Respuesta :
Answer:
No
Step-by-step explanation:
The given data is
Baseline After 20 days Difference d²
0.04 0.08 0.04 0.0016
0.02 0 -0.02 0.0004
0 1.1 1.1 1.21
0.02 0 -0.02 0.0004
0.37 1.12 0.75 0.5625
0.43 0.24 -0.19 0.0361
0.88 2.54 1.66 1.8106
d`= ∑di/ n = 1.66/ 6=0.2767
S²d= ∑(di-d)²/ n-1 = 1/n-1 [ ∑di² -(∑di)²/n]
= 1/5 [ 1.8106 - (1.66/ 6)²]
= 0.3468
Sd= √ 0.3468= 0.5889
t= d`/ sd/√n
t= 0.2767/0.5889 /√6
t= 0.2767/0.2404
t= 1.1509
t (∝,n-1) = t (0.025,5)= 2.571
The critical region is t > t(0.025,5)
The calculated value of t = 1.1509 is less than t (0.025,5)=2.571 and falls within the critical region therefore data does not give evidence at the 5 % level that the mean count of T cells is higher after 20 days on blinatumomab.
Using the paired test statistic, the test statistic is less than the critical value Hence, we conclude that mean count does not change after 20 days.
Taking the mean difference of the cell counts :
- -0.04, 0.02, -1.1, 0.02, -0.75, 0.19, -1.66
Using a calculator :
- Mean difference, d = -0.474
- Standard deviation of difference, Sd = 0.706
- Sample size, n = 7
The Test statistic for a paired test :
- Test statistic = [d ÷ (Sd/√n)]
Hence,
Test statistic = [-0.474 ÷ (0.706/√7)]
Test statistic = -1.776
Decison Region :
Reject [tex] H_{0} [/tex] if |Test statistic | > Tcritical
Tcritical(df = 6, 0.05) = 1.943
Since ; 1.776 < 1.943 ; we fail to reject [tex] H_{0}[/tex] and conclude that mean count does not change after 20 days.
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