Question

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	0.32	0.42
After 20 days	0.18	0.37	1.2	0.15	1.02	0
Difference	0.14	0.35	1.2	0.13	0.7	-0.42

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

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_d< 0

Alternative hypothesis: u_d > 0

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).

s = sqrt [ (\sum (d_i - d)² / (n - 1) ]

s = 0.55324

SE = s / sqrt(n)

S.E = 0.2259

DF = n - 1 = 6 -1

D.F = 5

t = [ (x₁ - x₂) - D ] / SE

t = 1.55

where d_i is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 5 degrees of freedom is more extreme than 1.55; that is, less than - 1.55 or greater than 1.55.

Thus, the P-value = 0.182

Interpret results. Since the P-value (0.182) is greater than the significance level (0.05), we have to accept the null hypothesis.

Do not reject H₀. We do not have sufficient evidence at the 5% level that the mean count of T cells is higher after 20 days on blinatumomab.