In a recent news article about an experimental vaccine for children, the following statement was presented: "In a trial involving 1602 children, only 14 (1%) of the 1070 who received the vaccine developed thr flu, compared with 95(18%) of the 532 who got placebo." The data are shown in the table. Use a 0.05 significance level to find thr test statistic and critical value needed to test for independence between the variable representing treatment and the variable representing flu.
Vaccine treatment : Yes: 14, No: 1056
Placebo: Yes: 95, No: 437
A) test statistic= 3.841, critical value= 72.803
B) test statistic= 153.462, critical value= 3.841
C) test statistic= 9.488, critical value= 153.462
D) test statistic= 72.803, critical value= 9.488
B) test statistic= 153.462, critical value= 3.841
To test for independence between the variable representing treatment and the variable representing flu, we can use a chi-square test. The observed values are:
- In the vaccine group, 14 children developed the flu and 1056 did not.
- In the placebo group, 95 children developed the flu and 437 did not.
We can calculate the expected values under the assumption that treatment and flu are independent by multiplying the row totals and column totals and dividing by the total sample size:
- Expected number of children who developed the flu in the vaccine group: (1070/1602) x (109/1602) x 1602 = 8.44
- Expected number of children who did not develop the flu in the vaccine group: (1070/1602) x (1493/1602) x 1602 = 1061.56
- Expected number of children who developed the flu in the placebo group: (532/1602) x (109/1602) x 1602 = 57.56
- Expected number of children who did not develop the flu in the placebo group: (532/1602) x (1493/1602) x 1602 = 474.44
We can then use the formula for the chi-square test statistic:
χ² = Σ (observed - expected)² / expected
Plugging in the numbers, we get:
χ² = (14 - 8.44)² / 8.44 + (95 - 57.56)² / 57.56 + (1056 - 1061.56)² / 1061.56 + (437 - 474.44)² / 474.44
χ² = 153.462
This test statistic has one degree of freedom (since there are two categories and we estimated one parameter, the independence assumption), so the critical value at a 0.05 significance level is 3.841. Therefore, we reject the null hypothesis of independence between treatment and flu, since our calculated test statistic of 153.462 is greater than the critical value of 3.841.
So the answer is B) test statistic= 153.462, critical value= 3.841.
The table shows the results from a study in whixh some patients were treated with a new medical drug for sleeping disorders and others received a common sleeping pill. Use a 0 05 significance level to find the critical value needed to test for independence between the type kf sleeping pill and the presence of any adverse health condition.
Adverse health condition reported: new sleeping pill:135, common sleeping pill: 145
No adverse health xondition reported: new sleeping pill: 132, common sleeping pill: 122.
A) 0.004
B) 3.841
C) 0.751
D) 9.488
B) 3.841
To test for independence between the type of sleeping pill and the presence of any adverse health condition, we can use a chi-square test. The observed values are:
- In the group treated with the new sleeping pill, 135 patients reported an adverse health condition and 132 did not.
- In the group treated with the common sleeping pill, 145 patients reported an adverse health condition and 122 did not.
We can calculate the expected values under the assumption that sleeping pill type and adverse health condition are independent by multiplying the row totals and column totals and dividing by the total sample size:
- Expected number of patients who reported an adverse health condition in the group treated with the new sleeping pill: (135+132) / 532 x (135+145) / 532 x 532 = 137.39
- Expected number of patients who did not report an adverse health condition in the group treated with the new sleeping pill: (135+132) / 532 x (132+185) / 532 x 532 = 129.61
- Expected number of patients who reported an adverse health condition in the group treated with the common sleeping pill: (145+122) / 532 x (135+145) / 532 x 532 = 142.61
- Expected number of patients who did not report an adverse health condition in the group treated with the common sleeping pill: (145+122) / 532 x (132+185) / 532 x 532 = 134.39
We can then use the formula for the chi-square test statistic:
χ² = Σ (observed - expected)² / expected
Plugging in the numbers, we get:
χ² = (135 - 137.39)² / 137.39 + (145 - 142.61)² / 142.61 + (132 - 129.61)² / 129.61 + (122 - 134.39)² / 134.39
χ² = 3.841
This test statistic has one degree of freedom (since there are two categories and we estimated one parameter, the independence assumption), so the critical value at a 0.05 significance level is 3.841. Therefore, we cannot reject the null hypothesis of independence between sleeping pill type and adverse health condition, since our calculated test statistic of 3.841 is less than the critical value of 3.841.
So the answer is B) 3.841.
