Question

Problem 1 (6 pointe study of the amount of The manager of Pred s Grocery is making a money customers spend in the store. A sample of 10 weekday morning shoppers and 15 Saturday mornings the following amounts apent: Meekday $18.88 turday $11.54 24.76 Average of Weekdays: 50.347 Average of Saturday: 57.36 Standará Deviation of Reekday: 24.258 Standard Deviation of Saturday: 22.344 24.33 27.26 35.79 42.31 53.77 62.94 73.59 76.51 88.09 5.78 36.87 46.78 56.04 58.45 60.98 62.67 66.89 71-65 75.61 81.87 84-71 85.80 than weekday shoppera. Use the Student t Test and a s significance level Owner Fred Snead contends that Saturday morning shoppers spend more a. State the null and alternate hypotheses H1: Ho: b. State the decision rule. c. Compute the value of the test statistic

Answer 1

Data:     

n1 = 15    

n2 = 10    

x1-bar = 57.36    

x2-bar = 50.347    

s1 = 22.344    

s2 = 24.258    

Hypotheses:    

Ho: μ1 ≤ μ2    

Ha: μ1 > μ2    

Decision Rule:    

α = 0.05    

Degrees of freedom = 15 + 10 - 2 = 23

Critical t- score = 1.71387152   

Reject Ho if t > 1.71387152   

Test Statistic:    

Pooled SD, s = √[{(n1 - 1) s1^2 + (n2 - 1) s2^2} / (n1 + n2 - 2)] = √(((15 - 1) * 22.344^2 + (10 - 1) * 24.258^2)/(15 + 10 - 2)) = 23.11184129

SE = s * √{(1 /n1) + (1 /n2)} = 23.1118412902583 * √((1/15) + (1/10)) = 9.435369696

t = (x1-bar -x2-bar)/SE = (57.36 - 50.347)/9.43536969622006 = 0.743267114

p- value = 0.23242257    

Decision (in terms of the hypotheses):

Since 0.74326711 < 1.713871517 we fail to reject Ho

Conclusion (in terms of the problem):

There is no sufficient evidence that Saturday morning shoppers spend more than week day shoppers.