Question

A) Suppose a marketing research team is interested in comparing prices at two grocery chains: Sobeys and Metro. In a random sample of 10 Sobeys grocery stores, the researchers found that a standardized weekly shopping list had a sample mean price of $121.92, and a sample standard deviation of $1.40. In a random sample of 10 Metro stores, a standardized weekly shopping list had a sample mean price of $114.81 with a sample standard deviation of $1.84. Let the Sobeys prices be population 1, and let the Metro prices be population 2.

Calculate the point estimate for a 95% confidence interval for the difference in the population mean weekly shopping price between the two grocery stores.

B) Suppose a bank manager has developed a new system to reduce customer wait time at branches for teller services. The bank manager hopes that the new system will have a population mean wait time that is less than the population mean time of the old system. She obtains a sample of 100 customer wait times from the old system, and 50 customer wait times from the new system. She calculates the sample mean wait time under the new system to be 5.65 minutes with a sample standard deviation of 2 minutes, and the sample mean wait time under the old system was 8 minutes with a sample standard deviation of 4 minutes. Using degrees of freedom equal to 75:

Use Welch’s method to calculate a 99% confidence interval for the difference between the population means (let the old wait times be population

C) Enter the margin of error for your confidence interval below, rounded to 2 decimal places

Answer 1

a)

Sample #1   ---->   1
mean of sample 1,    x̅1=   121.920
standard deviation of sample 1,   s1 =    1.4000
size of sample 1,    n1=   10
      
Sample #2   ---->   2
mean of sample 2,    x̅2=   114.810
standard deviation of sample 2,   s2 =    1.8400
size of sample 2,    n2=   10

α=0.05

Degree of freedom, DF=   n1+n2-2 =    18              
t-critical value =    t α/2 =    2.1009   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    1.6349              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    0.7311              
margin of error, E = t*SE =    2.1009   *   0.73   =   1.53606  
                      
difference of means =    x̅1-x̅2 =    121.9200   -   114.810   =   7.1100
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    7.1100   -   1.5361   =   5.574
Interval Upper Limit=   (x̅1-x̅2) + E =    7.1100   +   1.5361   =   8.646

b)

Sample #1   ---->   1
mean of sample 1,    x̅1=   8.00
standard deviation of sample 1,   s1 =    4
size of sample 1,    n1=   100
      
Sample #2   ---->   2
mean of sample 2,    x̅2=   5.650
standard deviation of sample 2,   s2 =    2.00
size of sample 2,    n2=   50

α=0.01

Degree of freedom, DF=       75          
t-critical value =    t α/2 =    2.643   (excel formula =t.inv(α/2,df)      
                  
                  
                  
std error , SE =    √(s1²/n1+s2²/n2) =    0.490          
margin of error, E = t*SE =    2.643   *   0.490   =   1.294792
                  
difference of means = x̅1-x̅2 =    8.0000   -   5.650   =   2.3500
confidence interval is                   
Interval Lower Limit = (x̅1-x̅2) - E =    2.3500   -   1.295   =   1.055
Interval Upper Limit = (x̅1-x̅2) + E =    2.3500   -   1.295   =   3.645

c)

margin of error, E = t*SE =    2.643   *   0.490   =   1.29