Question

The normal scores test for non-normality was conducted on the data in question 2 above. The
correlation between the ordered sample and the normal scores was found to be r = 0.9454. By
hand conduct the normal scores test at α = 0.10 [i.e. state Ho and Ha, set up the appropriate
rejection region and make your decision].

Answer 1

SOLUTION:

4) Assumed value is sample size (n)=15,
because not given in the data,
Given that,
value of r =0.9454
number (n)=15
null, Ho: p =0
alternate, Hl: p!=0
level of significance, a = 0.1
from standard normal table, two tailed t a/2 =1.771
since our test is two-tailed
reject Ho, if to < -1.771 OR if to > 1.771
we use test statistic (t) = r / sqrt(1,^2/(n-2))
to=0.9454/(sgrt( ( 1-0.9454,2 )/115-21)
to =10.459
Ito 1 =10.459
critical value
the value of It al at los 0.1% is 1.771
we got Ito' =10.459 &Ital =1.771
make decision
hence value of Ito I > I t al and here we reject Ho
ANSWERS
null, Ho: p =0
alternate, Hl: p!=0
test statistic: 10.459
critical value: -1.771 , 1.771
decision: reject Ho
we have enough evidence to support the claim that the relationship (correlation) between the ordered sample and normal scores.