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.
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 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
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.
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
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 determine the critical value for the test, we need to use the chi-square distribution table with (r-1) × (c-1) degrees of freedom, where r is the number of rows and c is the number of columns in the table.
In this case, we have 2 rows (adverse effects and no effect) and 3 columns (placebo, 1332 mg of drug, and 1998 mg of drug). So, the degrees of freedom would be (2-1) × (3-1) = 1 × 2 = 2.
Referencing the chi-square distribution table with 2 degrees of freedom at a significance level of 0.01, we find the critical value to be 9.210 (option B).
To test the claim that experiencing an adverse reaction in the digestive system is independent of the treatment group, we will compare the calculated test statistic to the critical value.
Since the test statistic is 42.568 and the critical value is 9.210, the test statistic is greater than the critical value.
Therefore, we reject the null hypothesis and conclude that there is evidence to support the claim that experiencing an adverse reaction in the digestive system is not independent of the treatment group.
To test the claim that experiencing an adverse reaction in the digestive system is independent of the treatment group, we will perform a chi-square test of independence.
Step 1: State the hypotheses:
Null hypothesis (H0): There is no association between treatment group and experiencing an adverse reaction in the digestive system.
Alternative hypothesis (HA): There is an association between treatment group and experiencing an adverse reaction in the digestive system.
Step 2: Determine the significance level and find the critical value:
The significance level is given as 0.01. Since we have a 2x3 contingency table (2 rows for adverse effects and no effects, 3 columns for treatment groups), we need to find the critical value in the chi-square distribution table at (2-1) * (3-1) = 2 degrees of freedom and a significance level of 0.01. Looking up the critical value in the chi-square distribution table, we find that it is 9.210.
Step 3: Calculate the test statistic:
The test statistic for a chi-square test of independence is calculated using the formula:
χ^2 = Σ [(Oij - Eij)^2 / Eij]
Where Oij represents the observed frequency in each cell and Eij represents the expected frequency in each cell under the assumption of independence.
In this case, the observed frequencies are given in the table, and the expected frequencies can be calculated as the row total multiplied by the column total divided by the grand total.
For example, the expected frequency for the first cell (adverse effects on placebo) can be calculated as:
E11 = (344 + 1362) * (344 + 89 + 1362 + 1332 + 1998 + 774) / (344 + 89 + 8 + 1362 + 774 + 71 + 1362)
Calculating the expected frequencies for all cells and substituting into the formula, we can calculate the test statistic as 42.568.
Step 4: Compare the test statistic with the critical value:
Since the test statistic (42.568) is greater than the critical value (9.210), we can reject the null hypothesis.
Step 5: State the initial conclusion:
Based on the test result, there is evidence to suggest that experiencing an adverse reaction in the digestive system is not independent of the treatment group.
Therefore, the correct answer is:
B) 9.210; reject the null hypothesis.