Question

Independent-samples t-test

A team of investigators has good reason to believe that a small dosage of a certain drug changes the speed with which people can make decisions. They decide to try to confirm this by carrying out an experiment in which the decision times of 14 people who have taken the drug are compared with those of a control group of 14 other people who have performed the task under a placebo condition. The experimenters expect that the average decision time of the experimental group will differ from that of the placebo group. The results are shown the following Table
Drug group
Case: 1 2 3 4 5 6 7 8 9 10 11 12 13 14
Time:390 494 386 323 660 406 345 425 421 407 386 550 470 393

Placibo Group
Case: 15 16 17 18 19 20 21 22 23 24 25 26 27 28
Time: 446 749 599 460 390 477 556 440 471 501 492 392 578 398

Answer 1

Conditions

Independent-samples t-test

A team of investigators has good reason to believe that a small dosage of a certain drug changes the speed with which people can make decisions. They decide to try to confirm this by carrying out an experiment in which the decision times of 14 people who have taken the drug are compared with those of a control group of 14 other people who have performed the task under a placebo condition. The experimenters expect that the average decision time of the experimental group will differ from that of the placebo group. The results are shown the following Table

Drug group

Case: 1 2 3 4 5 6 7 8 9 10 11 12 13 14

Time: 390 494 386 323 660 406 345 425 421 407 386 550 470 393

Placibo Group

Case: 15 16 17 18 19 20 21 22 23 24 25 26 27 28

Time: 446 749 599 460 390 477 556 440 471 501 492 392 578 398

Solution

For this test, the null hypothesis is that the means of samples are equal:

$$H _ {0}: M _ {1} = M _ {2}$$ $$H _ {\alpha}: M _ {1} \neq M _ {2}$$ $$t = \frac {\bar {X} _ {1} - \bar {X} _ {2}}{S _ {X _ {1} X _ {2}} \sqrt {\frac {1}{n _ {1}} + \frac {1}{n _ {2}}}},$$ $$S _ {X _ {1} X _ {2}} = \sqrt {\frac {1}{2} \left(S _ {X _ {1}} ^ {2} + S _ {X _ {2}} ^ {2}\right)}$$ $$S _ {X _ {1}} ^ {2} = \frac {\sum_ {i = 1} ^ {6} \left(X _ {1} - \bar {X} _ {1}\right) ^ {2}}{n}$$ $$S _ {X _ {2}} ^ {2} = \frac {\sum_ {i = 1} ^ {6} \left(X _ {2} - \bar {X} _ {2}\right) ^ {2}}{n}$$

For this example:

$$t = 1.817424$$

The degrees of freedom:

$$k = 14 + 14 - 2 = 26$$

For these degrees of freedom the t-criteria value is:

2.059 – for p=0.95

We can make a conclusion, that with probability 95% there is no difference between 2 groups. H0 is approved