The current rate of success is 70% for 200 students. When the new test is implemented, the rate increases to 85% for the 50 students who pass the test and are allowed into the program. Is this difference significant?
To determine if the difference in success rates is significant, we need to compare the two rates statistically. First, let's calculate the number of students who succeeded in each scenario.
For the current rate of success, 70% of 200 students succeeded. This equals (70/100) * 200 = 140 students.
For the new rate of success, 85% of the 50 students who passed succeeded. This equals (85/100) * 50 = 42.5 students. Since we cannot have a fraction of a student, we will round this down to the nearest whole number. Thus, 42 students succeeded.
Now, we can compare the success rates: 140 students succeeded out of 200 (70%) before the new test, and 42 students succeeded out of 50 (85%) after the new test.
To determine if the difference is significant, we can conduct a hypothesis test using a significance level (usually denoted by alpha). Let's assume a significance level of 0.05 (5%). We will perform a two-proportion z-test to compare the success rates.
The null hypothesis (H0) is that there is no significant difference between the two success rates. The alternative hypothesis (Ha) is that there is a significant difference between the two success rates.
Using a two-proportion z-test, we can calculate the z-test statistic using the following formula:
z = ((p1 - p2) - 0) / sqrt((p1 * (1 - p1) / n1) + (p2 * (1 - p2) / n2))
where p1 and p2 are the success rates, and n1 and n2 are the sample sizes.
Plugging in the values:
p1 = 140/200 = 0.7
p2 = 42/50 = 0.85
n1 = 200
n2 = 50
Calculating the z-test statistic:
z = ((0.7 - 0.85) - 0) / sqrt((0.7 * (1 - 0.7) / 200) + (0.85 * (1 - 0.85) / 50))
Simplifying the calculation:
z = (-0.15) / sqrt(0.00105 + 0.007975)
z ≈ -3.131
Next, we calculate the critical z-value at the 5% significance level. For a two-tailed test, the critical z-value is ±1.96.
Since -3.131 < -1.96, we reject the null hypothesis because the z-test statistic falls in the critical region.
Therefore, we can conclude that the difference in success rates is significant based on the given data at a 5% significance level.