Question

Solve problem #1 by hand using the 7-step method. To receive full credit, show all hand calculations.

1. Rats who grow up in an uncrowded environment (1 rat per cage) live a mean number of 38 months.
A psychologist wants to know the effect of crowding on life expectancy of rats. A group of 31 rats
are raised in crowded conditions (3 rats per cage) and their age at the time of their natural death is recorded.

The data are: 43 39 38 37 36 39 35 36 40 40 41 36 38 39 41 39
            41 40 37 35 38 38 39 37 43 39 37 38 40 39 32

a) Use the 7 step method at = .05 to test the psychologist's prediction.

Answer 1

Step 1: Null hypothesis states that the mean life of rats who grow up in croweded conditions is 38 months

Ho:

Step 2: Alternative hypothesis states taht the mean life of rats is not 38 months.

Ha:

Step 3: Level of significance = 0.05

Step 4: Sample data

sample mean = sum of all terms / no of terms = 1190/ 31 = 38.3871

sample standard deviation = s

data	data-mean	(data - mean)2
43	4.6129	21.27884641
39	0.6129	0.37564641
38	-0.3871	0.14984641
37	-1.3871	1.92404641
36	-2.3871	5.69824641
39	0.6129	0.37564641
35	-3.3871	11.47244641
36	-2.3871	5.69824641
40	1.6129	2.60144641
40	1.6129	2.60144641
41	2.6129	6.82724641
36	-2.3871	5.69824641
38	-0.3871	0.14984641
39	0.6129	0.37564641
41	2.6129	6.82724641
39	0.6129	0.37564641
41	2.6129	6.82724641
40	1.6129	2.60144641
37	-1.3871	1.92404641
35	-3.3871	11.47244641
38	-0.3871	0.14984641
38	-0.3871	0.14984641
39	0.6129	0.37564641
37	-1.3871	1.92404641
43	4.6129	21.27884641
39	0.6129	0.37564641
37	-1.3871	1.92404641
38	-0.3871	0.14984641
40	1.6129	2.60144641
39	0.6129	0.37564641
32	-6.3871	40.79504641

Step 5: Test statistics

Assuming that the data is normally distributed and as the population sd is not given, we will calculate t stat

P value

P value = TDIST (t statistics, df, 2) = TDIST(0.918, 30, 2)= 0.3659

Step 6: Rejection region

df = 31-1 = 30

level of significane = 30

The t-critical values for a two-tailed test, for a significance level of α=0.05

tc = −2.042 and tc = 2.042

Step 7: Conclusion

As the t stat (0.918) does not fall in the rejection area, we fail to reject the Null hypothesis.

Also as the p value (0.3659) is greater than level of significane (0.05), we fail to reject the Null hypothesis.