Question

HW#9 - Chapter 18 (20pts for turned in on time) Paired t test is when two dependent groups are compared with each other, and to perform a paired t test, a Difference will be calculated first between each pair of the individuals of the two groups, and then a one-sample t test will be conducted on the Difference. 5. (10pts)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 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. You are asked if the number of T cells after 20 days is higher. Baseline 0.04 0.02 0.00 0.02 0.38 0.33 After 20 0.28 0.47 1.30 0.25 1 .22 10.44 0.44 days X Difference 0.24 0.45 1.30 0.23 0.84 0.11 4- Но, и о ни со 0 0 . C b. c- rejection when s = 10.25566667 Ho P20.05 S= 0.2258 t = 0.1366-4866 -0.35 Df=5 0.23 0.35/50.2258 t-0.350 (2.449 +-3.8

Answer 1

Null Hypothesis(H0): $\mu_d$ $\leq$ 0

Alternative Hypothesis(H1): $\mu_d$ > 0 (one tailed: right-tailed test)

($\mu_d$ =Population mean difference)

Sample size, n =6

Sample mean difference, $\bar {d}$ =(0.24+0.45+.....+0.11)/6 =0.5283

Sample std.deviation of differences, s =0.457358

Test statistic, t =($\bar {d}$ - 0)/(s/$\sqrt {n}$​​​​​​) =(0.5283 - 0)/(0.457358/$\sqrt {6}$​​​​​​) =2.829

For one tailed test, for t =2.829, at n-1 =5 degrees of freedom, the p-value =0.0183

Decision criteria: Reject H0 if p-value < significance level.

Let the significance level be 0.05

Conclusion: Since p-value of 0.0183 < 0.05, we reject the null hypothesis (H0) at 5% significance level. Thus, we have a sufficient statistical evidence to claim that the number of T cells after 20 days is higher.