
MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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Transcribed Image Text:2. Some scientists believe that the presence of antibody to milk protein lowers the chance of survival after a heart attack.
To explore this belief, a medical team took blood specimens from 335 male heart-attack patients and tested them for the
presence of the antibody.
The patients were then followed to determine whether they lived for 6 months following their heart attack. The results are
presented in Table 2. Suppose we want to answer the following question: "Test, at the 5% level of significance, whether
there is an association between the presence of antibody to milk protein and survival after a heart attack."
Table 2: Survival vs Antibody to Milk Protein contingency table
Survival
Antibody to Milk Protein
Alive
Died
Total
Present
24
52
76
Absent
126
133
259
Total
150
185
335
SPSS produced the information needed to perform the hypothesis test in Table 3.
Table 3: Chi-Square Tests Results
Chi-Square Tests
Asymptotic
Significance
(2-sided)
Exact Sig. (2-
sided)
Exact Sig. (1-
sided)
Value
df
Pearson Chi-Square
6.924
1
.009
Continuity Correction
6.251
.012
Likelihood Ratio
7.089
.008
Fisher's Exact Test
.009
.006
N of Valid Cases
335
a. O cells (0.0%) have expected count less than 5. The minimum expected count is 34.03.
b. Computed only for a 2x2 table

Transcribed Image Text:e. Is the requirement that the expected frequency in each cell is 5 or greater satisfied? Explain.
Statement a.,
under the Chi-Square Test table, states that there are no expected frequencies less than 5. So the
requirement that the expected frequency in each cell is 5 or greater is satisfied.
O Statement b., under the Chi-Square Test table, states that the continuity correction is computed only for a 2X2 table.
So the requirement that the expected frequency in each cell is 5 or greater is not satisfied.
Statement a., under the Chi-Square Test table, states that there are no expected frequencies less than 5. So the
requirement that the expected frequency in each cell is 5 or greater is not satisfied.
Statement b., under the Chi-Square Test table, states that the continuity correction is computed only for a 2X2 table.
So the requirement that the expected frequency in each cell is 5 or greater is satisfied.
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