Question

.Finance course grade information regarding male and female students of a large university is shown below (?_"and ?_!unknown) and assumed unequal).

	Female	Male
Sample Size	24	28
Sample Mean Grade	29.8	27.3
Sample Variance	6,554	3,276

1. Test whether or not the average grade of females is significantly greater than that of males at ⍺=0,05 % significance level?
2. Develop 90 % confidence interval estimate of the difference between grade of the female and male students.

Not: You have to draw the table to see which area that you have to calculate

Answer 1

SOLUTION:

a)

Ho :   µ1 - µ2 =   0          
Ha :   µ1-µ2 >   0          
                  
Level of Significance ,    α =    0.05          
                  
Sample #1   ---->   1          
mean of sample 1,    x̅1=   29.80          
standard deviation of sample 1,   s1 =    80.95677859          
size of sample 1,    n1=   24          
                  
Sample #2   ---->   2          
mean of sample 2,    x̅2=   27.300          
standard deviation of sample 2,   s2 =    57.24          
size of sample 2,    n2=   28          
                  
difference in sample means = x̅1-x̅2 =    29.800   -   27.3000   =   2.5000
                  
std error , SE =    √(s1²/n1+s2²/n2) =    19.7505          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   2.5000   /   19.7505   ) =   0.1266
                  

                  
t-critical value , t* =        1.6839   (excel function: =t.inv(α,df)      
Decision:   | t-stat | < | critical value |, so, Do not Reject Ho              
p-value =        0.44995   [excel function: =T.DIST.RT(t stat,df) ]      
Conclusion:     p-value>α , Do not reject null hypothesis              

There is no enough evidence to support the claim

b)

Degree of freedom, DF=       40          
t-critical value =    t α/2 =    1.684   (excel formula =t.inv(α/2,df)      
                  
                  
                  
std error , SE =    √(s1²/n1+s2²/n2) =    19.751          
margin of error, E = t*SE =    1.684   *   19.751   =   33.256946
                  
difference of means = x̅1-x̅2 =    29.8000   -   27.300   =   2.5000
confidence interval is                   
Interval Lower Limit = (x̅1-x̅2) - E =    2.5000   -   33.257   =   -30.757
Interval Upper Limit = (x̅1-x̅2) + E =    2.5000   -   33.257   =   35.757

(OR) TRY THIS ANSWER

(a)

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(b)

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