Question

Sample 1 Sample 2 n1 = 15 n2 = 13 x1 =54 x2 = s1 =39 77 s2=46

Answer 1

a)

mean of sample 1,    x̅1=   5.4
standard deviation of sample 1,   s1 =    3.9000
size of sample 1,    n1=   15
      
mean of sample 2,    x̅2=   7.7
standard deviation of sample 2,   s2 =    4.6000
size of sample 2,    n2=   13
      
difference in sample means =    x̅1-x̅2 =    -2.3000
      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    4.2375
      
std error , SE =    Sp*√(1/n1+1/n2) =    1.6057
      
t-statistic =    ((x̅1-x̅2)-µd)/SE =    -1.43(answer)

Degree of freedom, DF=   n1+n2-2 =    26

p-value =        0.164

p-value>α , Do not reject null hypothesis

answer is : do not reject Ho, there is insufficient evidence.........

------------------------------------------

b)

Degree of freedom, DF=   n1+n2-2 =    26  
t-critical value =    t α/2 =    1.7056   (excel formula =t.inv(α/2,df)
          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    4.2375  
          
std error , SE =    Sp*√(1/n1+1/n2) =    1.6057  
margin of error, E =    t*SE =    2.7387  
          
difference of means =    x̅1-x̅2 =    -2.30

  
confidence interval is           
Interval Lower Limit=   (x̅1-x̅2) - E =    -5.04
Interval Upper Limit=   (x̅1-x̅2) + E =    0.44

answers:

1) inside

2) insufficient

3) includes