Question

You may need to use the appropriate appendix table or technology to answer this question.

The following results come from two independent random samples taken of two populations.

Sample 1	Sample 2
n1 = 50	n2 = 35
x1 = 13.6	x2 = 11.6
σ1 = 2.5	σ2 = 3

Provide a 90% confidence interval for the difference between the two population means. (Use x₁ − x₂.Round your answers to two decimal places.)

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Provide a 95% confidence interval for the difference between the two population means. (Use x₁ − x₂. Round your answers to two decimal places.)

( ) to ( )

Answer 1

sample #1   ------->  
mean of sample 1,    x̅1=   13.6000
population std dev of sample 1,   σ1 =    2.5
size of sample 1,    n1=   50
      
sample #2   --------->  
mean of sample 2,    x̅2=   11.6000
population std dev of sample 2,   σ2 =    3
size of sample 2,    n2=   35

a)

Level of Significance ,    α =    0.1          
z-critical value =    Z α/2 =    1.645   [excel function =normsinv(α/2) ]      
                  
std error , SE =    √(σ1²/n1+σ2²/n2) =    0.6182          
margin of error, E = Z*SE =    1.645   *   0.618   =   1.0168
                  
difference of means = x̅1 - x̅2 =    13.6   -   11.6   =   2.000
confidence interval is                   
Interval Lower Limit= (x̅1 - x̅2) - E =    2.000   -   1.017   =   0.98
Interval Upper Limit= (x̅1 - x̅2) + E =    2.000   +   1.017   =   3.02

b)

Level of Significance ,    α =    0.05          
z-critical value =    Z α/2 =    1.960   [excel function =normsinv(α/2) ]      
                  
std error , SE =    √(σ1²/n1+σ2²/n2) =    0.6182          
margin of error, E = Z*SE =    1.960   *   0.618   =   1.2116
                  
difference of means = x̅1 - x̅2 =    13.6   -   11.6   =   2.000
confidence interval is                   
Interval Lower Limit= (x̅1 - x̅2) - E =    2.000   -   1.212   =   0.79

Interval Upper Limit= (x̅1 - x̅2) + E =    2.000   +   1.212   =   3.21