Question

You are provided with the following dataset. Come up with a research question and make a prediction (hypothesis). Label each of x and y with an appropriate variable name relevant to your hypothesis. If you want, you can add a new variable to the data such as gender, age, etc, to make your hypothesis interesting however it's not mendatory and it won't lead to additional credits. Use SPSS to perform the following.

Clearly state your hypothesis.

What is the shape of each distribution? Is each group normally distributed? Test for normality.  Is there any outlier in the distribution?

Perform an appropriate analysis using this data to test your hypothesis. Is your hypothesis supported? Report your finding.

X	Y
21	16
21	18
22	20
22	18
23	18
23	22
24	22
24	54
25	24
25	26
26	22
26	28
27	28
27	30

Answer 1

shape of distribution is normal distributed.
each group normally distributed
there is no any outlier in the distribution
Given that,
mean(x)=24
standard deviation , s.d1=2.0755
number(n1)=14
y(mean)=24.7143
standard deviation, s.d2 =9.466
number(n2)=14
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.16
since our test is two-tailed
reject Ho, if to < -2.16 OR if to > 2.16
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =24-24.7143/sqrt((4.3077/14)+(89.60516/14))
to =-0.276
| to | =0.276
critical value
the value of |t α| with min (n1-1, n2-1) i.e 13 d.f is 2.16
we got |to| = 0.27579 & | t α | = 2.16
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 != -0.2758 ) = 0.787
hence value of p0.05 < 0.787,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -0.276
critical value: -2.16 , 2.16
decision: do not reject Ho
p-value: 0.787
we do not have enough evidence to support the claim that difference of means