Question

	%Lysis
	COMP 250	COMP 100	COMP 20
n=1	87.77	81.73	65.8
n=2	81.71	50.45	63.7
n=3	86.64	55.38	59.4
n=4	90.03	82.39	71.67
AVERAGE	87.5	68.4	66.1
STD DEV	4.0	15.6	5.4
SEM	2.5	8.3	3.2

4.Perform a t-test comparing COMP250 to COMP100 and COMP250 to COMP20. A p value when p<0.05 is statistically significant and p value when p>0.05 is not significant. Include all p values and comment on their significance.

5. In your discussion briefly comment on the following:

Which amount (μl) of complement gives the highest and lowest degree of lysis?

Discuss variability between the experiments?

What do you think is the reason for variability in this experiment

Answer 1

4)

t-test comparing COMP250 to COMP100

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   87.50                  
standard deviation of sample 1,   s1 =    4.00                  
size of sample 1,    n1=   4                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   68.40                  
standard deviation of sample 2,   s2 =    15.60                  
size of sample 2,    n2=   4                  
                          
difference in sample means =    x̅1-x̅2 =    87.5000   -   68.4   =   19.10  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    11.3877                  
std error , SE =    Sp*√(1/n1+1/n2) =    8.0523                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   19.1000   -   0   ) /    8.05   =   2.372
                          
Degree of freedom, DF=   n1+n2-2 =    6                  

p-value =        0.0554 (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value>α , Do not reject null hypothesis  

not significant

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

COMP250 to COMP20

Ho :   µ1 - µ2 =   0                  
Ha :   µ1-µ2 ╪   0                  
                          
Level of Significance ,    α =    0.05                  
                          
Sample #1   ---->   sample 1                  
mean of sample 1,    x̅1=   87.50                  
standard deviation of sample 1,   s1 =    4.00                  
size of sample 1,    n1=   4                  
                          
Sample #2   ---->   sample 2                  
mean of sample 2,    x̅2=   66.10                  
standard deviation of sample 2,   s2 =    5.40                  
size of sample 2,    n2=   4                  
                          
difference in sample means =    x̅1-x̅2 =    87.5000   -   66.1   =   21.40  
                          
pooled std dev , Sp=   √([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) =    4.7518                  
std error , SE =    Sp*√(1/n1+1/n2) =    3.3601                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   21.4000   -   0   ) /    3.36   =   6.369
                          
Degree of freedom, DF=   n1+n2-2 =    6                  
  
p-value =        0.000704   (excel function: =T.DIST.2T(t stat,df) )              
Conclusion:     p-value <α , Reject null hypothesis  

SIGNIFICANT

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

5)

COMP250 gives the highest degree of lysis (90.03)

COMP100 gives lowest degree of lysis (50.45)

There is variability between the experiments as shown in the above t test

the result of t test is significant showing the variability

thanks  
you did not provide complete question still i tried so revert back for doubt