Question

Do a two-sample test for equality of means assuming unequal variances. Calculate the p-value using Excel.

(a-1)	Comparison of GPA for randomly chosen college juniors and seniors:
	x⎯⎯1x¯1 = 4, s1 = .20, n1 = 15, x⎯⎯2x¯2 = 4.25, s2 = .30, n2 = 15, α = .025, left-tailed test.
	(Negative values should be indicated by a minus sign. Round down your d.f. answer to the nearest whole number and other answers to 4 decimal places.)

d.f.
t-calculated
p-value

(a-2)	Based on the above data choose the correct decision.
	Reject the null hypothesis Do not reject the null hypothesis

(b-1)	Comparison of average commute miles for randomly chosen students at two community colleges:
	x⎯⎯1x¯1 = 17, s1 = 5, n1 = 22, x⎯⎯2x¯2 = 21, s2 = 7, n2 = 19, α = .05, two-tailed test.
	(Negative values should be indicated by a minus sign. Round down your d.f. answer to the nearest whole number and other answers to 4 decimal places.)

d.f.
t-calculated
p-value

(b-2)	Based on the above data choose the correct decision.
	Do not reject the null hypothesis Reject the null hypothesis

(c-1)	Comparison of credits at time of graduation for randomly chosen accounting and economics students:
	x⎯⎯1x¯1 = 141, s1 = 2.8, n1 = 12, x⎯⎯2x¯2 = 138, s2 = 2.7, n2 = 17, α = .05, right-tailed test.
	(Negative values should be indicated by a minus sign. Round down your d.f. answer to the nearest whole number and other answers to 4 decimal places.)

d.f.
t-calculated
p-value

(c-2)	Based on the above data choose the correct decision.
	Reject the null hypothesis Do not reject the null hypothesis

Answer 1

1)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 <   0                  
                          
Level of Significance ,    α =    0.025                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   4.00                  
standard deviation of sample 1,   s1 =    0.20                  
size of sample 1,    n1=   15                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   4.25                  
standard deviation of sample 2,   s2 =    0.30                  
size of sample 2,    n2=   15                  
                          
difference in sample means =    x̅1-x̅2 =    4.0000   -   4.3   =   -0.25  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    0.2550                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.0931                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -0.2500   -   0   ) /    0.09   =   -2.6854
                          
Degree of freedom, DF=   n1+n2-2 =    28                 

p-value =        0.006019   [ excel function: =T.DIST(t stat,df) ]               
Conclusion:     p-value <α , Reject null hypothesis                     
  

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2)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   17.00                  
standard deviation of sample 1,   s1 =    5.00                  
size of sample 1,    n1=   22                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   21.00                  
standard deviation of sample 2,   s2 =    7.00                  
size of sample 2,    n2=   19                  
                          
difference in sample means =    x̅1-x̅2 =    17.0000   -   21.0   =   -4.00  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    6.0064                  
std error , SE =    Sp*√(1/n1+1/n2) =    1.8811                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -4.0000   -   0   ) /    1.88   =   -2.1264
                          
Degree of freedom, DF=   n1+n2-2 =    39                  

p-value =        0.0399 (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value <α , Reject null hypothesis                     
                   

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3)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 >   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   141.00                  
standard deviation of sample 1,   s1 =    2.80                  
size of sample 1,    n1=   12                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   138.00                  
standard deviation of sample 2,   s2 =    2.70                  
size of sample 2,    n2=   17                  
                          
difference in sample means =    x̅1-x̅2 =    141.0000   -   138.0   =   3.00  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    2.7412                  
std error , SE =    Sp*√(1/n1+1/n2) =    1.0335                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   3.0000   -   0   ) /    1.03   =   2.9027
                          
Degree of freedom, DF=   n1+n2-2 =    27                  

p-value =        0.0036 [excel function: =T.DIST.RT(t stat,df) ]              
Conclusion:     p-value <α , Reject null hypothesis                      

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