Question

Freud experimented with randomly selected rats and recorded the number of hours it took mice to learn to tap on a lever to get water. Let ?̅ = male mice. Here are his findings:

Hours to learn how to tap on lever

Female Rats - 47.9, 69.2, 46.2, 56.3, 57.6, 55.5, 60.1, 65.8

Male Rats - 42.5, 70.8, 55.1, 50.4, 65.0, 66.9, 64.2, 64.7

a) State the conditions and verify they are met.

           b) Find the indicated values for a 87% confidence interval of the difference in hours to learn. Interpret your results.

c) Now perform an appropriate hypothesis test, and state your conclusion.

Answer 1

a)

Population variances are equal and unknown.

	Sample 1:
Sample Standard deviation,	s₁ =	7.906552
Sample size,	n₁ =	8
	Sample 2:
Sample Standard deviation,	s₂ =	9.646465
Sample size,	n₂ =	8
	α =	0.05

Null and alternative hypothesis:  
Hₒ : σ₁ = σ₂  
H₁ : σ₁ ≠ σ₂  
Test statistic:  
F = s₁² / s₂² = 7.90655243633852² / 9.64646493355389² =    0.6718
Degree of freedom:  
df₁ = n₁-1 =    7
df₂ = n₂-1 =    7
Critical value(s):  
Lower tailed critical value, FL = F.INV(0.13/2, 7, 7) =    0.2953
Upper tailed critical value, FU = F.INV(1-0.13/2, 7, 7) =    3.3860
P-value :  
P-value = 2*F.DIST.RT(0.6718, 7, 7) =    1.3873
Conclusion:  
As p-value > α, we fail to reject the null hypothesis.  

b)

Degree of freedom, DF=   n1+n2-2 =    14              
t-critical value =    t α/2 =    1.6087   (excel formula =t.inv(α/2,df)          
                      
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    8.8195              
                      
std error , SE =    Sp*√(1/n1+1/n2) =    4.4098              
margin of error, E = t*SE =    1.6087   *   4.4098   =   7.0938  
                      
difference of means =    x̅1-x̅2 =    57.3250   -   59.950   =   -2.6250
confidence interval is                       
Interval Lower Limit=   (x̅1-x̅2) - E =    -2.6250   -   7.0938   =   -9.7188
Interval Upper Limit=   (x̅1-x̅2) + E =    -2.6250   +   7.0938   =   4.4688

3)

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.13                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   57.33                  
standard deviation of sample 1,   s1 =    7.91                  
size of sample 1,    n1=   8                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   59.95                  
standard deviation of sample 2,   s2 =    9.65                  
size of sample 2,    n2=   8                  
                          
difference in sample means =    x̅1-x̅2 =    57.3250   -   60.0   =   -2.62  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    8.8195                  
std error , SE =    Sp*√(1/n1+1/n2) =    4.4098                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   -2.6250   -   0   ) /    4.41   =   -0.595
                          
Degree of freedom, DF=   n1+n2-2 =    14                  
t-critical value , t* =        1.6087   (excel formula =t.inv(α/2,df)              
Decision:   | t-stat | < | critical value |, so, Do not Reject Ho                      
p-value =        0.561162   (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value>α , Do not reject null hypothesis                      
                          
There is not enough evidence to prove that male and female mice differe.

Please revert back in case of any doubt.

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