In: Statistics and Probability
%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
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