Question

2Upts Tor turned in on time 4. (16pt JThere is evidence that cytotonic T lymphocytes (T cels) participate in controling tumor growth and that they can be harnessed to use the body's immune system to treat cancer. One study investigated the use of T cell-engaging antibody, blinatumomab, to recruilt T cells to control tumor growth. Oe important factor in the research is to see whether T cell will be growing by the use of blinatumomab. 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 You are asked if the number of T cells after 20 days is higher than the baseline. Baseline 0.0 After 20 0.28 days 0.02 0.47 0.00 130 0.020.38 0.25 1.22 0.33 0.44 (1) (2pts) There are two groups we need to compare for the study: Baseline T cell level and After 20 days T cell level for patients. Are those two groups dependent or independent? Based on your answer, what inference procedure should we apply for this research (2) (12)Please perform the inference you decided in (1), and make sure to fo ow the 5-step procedure for any hypothesis test. 3) (2pts)Based on your conclusion in (21. what ind of error could you make? Explain the type of error using the context words for this research

Answer 1

The paired sample t-test, sometimes called the dependent sample t-test, is a statistical procedure used to determine whether the mean difference between two sets of observations is zero.

we will set up the null hypothesis that

$H_{0} : \mu _{1} = \mu _{2}$ Or The two data sets are independent.

$H_{1}:\mu _{1}\geq\mu _{2}$ Or The two data sets are dependent.

under the H₀ the test statistics is

$t=\frac{|\bar{d}|}{S.E(\bar{d})}\sim t_{n-1}$

where, $\bar{d}=\sum \frac{d_{i}}{n}$ , $S.D=\sqrt{\sum \frac{(d_{i}-\bar{d})^{2}}{n}}$ &   $S.E=\frac{S.D}{\sqrt{n}}$

Baseline (Xi)	After 20 days (Yi)	$d_{i}=(x_{i}-y_{i})$	$(d_{i}-\bar{d})^{2}$
0.04	0.28	-0.24	0.083136111
0.02	0.47	-0.45	0.006136111
0.00	1.30	-1.30	0.595469444
0.02	0.25	-0.23	0.089002778
0.38	1.22	-0.84	0.097136111
0.33	0.44	-0.11	0.175002778
		$\sum$ = -3.17	$\sum$ = 1.045883333

$\bar{d}=\sum \frac{d_{i}}{n}=\frac{-3.17}{6}=-0.53$

$S.D=\sqrt{\sum \frac{(d_{i}-\bar{d})^{2}}{n}}=1.0458$

$S.E=\frac{S.D}{\sqrt{n}} = 0.1704$

$t=\frac{|\bar{d}|}{S.E(\bar{d})} = \frac{0.53}{0.1704} = 3.10$

and corresponding p.value = 0.01328

Since calculated p.value = 0.01328 < 0.05 we reject null hypothesis and accept alternate hypothesis.