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e. Is the requirement that the expected frequency in each cell is 5 or greater satisfied? Explain. O 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. O 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. 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 satisfied.

e. Is the requirement that the expected frequency in each cell is 5 or greater satisfied? Explain. O 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. O 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. 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 satisfied.

MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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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
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.
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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