Factor V is a protein involved in the forming of blood clots. The higher the lev
ID: 3338783 • Letter: F
Question
Factor V is a protein involved in the forming of blood clots. The higher the level of factor V, the faster the blood clots. A Blood Transfusions Service is interested in the effects of sterilization of blood plasma because factor V is unstable and may break down during sterilization. The table below gives measured levels of factor V in blood samples from 16 donors. Both pre- and post-sterilization measurements are given for each person. Test whether the mean factor V levels are different post-sterilization. Run the test at a 5% level of significance. Give each of the following to receive full credit: 1) the appropriate null and alternative hypotheses; 2) the appropriate test; 3) the decision rule; 4) the calculation of the test statistic; and 5) your conclusion including a comparison to alpha or the critical value. You MUST show your work to receive full credit. Partial credit is available.
Factor V measurements
Donor number
Pre-Sterilization
Post-Sterilization
1
1023
916
2
1050
1030
3
933
923
4
849
892
5
750
628
6
855
759
7
947
828
8
978
784
9
890
809
10
829
773
Donor number
Pre-Sterilization
Post-Sterilization
1
1023
916
2
1050
1030
3
933
923
4
849
892
5
750
628
6
855
759
7
947
828
8
978
784
9
890
809
10
829
773
Explanation / Answer
Given that,
null, H0: Ud = 0
alternate, H1: Ud <> 0
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.262
since our test is two-tailed
reject Ho, if to < -2.262 OR if to > 2.262
we use Test Statistic
to= d/ (S/n)
where
value of S^2 = [ di^2 – ( di )^2 / n ] / ( n-1 ) )
d = ( Xi-Yi)/n) = 76.2
We have d = 76.2
pooled variance = calculate value of Sd= S^2 = sqrt [ 99392-(762^2/10 ] / 9 = 67.76
to = d/ (S/n) = 3.56
critical Value
the value of |t | with n-1 = 9 d.f is 2.262
we got |t o| = 3.56 & |t | =2.262
make Decision
hence Value of | to | > | t | and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 3.556 ) = 0.0062
hence value of p0.05 > 0.0062,here we reject Ho
ANSWERS
---------------
null, H0: Ud = 0
alternate, H1: Ud <> 0
test statistic: 3.56
critical value: reject Ho, if to < -2.262 OR if to > 2.262
decision: Reject Ho
p-value: 0.0062
we have evidence that both result is differ to each other
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