Question

A study examines whether an enriched vs. normal rearing environment changes social behavior in rats. Two rats from each of eight litters are randomly assigned to an enriched vs standard environment. The scores of their social behavior are assessed by a standard rating scale, and are as follows:

Litter #	Enriched Environment	Standard Environment
1	23	24
2	19	26
3	26	19
4	20	18
5	35	30
6	40	36
7	25	20
8	28	21

a) Conduct a matched-pair t test to test the null hypothesis that the mean difference in sociability equals 0.

b)The researcher conducted a matched-pair t test because she anticipated that the scores of rats reared in the same litter are not independent of one another. Test this hypothesis by computing the Pearson correlation between the enriched and standard scores and whether it is significantly different from 0.

Answer 1

Hellow,

a. Calculate matched - Pair t test by using Excel,

t - Statistics = 1.640

And, the null hypothesis that the mean difference in sociability equals 0 then use two tail test with Alpha = 0.05 level of Significance,

There for,

P- Value = 0.145

Conclusion: P-Value (0.145) is Greater than 0.05 then accept null hypothesis. That is, the mean difference in sociability equals 0.

b. Test the hypothesis by computing the Pearson correlation:

Null hypothesis: Ho: r = 0 (No Correlation).

Alternative hypothesis: H1: r is not equal to 0 (Correlation).

Test statistic for Pearson correlation is,

t =   where, r = Pearson correlation,

Pearson correlation = r = 0.762 (Which is calculated by using Excel).

Therefor by using above formula,

t = 0.762 * sqrt(8-2) / sqrt(1 - 0.7622)

= 0.762 * 2.449 / sqrt(1 - 0.580)

= 1.866 / 0.648

t = 2.879.

Then, Calculate Critical value for This statistic with n-2 (i.e. 6) df and alpha = 0.05 by using excel,

Critical Value = 2.44.

Conclussion: Calculated value (2.879) is greater than Critical value (2.44) then we reject the null hypothesis.

There is a correlation between enriched and standard scores