Question

A cell phone company offers two plans to its subscribers. At the time new subscribers sign up, they are asked to provide some demographic information. The mean yearly income for a sample of 38 subscribers to Plan A is $58,500 with a standard deviation of $8,000. This distribution is positively skewed; the coefficient of skewness is not larger. For a sample of 42 subscribers to Plan B, the mean income is $59,100 with a standard deviation of $9,500.

At the 0.02 significance level, is it reasonable to conclude the mean income of those selecting Plan B is larger?

1. State the decision rule. (Negative answer should be indicated by a minus sign. Round the final answer to 3 decimal places.)

Reject H₀ if t > _________ .

2. Compute the value of the test statistic. (Negative answer should be indicated by a minus sign. Round the final answer to 3 decimal places.)

Value of the test statistic __________.

3. What is the p-value? (Round the final answer to 4 decimal places.)

P-value ____________.

Answer 1

Ho :   µ1 - µ2 =   0
Ha :   µ1-µ2 <   0

1)

      
Level of Significance ,    α =    0.05

Degree of freedom, DF=   n1+n2-2 =    77      
t-critical value , t* =        -2.0891 (excel function: =t.inv(α,df)  

Decision: reject Ho if t-stat < critical value = -2.0891

2)

Sample #1   ---->   A                  
mean of sample 1,    x̅1=   58500.00                  
standard deviation of sample 1,   s1 =    8000.00                  
size of sample 1,    n1=   38                  
                          
Sample #2   ---->   B                  
mean of sample 2,    x̅2=   59100.00                  
standard deviation of sample 2,   s2 =    9500.00                  
size of sample 2,    n2=   42                  
                          
difference in sample means =    x̅1-x̅2 =    58500.0000   -   59100.0   =   -600.00  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    N/A                  
std error , SE =    Sp*√(1/n1+1/n2) =    1957.8100                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -600.0000   -   0   ) /    1957.81   =   -0.306

3)

Degree of freedom, DF=   n1+n2-2 =    77
p-value =        0.3800    [ excel function: =T.DIST(t stat,df) ]