D) 5.991; reject the null hypothesis.
To test the claim of independence, we need to use a chi-square test for independence. The null hypothesis is that there is no association between treatment group and experiencing adverse effects in the digestive system, while the alternative hypothesis is that there is an association between them.
We can calculate the expected frequencies for each cell assuming independence using the formula: (row total x column total) / grand total.
The expected frequencies are:
Placebo: 432.97, 1253.99, 89.04
1332 mg of drug: 982.55, 2841.71, 198.74
1998 mg of drug: 89.48, 259.30, 18.22
We can then calculate the chi-square test statistic using the formula:
χ2 = Σ (observed frequency - expected frequency)2 / expected frequency
χ2 = (344-432.97)2 / 432.97 + (1362-1253.99)2 / 1253.99 + (89-89.04)2 / 89.04 + (89-982.55)2 / 982.55 + (774-2841.71)2 / 2841.71 + (259-89.48)2 / 89.48 + (1332-89.04)2 / 89.04 + (71-198.74)2 / 198.74 + (8-18.22)2 / 18.22
χ2 = 42.568
Using a chi-square distribution table with (3-1) x (3-1) = 4 degrees of freedom and a significance level of 0.01, we find the critical value to be 9.210.
Since the calculated test statistic (42.568) is greater than the critical value (9.210), we reject the null hypothesis and conclude that there is significant evidence to suggest that experiencing an adverse reaction in the digestive system is not independent of the treatment group.
A new drug was designed to help patients continue their abstinence from the use of alcohol. Adverse reactions to the drug have been studied in clinical trials; the table summarizes results for digestive system effects among patients from different treatment groups. A test statistic was found to be 42.568. Use a 0.01 significance level to find the critical value and test the claim that experiencing an adverse reaction in the digestive system is independent of the treatment group. State the initial conclusion.
Adverse effects on the digestive system: placebo: 344, 1332 mg of drug: 89, 1998 mg of drug: 8.
No effect on the digestive system: placebo: 1362, 1332 mg of drug: 774, 1998 mg of drug: 71.
A) 5.991; fail to reject the null hypothesis.
B) 9.210; reject the null hypothesis.
C) 9.210; fail to reject the null hypothesis.
D) 5.991; reject the null hypothesis.
5 answers
A new medical drug used as an aid for those who want to stop smoking is tested. The adverse reaction of nausea has been studied in clinical trials, and the table summarizes results. Find the test statistic needed to test the claim that nausea is independent of whether the subject took a placebo or the new drug.
Nausea: placebo: 10, new medical drug: 145.
No nausea: placebo: 795, new medical drug: 122.
A) 6.635
B) 116.395
C) 97.254
D) 456.473
Nausea: placebo: 10, new medical drug: 145.
No nausea: placebo: 795, new medical drug: 122.
A) 6.635
B) 116.395
C) 97.254
D) 456.473
B) 116.395
To test the claim of independence, we need to use a chi-square test for independence. The null hypothesis is that there is no association between taking the drug and experiencing nausea, while the alternative hypothesis is that there is an association between them.
The expected frequencies for each cell assuming independence are:
Placebo: (10+795) x (10+145+795+122) / (10+145+795+122) = 417.73
New medical drug: (145+122) x (10+145+795+122) / (10+145+795+122) = 797.27
We can then calculate the chi-square test statistic using the formula:
χ2 = Σ (observed frequency - expected frequency)2 / expected frequency
χ2 = (10-417.73)2 / 417.73 + (795-417.73)2 / 417.73 + (145-797.27)2 / 797.27 + (122-797.27)2 / 797.27
χ2 = 116.395
Using a chi-square distribution table with 1 degree of freedom and a significance level of 0.05, we cannot find the exact critical value of the test statistic. However, we know it is greater than 3.84 (the critical value for a chi-square distribution with 1 degree of freedom and a significance level of 0.05), but less than 6.635 (the critical value for a chi-square distribution with 1 degree of freedom and a significance level of 0.01).
Since the calculated test statistic (116.395) falls outside the range of possible critical values, we can safely reject the null hypothesis and conclude that there is significant evidence to suggest that experiencing nausea is not independent of taking the new medical drug.
To test the claim of independence, we need to use a chi-square test for independence. The null hypothesis is that there is no association between taking the drug and experiencing nausea, while the alternative hypothesis is that there is an association between them.
