Answer to Question #155341 in Statistics and Probability for Haider

Question

Association of attributes :

The patients reactions to the treatment are recorded in the following table. Test the hypothesis that the drug is no better than sugar pills foru cring colds. Let a =0.05.

Category

Drug, Sugar

Helped

52, 44

Harmed

10, 12

No Effect

20, 26

Answer 1

The expected frequencies are

$$\frac{row\ total*column\ total}{grand\ total}$$

$$Drug.helped= \frac{(52+44)(52+10+20)}{52+10+20+44+12+26}= 48$$

$$Drug.harmed= \frac{(10+12)(52+10+20)}{52+10+20+44+12+26}= 11$$

$$Drug.no.effect= \frac{(20+26)(52+10+20)}{52+10+20+44+12+26}= 23$$

$$Sugar.helped= \frac{(52+44)(44+12+26)}{52+10+20+44+12+26}= 48$$

$$Sugar.harmed= \frac{(10+12)(44+12+26)}{52+10+20+44+12+26}= 11$$

$$Sugar.no.effect= \frac{(20+26)(44+12+26)}{52+10+20+44+12+26}= 23$$

The degrees of freedom for the chi square coefficient is

$$(3-1)(2-1)=2$$

The chi square test statistic is given by

$$X^2=\sum\frac{(observed-expected)^2}{expected}$$

$$X^2=\frac{(52-48)^2}{48}+\frac{(10-11)^2}{11}+\frac{(20-23)^2}{23}+ \frac{(44-48)^2}{48}+\frac{(12-11)^2}{11}+\frac{(26-23)^2}{23}$$

$$X^2=1.6311$$

The p value corresponding to the chi square score of $1.6311$ and a level of significance of$0.05$ is $0.4424$ . We therefore conclude that there is no significant evidence to conclude that there is a difference in sugar and the drug.