Question

Use the given statistics to complete parts​ (a) and​ (b). Assume that the populations are normally distributed. ​(a) Test whether mu 1greater thanmu 2 at the alphaequals0.05 level of significance for the given sample data. ​(b) Construct a 90​% confidence interval about mu 1minusmu 2. Population 1 Population 2 n 20 23 x overbar 50.7 46.9 s 4.8 12.8 ​(a) Identify the null and alternative hypotheses for this test. A. Upper H 0​: mu 1not equalsmu 2 Upper H 1​: mu 1equalsmu 2 B. Upper H 0​: mu 1equalsmu 2 Upper H 1​: mu 1not equalsmu 2 C. Upper H 0​: mu 1greater thanmu 2 Upper H 1​: mu 1equalsmu 2 D. Upper H 0​: mu 1equalsmu 2 Upper H 1​: mu 1less thanmu 2 E. Upper H 0​: mu 1less thanmu 2 Upper H 1​: mu 1equalsmu 2 F. Upper H 0​: mu 1equalsmu 2 Upper H 1​: mu 1greater thanmu 2 Your answer is correct. Find the test statistic for this hypothesis test. nothing ​(Round to two decimal places as​ needed.)

Need P-Value

Conclusion for this hypothesis test

And the answer for part B

Answer 1

a)

Ho :   µ1 = µ2
Ha :   µ1 > µ2

Sample #1   ---->   
mean of sample 1,    x̅1=   50.70          
standard deviation of sample 1,   s1 =    4.8          
size of sample 1,    n1=   20          
                  
Sample #2   ---->     
mean of sample 2,    x̅2=   46.900          
standard deviation of sample 2,   s2 =    12.80          
size of sample 2,    n2=   23          
                  
difference in sample means = x̅1-x̅2 =    50.700   -   46.9000   =   3.8000
                  
std error , SE =    √(s1²/n1+s2²/n2) =    2.8767          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   3.8000   /   2.8767   ) =   1.32

p-value =        0.0986 [excel function: =T.DIST.RT(t stat,df) ]
Conclusion:     p-value>α , Do not reject null hypothesis      

There is not enough evidence to conclude that µ1 is greater than µ 2

b)

α=0.10

Degree of freedom, DF=       28          
t-critical value =    t α/2 =    1.701   (excel formula =t.inv(α/2,df)      
                  
                  
                  
std error , SE =    √(s1²/n1+s2²/n2) =    2.877          
margin of error, E = t*SE =    1.701   *   2.877   =   4.8937
                  
difference of means = x̅1-x̅2 =    50.7000   -   46.900   =   3.8000
confidence interval is                   
Interval Lower Limit = (x̅1-x̅2) - E =    3.8000   -   4.894   =   -1.0937
Interval Upper Limit = (x̅1-x̅2) + E =    3.8000   -   4.894   =   8.6937

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