The expected frequencies for each cell assuming independence are:
Placebo: (10+795) x (10+145+795+122) / (10+145+795+122) = 417.73
New medical drug: (145+122) x (10+145+795+122) / (10+145+795+122) = 797.27
We can then calculate the chi-square test statistic using the formula:
χ2 = Σ (observed frequency - expected frequency)2 / expected frequency
χ2 = (10-417.73)2 / 417.73 + (795-417.73)2 / 417.73 + (145-797.27)2 / 797.27 + (122-797.27)2 / 797.27
χ2 = 116.395
Using a chi-square distribution table with 1 degree of freedom and a significance level of 0.05, we cannot find the exact critical value of the test statistic. However, we know it is greater than 3.84 (the critical value for a chi-square distribution with 1 degree of freedom and a significance level of 0.05), but less than 6.635 (the critical value for a chi-square distribution with 1 degree of freedom and a significance level of 0.01).
Since the calculated test statistic (116.395) falls outside the range of possible critical values, we can safely reject the null hypothesis and conclude that there is significant evidence to suggest that experiencing nausea is not independent of taking the new medical drug.
Clinical trials studied adverse reactions to a medical drug used to lower cholesterol in patients. The table summarizes the results for infections in patients from different treatment groups. Use a 0.05 significance level to test the claim that getting an infection is independent of the treatment. Find the test statistic and the critical value. State the conclusion.
Infection: placebo: 27, 10 mg of drug:89, 40 mg of drug: 8, 80 mg of drug: 7.
No infection: placebo:243, 10 mg of drug: 774, 40 mg of drug: 71, 80 mg of drug:87.
A) test statistic =18.254, critical value =7.815
B) test statistic =0.773, critical value =15.507
C) test statistic =18.254, critical value =15.507
D) test statistic =0.773, critical value =7.815
Infection: placebo: 27, 10 mg of drug:89, 40 mg of drug: 8, 80 mg of drug: 7.
No infection: placebo:243, 10 mg of drug: 774, 40 mg of drug: 71, 80 mg of drug:87.
A) test statistic =18.254, critical value =7.815
B) test statistic =0.773, critical value =15.507
C) test statistic =18.254, critical value =15.507
D) test statistic =0.773, critical value =7.815
A) test statistic =18.254, critical value =7.815
To test the claim of independence, we need to use a chi-square test for independence. The null hypothesis is that there is no association between treatment group and getting an infection, while the alternative hypothesis is that there is an association between them.
We can calculate the expected frequencies for each cell assuming independence using the formula: (row total x column total) / grand total.
The expected frequencies are:
Placebo: 56.40, 196.61, 21.06, 16.93
10 mg of drug: 236.89, 826.56, 88.35, 71.20
40 mg of drug: 29.35, 102.17, 10.91, 8.77
80 mg of drug: 31.35, 109.66, 11.68, 9.40
We can then calculate the chi-square test statistic using the formula:
χ2 = Σ (observed frequency - expected frequency)2 / expected frequency
χ2 = (27-56.40)2 / 56.40 + (243-56.40)2 / 56.40 + (89-236.89)2 / 236.89 + (774-826.56)2 / 826.56 + (8-29.35)2 / 29.35 + (71-88.35)2 / 88.35 + (7-31.35)2 / 31.35 + (87-109.66)2 / 109.66
χ2 = 18.254
Using a chi-square distribution table with (4-1) x (2-1) = 3 degrees of freedom and a significance level of 0.05, we find the critical value to be 7.815.
Since the calculated test statistic (18.254) is greater than the critical value (7.815), we reject the null hypothesis and conclude that there is significant evidence to suggest that getting an infection is not independent of the treatment group.
To test the claim of independence, we need to use a chi-square test for independence. The null hypothesis is that there is no association between treatment group and getting an infection, while the alternative hypothesis is that there is an association between them.
We can calculate the expected frequencies for each cell assuming independence using the formula: (row total x column total) / grand total.
The expected frequencies are:
Placebo: 56.40, 196.61, 21.06, 16.93
10 mg of drug: 236.89, 826.56, 88.35, 71.20
40 mg of drug: 29.35, 102.17, 10.91, 8.77
80 mg of drug: 31.35, 109.66, 11.68, 9.40
We can then calculate the chi-square test statistic using the formula:
χ2 = Σ (observed frequency - expected frequency)2 / expected frequency
χ2 = (27-56.40)2 / 56.40 + (243-56.40)2 / 56.40 + (89-236.89)2 / 236.89 + (774-826.56)2 / 826.56 + (8-29.35)2 / 29.35 + (71-88.35)2 / 88.35 + (7-31.35)2 / 31.35 + (87-109.66)2 / 109.66
χ2 = 18.254
Using a chi-square distribution table with (4-1) x (2-1) = 3 degrees of freedom and a significance level of 0.05, we find the critical value to be 7.815.
Since the calculated test statistic (18.254) is greater than the critical value (7.815), we reject the null hypothesis and conclude that there is significant evidence to suggest that getting an infection is not independent of the treatment group.