Step 1: Formulate the null hypothesis or Ho. Step 2: Formulate the alternative H
ID: 3313998 • Letter: S
Question
Step 1: Formulate the null hypothesis or Ho.
Step 2: Formulate the alternative Hypothesis or H1
Q2LOHI Frequency of visiting rtz in F17 * Q19GROUP rtz Affiliation Crosstabulation
Q19GROUP rtz Affiliation
Total
1 Students
2 Non- Students
Q2LOHI Frequency of visiting rtz in F17
1 Low
Count
264
217
481
% within Q19GROUP rtz Affiliation
41.5%
55.8%
46.9%
2 High
Count
372
172
544
% within Q19GROUP rtz Affiliation
58.5%
44.2%
53.1%
Total
Count
636
389
1025
% within Q19GROUP rtz Affiliation
100.0%
100.0%
100.0%
Chi-Square Tests
Value
df
Asymptotic Significance (2-sided)
Exact Sig. (2-sided)
Exact Sig. (1-sided)
Pearson Chi-Square
19.748a
1
.000
Continuity Correctionb
19.179
1
.000
Likelihood Ratio
19.771
1
.000
Fisher's Exact Test
.000
.000
Linear-by-Linear Association
19.728
1
.000
N of Valid Cases
1025
a. 0 cells (0.0%) have expected count less than 5. The minimum expected count is 182.55.
b. Computed only for a 2x2 table
Q2LOHI Frequency of visiting rtz in F17 * Q19GROUP rtz Affiliation Crosstabulation
Q19GROUP rtz Affiliation
Total
1 Students
2 Non- Students
Q2LOHI Frequency of visiting rtz in F17
1 Low
Count
264
217
481
% within Q19GROUP rtz Affiliation
41.5%
55.8%
46.9%
2 High
Count
372
172
544
% within Q19GROUP rtz Affiliation
58.5%
44.2%
53.1%
Total
Count
636
389
1025
% within Q19GROUP rtz Affiliation
100.0%
100.0%
100.0%
Explanation / Answer
Step - 1: Null Hypothesis: Affiliation and Frequency of Visiting rtz in F17 are independent.
Step - 2: Alternative Hypothesis: Affiliation and Frequency of Visiting rtz in F17 are not independent.
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