Question

A market research firm supplies manufacturers with estimates of the retail sales of their products from samples of retail stores. Marketing managers are prone to look at the estimate and ignore sampling error. An SRS of 29 stores this year shows mean sales of 61 units of a small appliance, with a standard deviation of 6.2 units. During the same point in time last year, an SRS of 17 stores had mean sales of 53.994 units, with standard deviation 7.7 units. An increase from 53.994 to 61 is a rise of about 11%.

1. Construct a 99% confidence interval estimate of the difference μ1−μ2, where μ1 is the mean of this year's sales and μ2 is the mean of last year's sales.

(a) <(μ1−μ2)<

(b) The margin of error is .

2. At a 0.01 significance level, is there sufficient evidence to show that sales this year are different from last year?

A. No
B. Yes

Answer 1

1)

a)

Sample #1   ---->   this year
mean of sample 1,    x̅1=   61.000
standard deviation of sample 1,   s1 =    6.20
size of sample 1,    n1=   29
      
Sample #2   ---->   last year
mean of sample 2,    x̅2=   53.994
standard deviation of sample 2,   s2 =    7.70
size of sample 2,    n2=   17

DF = min(n1-1 , n2-1 )=   16              
t-critical value , t* =    2.9208   (excel formula =t.inv(α/2,df)          
                  
       2.921   (excel formula =t.inv(α/2,df)      
                  
std error , SE =    √(s1²/n1+s2²/n2) =    2.194          
margin of error, E = t*SE =    2.9208   *   2.194   =   6.4079
                  
difference of means = x̅1-x̅2 =    61.0000   -   53.994   =   7.0060
99% confidence interval is                   
Interval Lower Limit = (x̅1-x̅2) - E =    7.0060   -   6.4079   =   0.5981
Interval Upper Limit = (x̅1-x̅2) + E =    7.0060   -   6.4079   =   13.4139

b)

margin of error, E =   6.4079

2)

since confidence interval do not include 0, so, test is significant

Yes.

At a 0.01 significance level, there is sufficient evidence to show that sales this year are different from last year