2.Alcohol affects spatial ability in rats. Researchers have rat run through a maze looking for cheese, first sober, then drunk, and count the number of errors they made.

rat	Sober	Drunk	D(difference)
1	2	4
2	1	4
3	2	4
4	1	3
5	3	5
6	4	6
7	3	5
8	2	4
9	4	5

State your hypotheses:

Find critical value at =0.05

Calculatet

Conclusion.

One answer from tutor:

3. Calculated Value

Variance (s2 ) = SS/ n - 1 = 38/8 = 4.75

Standard deviation (s) = s2 = 4.75 = 2.18

SEM = s2 / n = 4.75/9 = 0.73

t = MD - 0/ SEM = 2 - 0 / 0.73 = 2.74

Thus, t = 2.74

4. Decision/ Conclusion

Since the obtained value of t > the critical value of t, it falls in the region of rejection. Thus, the H0 can be rejected.

It may be concluded that there are significant differences in the number of errors made by the rats before and after alcohol consumption such that there are higher errors made after alcohol.

Second answer from tutor:

t = (xd - d)/[sd/n]

xd =-2

sd=0.5

t = (-2 - 0)/[0.5/9]

t = -2/0.166667

t = -12.0

Decision rule; reject H0 if t < -2.306

Conclusion;

We reject H0because -12.00 < -2.306

Hence there is no enough evidence to show that alcohol affects spatial ability in rats.