(20.48) There is evidence that cytotoxic T lymphocytes (T cells) participate in controlling tumor growth and that they can be harnessed to use the body's immune system to treat cancer. One study investigated the use of a T cell-engaging antibody, blinatumomab, to recruit T cells to control tumor growth. The data below are T cell counts (1000 per microliter) at baseline (beginning of the study) and after 20 days on blinatumomab for 6 subjects in the study. The difference (after 20 days minus baseline) is the response variable. Baseline 0.04 0.02 0 0.02 0.37 0.43 After 20 days 0.08 0 1.1 0 1.12 0.24 Difference 0.04 -0.02 1.1 -0.02 0.75 -0.19 Do the data give evidence at the 5 % level that the mean count of T cells is higher after 20 days on blinatumomab? The test statistic is t = (±0.001) Yes No

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 :

Using a calculator :

The Test statistic for a paired test :

Hence,

Test statistic = [-0.474 ÷ (0.706/√7)]

Test statistic = -1.776

Decison Region :

Reject if |Test statistic | > Tcritical

Tcritical(df = 6, 0.05) = 1.943

Since ; 1.776 < 1.943 ; we fail to reject and conclude that mean count does not change after 20 days.

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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.

answer: bearing in mind that a rectangle has a perimeter of length + length + width + width.

step-by-step explanation:

p = perimeter

l = length

x=width

p = 2l + 2x

p = 2(l+x)     < ==== perimeter of the base of the box.

.95

step-by-step explanation: