The table represents a survey of a random sample of 229 residents on whether nam
ID: 2907567 • Letter: T
Question
The table represents a survey of a random sample of 229 residents on whether names of convicted sexual offenders should be publicized.
Test hypothesis of no difference between males and females on the question of whether sex offenders' names should be publicized on their release from jail. Use 0.05 significance level.
1. (Multiple Choice) What are the null and alternative hypotheses?
(a) H0: Dependence; H1: Independence
(b) H0: u1=µ2; H1: µ1?µ2
(c) H0: µ1?µ2 ; H1: u1=µ2
(d) H0: Independence; H1: dependence
2. Calculate the test statistic.
3. (Multiple Choice) Do you accept or reject the null hypothesis?
(a) accept the null hypothesis and conclude there is a relationship btwn the variables
(b) accept the null hypothesis and conclude there is no relation btwn the variables
(c) reject the null hypothesis and conclude there is a relation btwn the variables
(d) reject the null hypothesis and conclude there is no relation btwn the variables
4. What is the critical value?
5. Compute Phi-adjusted.
Male Female Yes 41 85 No 59 44Explanation / Answer
2)
using chi-square method
TS = 14.10323
3)
since TS> critical value (3.84)
hence
we reject the null hypothesis
c) is correct
4) critical value = 3.84
5)
phi-value = sqrt(14.10323/229)
= 0.24816
c1 c2 total r1 41 85 126 r2 59 44 103 total 100 129 229 Oi 41 85 59 44 Ei 55.02183 70.97817 44.97817 58.02183 TS (O--Ei)^2/Ei 3.573342 2.770033 4.371273 3.388584 14.10323 critical value 3.84 p-value 0.000173046 phi 0.248165591 V 0.488914932Related Questions
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