Question

1. Consider the hypothesis test Ho: u1=p2 against H1: u 17 M2. Suppose that sample sizes are n1 = 13, n2 = 10 x1=4.7, x2 =6.8, s1= 2 and s2 =2.5. Assume that the data are randomly drawn from two independent Normal distributions (a) Confirm that it is reasonable to assume 01 2 = 022 by completing the steps i. through v. below. Use a = 0.05. HO: 01 2 = 02 20:01 2 * 022 i. Test Statistic ii. Rejection Criterion: ii. Decision (reject or do not reject the null hypothesis): iv. The probability you made an incorrect decision: v. Conclusion (state in the terms of this test): (b) Complete the steps i. through to test the hypothesis stated in number 1 using a = 0.05 and the fact that the populations variances can be assumed to be equal. HQ: u1=p2 H1: 217 p2 i. Test Statistic ii. Rejection Criterion: iii. Decision (reject or do not reject the null hypothesis): iv. The probability you made an incorrect decision: v. Conclusion (state in the terms of this test): — (c) Derive a two-sided 95% confidence interval. Interpret the confidence interval and explain how it confirms your results in part (b).

Answer 1

1)

Hₒ : σ₁ = σ₂  
H₁ : σ₁ ≠ σ₂  
Test statistic:  
F = s₁² / s₂² = 2² / 2.5² =    0.64
Degree of freedom:  
df₁ = n₁-1 =    12
df₂ = n₂-1 =    9
Critical value(s):  
Lower tailed critical value, FL = F.INV(0.05/2, 12, 9) =    0.2910
Upper tailed critical value, FU = F.INV(1-0.05/2, 12, 9) =    3.8682
P-value :  
P-value = 2*F.DIST.RT(0.64, 12, 9) =    1.5366
Conclusion:  
As p-value > α, we fail to reject the null hypothesis.  

Conclusion: Both are equal

2)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   1                  
mean of sample 1,    x̅1=   4.700                  
standard deviation of sample 1,   s1 =    2.000                  
size of sample 1,    n1=   13                  
                          
Sample #2   ---->   2                  
mean of sample 2,    x̅2=   6.800                  
standard deviation of sample 2,   s2 =    2.500                  
size of sample 2,    n2=   10                  
                          
difference in sample means =    x̅1-x̅2 =    4.7000   -   6.8   =   -2.10  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    2.2281                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.9372                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -2.1000   -   0   ) /    0.94   =   -2.2408
                          
Degree of freedom, DF=   n1+n2-2 =    21                  
t-critical value , t* =        2.080   (excel formula =t.inv(α/2,df)              
Decision:   | t-stat | > | critical value |, so, Reject Ho                      
p-value =        0.035979   (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value <α , Reject null hypothesis                      
                          
There is enough evidence that both mean are different

c)

Degree of freedom, DF=   n1+n2-2 =    21              
t-critical value =    t α/2 =    2.0796   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    2.2281              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    0.9372              
margin of error, E = t*SE =    2.0796   *   0.94   =   1.95  
                      
difference of means =    x̅1-x̅2 =    4.7000   -   6.800   =   -2.1000
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    -2.1000   -   1.9490   =   -4.049
Interval Upper Limit=   (x̅1-x̅2) + E =    -2.1000   +   1.9490   =   -0.151

  

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