Chi-Squared Assignment in Statistical Psychology

This document provides a Chi-Squared Assignment in Statistical Psychology. It includes information on hypotheses, alpha level, degree of freedom, population size, chi-square calculation, p-value, and conclusions. The assignment also covers the test of independence and its results. The document is suitable for students studying statistical psychology.

STATISTICAL Psychology
Chi-Squared Assignment
[DATE]
Name, Id
Section

Question 1
(a) Belief in Free Will is the variable that would be tested in this Goodness of Fit test.
(b) Hypotheses and alpha level
Null hypothesis H0: The given data of budding psychologists from SFSU is consistent with
the general population.
Alternative hypothesis Ha: The given data of budding psychologists from SFSU is
inconsistent with the general population.
Level of significance (alpha) = 5%
(c) The degree of freedom for the chi -square statistic ( df )=k −1=2−1=1
(d) Population size would be same as total number of budding psychologists at SFSU which
means 1038.
(e) Chi square statistic is computed as shown below.
Observed frequency
Total
Belief in Free Will
Yes 618
No 420
Total 1038
Expected frequency
E(i) =N*P
Where,
N = 1038
P = Proportion of people who believed in free will = 48%
Total
Belief in Free Will
Yes 1038*0.48 = 498.24
No 1038*0.52 =539.76
Total 1038
Chi square calculation
1

X2 =∑ ( Observed −Expected ) 2
Expected
X2 = ( 618−498.24 )2
498.24 + ( 420−539.76 )2
539.76 =55.36
X2 =55.36
(f) The p value for (df=1, X2 =55.36) comes out to be 0.00001. The p value is clearly lower
than level of significance (alpha) which indicates that null hypothesis would be rejected
and alternative hypothesis would be accepted. Hence, it can be concluded that the given
data suggested that responses from budding psychologists at SFSU are different from
general population.
Question 2
(a) Belief in free will and belief in the paranormal amongst the budding psychologist at
SFSU would be considered for test of independence.
(b) Hypotheses and alpha level
Null hypothesis H0: Pij=Pi∗P j ∀ ( i , j ) = A , B , a ,b
Alternative hypothesis Ha: Pij ≠ Pi∗P j for at least one ( i, j ) =A , B , a , b
Where,
PA: The proportion of students who belief in paranormal
PB : The proportion of students who do not belief in paranormal
Pa : The proportion of students who belief in free will
Pb: The proportion of students who do not belief in free will
Level of significance (alpha) = 5%
(c) The degree of freedom for the chi -square statistic
2

( df )= ( c−1 ) ( r−1 )= ( 2−1 )∗(2−1)=1
(d) Population size would be equal as number of budding psychologists at SFSU which
means 1038.
(e) Test of independence through SPSS
3

(g) Chi square statistic is computed as shown below.
Observed frequency
Expected frequency
Belief in Paranormal
Yes No Total
Yes (744*618)/1038 = 443 (294*618)/1038 =175 618
Belief in
Free Will No (744*420)/1038 =301 (294*420)/1038 =119 420
Total 744 294 1038
Chi square calculation
X2 =∑ ( Observed −Expected ) 2
Expected
X2 = ( 446−443 )2
443 + ( 172−175 )2
175 + ( 298−301 )2
301 + ( 122−119 )2
119 =0.1773
X2 =0.1773
4

(h) The part of output generated from SPSS (symmetric measures) is highlighted below.
It can be seen from the above shown table that the value of Phi and Cramer’s V is 0.013
which is lower than 0.1 and hence, it can be concluded that effect size should be considered
as small.
(i) The p value comes out to be 0.6696 which is higher than level of significance (alpha=5%)
and hence, null hypothesis would not be rejected and alternative hypothesis would not be
accepted. Therefore, it can be concluded that responses from budding psychologist at
SFSU do not show any significant relationship between belief in free will and belief in
paranormal.
Question 3
Poster
5

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