Question

1. A polymer chemist studied the effect of relative humidity (RH) on the weight of molded parts made from a polymer material. The chemist believes that any weight differences greater than 5 will have a impact on the end use of the molded part. Use the data from the study, provided below, to state conclusions about the effect of relative humidity on the average and variability of the weight.

70% RH

22.5, 24.2, 24.5, 29.2, 27.5, 24.5, 26.5, 23.5, 25.5, 27.0, 23.5, 23.5, 25.5, 31.0, 24.0, 27.8, 23.5

20% RH

19.0, 18.0, 14.5, 18.5, 19.0, 19.5, 21.3, 18.3, 18.0, 18.5, 18.8, 17.9, 21.2, 20.6, 17.1, 19.5

Answer 1

	70%rh
Sample Variance,	s₁² =	5.4636
Sample size,	n₁ =	17
	20%rh
Sample Standard deviation,	s₂² =	2.6756
Sample size,	n₂ =	16
	α =	0.05

Hₒ : σ₁² = σ₂²  
H₁ : σ₁² ≠ σ₂²  
Test statistic:  
F = s₁² / s₂² =    2.0420
Degree of freedom:  
df₁ = n₁-1 =    16
df₂ = n₂-1 =    15

P-value :  
P-value = 2*F.DIST.RT(2.042, 16, 15) =    0.1746

p value > 0.05, do not reject Ho

so variances are equal

..................

2 mean t test with equal variances

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.05                  
                          
mean of sample 1,    x̅1=   25.51                  
standard deviation of sample 1,   s1 =    2.34                  
size of sample 1,    n1=   17                  
                          
mean of sample 2,    x̅2=   18.73                  
standard deviation of sample 2,   s2 =    1.64                  
size of sample 2,    n2=   16                  
                          
difference in sample means =    x̅1-x̅2 =    25.5118   -   18.7   =   6.78  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    2.0284                  
std error , SE =    Sp*√(1/n1+1/n2) =    0.7065                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   6.7805   -   0   ) /    0.71   =   9.597
                          
Degree of freedom, DF=   n1+n2-2 =    31                  
  
p-value =        0.000000   (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value <α , Reject null hypothesis