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 1 Likelihood Ratio 7.089 1 .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.

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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: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
a. For the test, how many degrees of freedom does the distribution of the test statistic, x?, have?
O df= 1
O df= 2
O df= 4
O df= 3
Transcribed Image Text:a. For the test, how many degrees of freedom does the distribution of the test statistic, x?, have? O df= 1 O df= 2 O df= 4 O df= 3
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