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 1010 stores this year shows mean sales of 8282 units of a small appliance, with a standard deviation of 9.29.2 units. During the same point in time last year, an SRS of 1414 stores had mean sales of 89.8289.82 units, with standard deviation 13.413.4 units. A decrease from 89.8289.82 to 8282 is a drop of about 10%.

1. Construct a 95% confidence interval estimate of the difference μ1−μ2μ1−μ2, where μ1μ1 is the mean of this year's sales and μ2μ2 is the mean of last year's sales.

(a) <(μ1−μ2)<<(μ1−μ2)<

(b) The margin of error is .

2. At a 0.050.05 significance level, is there sufficient evidence to show that sales this year are different from last year?

A. No
B. Yes

please show your work and what function to use on the calculator. thank you !

Answer 1

1)

a)

Sample #1   ---->   This year   
mean of sample 1,    x̅1=   82.00                  
standard deviation of sample 1,   s1 =    9.20                  
size of sample 1,    n1=   10                  
                          
Sample #2   ---->   Previois year
mean of sample 2,    x̅2=   89.82                  
standard deviation of sample 2,   s2 =    13.40                  
size of sample 2,    n2=   14                  
                          
difference in sample means =    x̅1-x̅2 =    82.0000   -   89.8   =   -7.82  

Degree of freedom, DF=   n1+n2-2 =    22              
t-critical value =    t α/2 =    2.0739   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    11.8629              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    4.9117              
margin of error, E = t*SE =    2.0739   *   4.9117   =   10.1863  
                      
difference of means =    x̅1-x̅2 =    82.0000   -   89.820   =   -7.8200
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    -7.8200   -   10.1863   =   -18.006
Interval Upper Limit=   (x̅1-x̅2) + E =    -7.8200   +   10.1863   =   2.366

b)

margin of error, E = t*SE =    2.0739   *   4.9117   =   10.186

2)

A. No

Because confidence interval contain the zero

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