
Thus, z(alpha/2) =    1.644853627          
Thus,              
Margin of error = z(alpha/2)*sp =    0.075562586          
lower bound = p^ - z(alpha/2) * sp =   0.479992969          
upper bound = p^ + z(alpha/2) * sp =    0.631118142          
              
Thus, the confidence interval is              
              
(   0.479992969   ,   0.631118142   ) [ANSWER]

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