In: Statistics and Probability
Dr. Shultz’s diet pills are supposed to cause significant weight loss. The table below shows the results of a recent study where 180 individuals took the diet pills and 120 did not. At the 5% significance level, test to determine if losing weight is dependent on taking the diet pills.
(a) Ho and Ha
(b) Test method, test statistic and p-value
(c) Statistical decision and case-specific conclusion
Diet Pills | No Diet Pills | Total | |
No weight loss | 80 | 20 | 100 |
Weight loss | 100 | 100 | 200 |
Total | 180 | 120 | 300 |
Here, we have to use chi square test for independence of two categorical variables.
Null hypothesis: H0: Losing weight is not dependent on taking the diet pills.
Alternative hypothesis: Ha: Losing weight is dependent on taking the diet pills.
We assume/given level of significance = α = 0.05
Test statistic formula is given as below:
Chi square = ∑[(O – E)^2/E]
Where, O is observed frequencies and E is expected frequencies.
E = row total * column total / Grand total
We are given
Number of rows = r = 2
Number of columns = c = 2
Degrees of freedom = df = (r – 1)*(c – 1) = 1*1 = 1
α = 0.05
Critical value = 3.841459
(by using Chi square table or excel)
Calculation tables for test statistic are given as below:
Observed Frequencies |
|||
Pills |
|||
Weight loss |
Diet |
No Diet |
Total |
NO |
80 |
20 |
100 |
Yes |
100 |
100 |
200 |
Total |
180 |
120 |
300 |
Expected Frequencies |
|||
Pills |
|||
Weight loss |
Diet |
No Diet |
Total |
NO |
60 |
40 |
100 |
Yes |
120 |
80 |
200 |
Total |
180 |
120 |
300 |
Calculations |
|
(O - E) |
|
20 |
-20 |
-20 |
20 |
(O - E)^2/E |
|
6.666667 |
10 |
3.333333 |
5 |
Test Statistic = Chi square = ∑[(O – E)^2/E] = 25
χ2 statistic = 25.00
P-value = 0.0000
(By using Chi square table or excel)
P-value < α = 0.05
So, we reject the null hypothesis
There is sufficient evidence to conclude that losing weight is dependent on taking the diet pills.