Question

 

In a consumer research study, several Meijer and Walmart stores were surveyed at random and the average basket price was recorded for each. It was found that the average basket price for 37 Meijer stores was $47.268 with a standard deviation of $10.83. Similarly, 48 Walmart stores had an average basket price of $67.244 with a standard deviation of $11.669. If a 99% confidence interval for the difference between the true average basket prices of Meijer versus Walmart is calculated, what is the margin of error? You can assume that the standard deviations of the two populations are statistically similar.

Question 7 options:

	1) 2.63636876
	2) 5.871
	3) 6.375
	4) 2.47488943
	5) 6.525

Answer 1

Level of Significance ,    α =    0.01
      
mean of sample 1,    x̅1=   47.2680
standard deviation of sample 1,   s1 =    10.8300
size of sample 1,    n1=   37
      
mean of sample 2,    x̅2=   67.244
standard deviation of sample 2,   s2 =    11.669
size of sample 2,    n2=   48
      
difference in sample means =    x̅1-x̅2 =    -19.9760

Degree of freedom, DF=   n1+n2-2 =    83  
t-critical value =    t α/2 =    2.6364   (excel formula =t.inv(α/2,df)
          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    11.3127  
          
std error , SE =    Sp*√(1/n1+1/n2) =    2.4749  
margin of error, E =    t*SE =    6.525

so, answer is option 5)