Question

Does watching Barney increase altruism in children? Amanda put nine kids in a room where they watched Barney and another nine kids is a quiet room without television. Then the kids filled out a questionnaire that determined their altruism score (which we can assume to be interval). The scores are:

Barney: 43, 55, 63, 57, 77, 65, 84, 73, 66

                Quiet:     67, 55, 74, 63, 56, 33, 45, 53, 47

What kind of Test it is ?

A: one sample z-test    B: one-sample t-test   C: t-test for the difference between means for two related samples  D: t-test for the difference between means for two independent samples with homogeneity of variance  E: t-test for the difference between means for two independent samples with heterogeneity of variance   F: a one sample z-test for proportions (or a chi-square goodness of fit)   G:  chi-square goodness of fit only (where a one sample z-test of proportions isn’t appropriate) H: a two-sample z-test for the difference between proportions (or a chi-square test of independence) I:  chi-square test of independence only (where a two-sample z-test for the difference between proportions isn’t appropriate.)  J:  simple regression             K: multiple regression L:  one-way ANOVA M: two-way ANOVA N: Mann-Whitney O:  Wilcoxon P: none of the above.

Answer 1

Given that,
mean(x)=64.7778
standard deviation , s.d1=12.357
number(n1)=9
y(mean)=54.7778
standard deviation, s.d2 =12.357
number(n2)=9
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.306
since our test is two-tailed
reject Ho, if to < -2.306 OR if to > 2.306
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =64.7778-54.7778/sqrt((152.69545/9)+(152.69545/9))
to =1.717
| to | =1.717
critical value
the value of |t α| with min (n1-1, n2-1) i.e 8 d.f is 2.306
we got |to| = 1.7167 & | t α | = 2.306
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.7167 ) = 0.124
hence value of p0.05 < 0.124,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 1.717
critical value: -2.306 , 2.306
decision: do not reject Ho
p-value: 0.124
we do not have enough evidence to support the claim that the difference between means for two related samples
option:C