Question

In a current study, a random sample of 20 teachers from Texas and an independent random sample of 20 teachers from Indiana have been asked to report their annual income. The data obtained are as follows:

	Annual income in dollars
Texas	36428, 47516, 30452, 33084, 49005, 27069, 30650, 37243, 32531, 37394, 48089, 38013, 44729, 25913, 41093 41199, 32775, 33773, 36110, 43329
Indiana	49281, 46645, 48434, 51386, 39959, 34413, 49065, 35413, 33934, 40901, 35575, 35369, 41076, 41661, 42299 33578, 42152, 39115, 30098, 42830

The population standard deviations for the annual incomes of teachers in Texas and in Indiana are estimated as $6200 and $6000, respectively. It is also known that both populations are approximately normally distributed. Construct a 95% confidence interval for the difference −μ1μ2 between the mean annual income of teachers from Texas (μ1) and the mean annual income of teachers from Indiana (μ2). Then complete the table below.Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places. (If necessary, consult a list of formulas.)

What is the lower limit of the 95% confidence interval? What is the upper limit of the 95% confidence interval?

Answer 1

sample #1   ------->              
mean of sample 1,    x̅1=   37319.75          
population std dev of sample 1,   σ1 =    6200          
size of sample 1,    n1=   20          
                  
sample #2   --------->              
mean of sample 2,    x̅2=   40659.2          
population std dev of sample 2,   σ2 =    6000          
size of sample 2,    n2=   20          
                  
difference in sample means = x̅1 - x̅2 =    37319.75   -   40659.2   =   -3339.45
Level of Significance ,    α =    0.05          
z-critical value =    Z α/2 =    1.9600   [excel function =normsinv(α/2) ]      
                  
std error , SE =    √(σ1²/n1+σ2²/n2) =    1929.2486          
margin of error, E = Z*SE =    1.9600   *   1929.249   =   3781.2577
                  
difference of means = x̅1 - x̅2 =    37319.75   -   40659.2   =   -3339.450
confidence interval is                   
Interval Lower Limit= (x̅1 - x̅2) - E =    -3339.450   -   3781.258   =   -7120.71
Interval Upper Limit= (x̅1 - x̅2) + E =    -3339.450   +   3781.258   =   441.81

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