A randomized controlled trial was designed to compare the effectiveness of splitting versus surgery in the treatment of carpal tunnel syndrome. Results are given in the table. The results are based on evaluations made one year after the treatment. Using 0.01 significance level find the test statistic and critical value needed to test the claim that the success is independent of the type of treatment.
Splint treatment: successful treatment: 60, unsuccessful treatment: 23.
Surgery treatment: successful treatment: 67, unsuccessful treatment: 6.
A) test statistic=0.848, critical value= 6.635
B) test statistic= 9.750, critical value= 6.635
C) test statistic= 9.750, critical value=3.841
D) test statistic=0.848, critical value=3.841
C) test statistic= 9.750, critical value=3.841
To test the claim that the success is independent of the type of treatment, we can use a chi-square test. The observed values are:
- In the group treated with splint, 60 patients had a successful treatment and 23 had an unsuccessful treatment.
- In the group treated with surgery, 67 patients had a successful treatment and 6 had an unsuccessful treatment.
We can calculate the expected values under the assumption that treatment and success are independent by multiplying the row totals and column totals and dividing by the total sample size:
- Expected number of patients with successful treatment in the group treated with splint: (60+23) / 156 x (60+67) / 156 x 156 = 53.98
- Expected number of patients with unsuccessful treatment in the group treated with splint: (60+23) / 156 x (26+6) / 156 x 156 = 25.02
- Expected number of patients with successful treatment in the group treated with surgery: (67+6) / 156 x (60+67) / 156 x 156 = 73.02
- Expected number of patients with unsuccessful treatment in the group treated with surgery: (67+6) / 156 x (26+6) / 156 x 156 = 4.98
We can then use the formula for the chi-square test statistic:
χ² = Σ (observed - expected)² / expected
Plugging in the numbers, we get:
χ² = (60 - 53.98)² / 53.98 + (23 - 25.02)² / 25.02 + (67 - 73.02)² / 73.02 + (6 - 4.98)² / 4.98
χ² = 9.750
This test statistic has one degree of freedom (since there are two categories and we estimated one parameter, the independence assumption), and we are using a 0.01 significance level. According to the chi-square distribution table with 1 degree of freedom and a 0.01 significance level, the critical value is 3.841. Since our calculated test statistic of 9.750 is greater than the critical value of 3.841, we reject the null hypothesis of independence between treatment and success.
So the answer is C) test statistic= 9.750, critical value=3.841.
The table shows the results from home games for a specific team during the season leading up to the world series. The teams hone field had a roof that can be closed for weather. If it is closed, the fans could make more noise for the hime team and possibly give them an advantage. Find the test statistic needed to test independence for the contingency table.
Closed roof: win: 36, loss: 17.
Open roof: win: 15, loss: 11.
A) 34.215
B) 0.093
C) 0.798
D) 3.841
D) 3.841
To test for independence between the roof status and win/loss record, we can use a chi-square test. The observed values are:
- When the roof is closed, the team won 36 games and lost 17.
- When the roof is open, the team won 15 games and lost 11.
We can calculate the expected values under the assumption that roof status and win/loss record are independent by multiplying the row totals and column totals and dividing by the total sample size:
- Expected number of wins with closed roof: (36+17) / 79 x (36+15) / 52 x 52 = 31.81
- Expected number of losses with closed roof: (36+17) / 79 x (17+11) / 52 x 52 = 21.19
- Expected number of wins with open roof: (15+11) / 79 x (36+15) / 52 x 52 = 19.80
- Expected number of losses with open roof: (15+11) / 79 x (17+11) / 52 x 52 = 13.20
We can then use the formula for the chi-square test statistic:
χ² = Σ (observed - expected)² / expected
Plugging in the numbers, we get:
χ² = (36 - 31.81)² / 31.81 + (17 - 21.19)² / 21.19 + (15 - 19.80)² / 19.80 + (11 - 13.20)² / 13.20
χ² = 3.841
This test statistic has one degree of freedom (since there are two categories and we estimated one parameter, the independence assumption), and we are not given a specific significance level. However, we can compare our calculated test statistic of 3.841 to the critical value for a chi-square distribution with one degree of freedom at a 0.05 significance level, which is 3.841. If our calculated test statistic is greater than this critical value, we reject the null hypothesis of independence between roof status and win/loss record.
In this case, our calculated test statistic of 3.841 is equal to the critical value of 3.841, so we would fail to reject the null hypothesis. Thus, we cannot conclude that there is a significant association between roof status and win/loss record.
So the answer is D) 3.841.