In: Statistics and Probability
A practice tutorial session was conducted before the exam. The examiner wants to know if the practice tutorial session helped in increasing the percentage of students who passed.
Using 5% critical probability, can the examiner conclude if the test is effective.
Passed |
Failed |
|
Didn’t take practice tutorial |
23 |
62 |
Took practice tutorials |
41 |
11 |
Let p1 and p2 denote the proportions of the students passing who didn't take practice tutorial and those who took practice tutorials respectively.
Let n1 and n2 denote the total number of students who didn't take practice tutorial and those who took practice tutorials respectively.
Here, p1 = 23 / (23 + 62) = 0.2706
p2 = 41 / (41 + 11) = 0.7885
n1 = 23 + 62 = 85
n2 = 41 + 11 = 52
The hypothesis is given as:
H0 : The percentage of students who passed remains the same for those who didn't take practice tutorial and those who took practice tutorials i.e p1 = p2
H1 : The percentage of students who passed who took practice tutorials is greater than those who didn't take practice tutorial i.e p1 < p2
The significance level is given as 0.05
We use two proportion z-test for the hypothesis testing.
The test statistic is Z = (p1 - p2) / SE
where SE = sqrt (p * (1 - p) * ((1 / n1) + (1/n2)))
and p = (p1 * n1 + p2 * n2) / (n1 + n2)
Therefore calculating all the required variables,
p = (0.2706 * 85 + 0.7885 * 52) / (85 + 52) = 0.4672
SE = sqrt (0.4672 * (1 - 0.4672) *((1 / 85) + (1 / 52))) = 0.0878
The test statistic is Z = (0.2706 - 0.7885) / 0.0878 = -5.8986
This is a one-tailed test and p-value is the probability that Z-table < -5.8986.
Using the standard normal table, P (Z < -5.8986) = 0
Since the p-value (0) is less than the significance level (0.05), we reject the null hypothesis.
Therefore, there is significant evidence that the percentage of students who passed by taking practice tutorial is greater than those who didn't take practice tutorials.