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 0.38 0.31
After 20 days 0.18 0.37 1.1 0 0 0.34
Difference 0.14 0.35 1.1 -0.02 -0.38 0.03
Do the data give evidence at the 2% level that the mean count of T cells is higher after 20 days on blinatumomab? The test statistic is t = (±0.001)?

Answer 1

Answer:

1) First of all we have to state the null hypothesis and also an alternative hypothesis.

Null hypothesis: ud< 0

Alternative hypothesis: ud > 0

It is important to note that these hypotheses are constituting a one-tailed test.

2) Second thing is to formulate an analysis plan. And for that, the significance level must be set to 0.05. And by Utilizing sample data, we will conduct a matched-pairs t-test of the null hypothesis.

3) Thirs thing is to analyze sample data. By utilizng the sample data, we calculate the following:

• standard deviation of the differences (s)
• standard error (SE) of the mean difference
• degrees of freedom (DF)
• t statistic test statistic (t).

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

s = 0.55324

Standard error = $s / sqrt(n)$

Standard error = 0.2259

Degree of freedom = n - 1 = 6 -1

DF = 5

t statistic test statistic (t). = $[ (x1 - x2) - D ] / SE$

t statistic test statistic (t). = 1.55

Here di shows observed difference for pair i

d shows the mean difference between sample pairs

D shows the hypothesized mean difference between population pairs

n shows the number of pairs.

Hence we now 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; i-e less than - 1.55 or greater than 1.55.

Hence, the P-value = 0.182

Now interpret the results. As we know the P-value (0.182) is > (0.05) the significance level , Null hypothesis is accpeted.

We cannot reject H0. Which means that we are lacking evidence at the 5% level that the mean count of T cells is higher after 20 days on blinatumomab.