Question 1083205: Please suggest the null and alternative hypothesis for the following problem.
At present, many universities in Australia are adopting the practice of having lecture recordings automatically available to students. A university lecturer is trying to investigate whether having lecture recordings available to students has significantly decreased the proportion of students passing her course. When lecture recordings were not provided to students, the proportion of students that passed her course was 80%. The lecturer takes a random sample of 25 students, when lecture recordings are offered to students, and finds that 11 students have passed the course. Is there significant evidence to support this university lecturer’s claim? Use α = 0.01
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! you are testing to see if having lecture recordings available to students has significantly decreased the proportion of students passing her course.
therefore, the null hypothesis should be that it does not significantly decrease the proportion of students passing her course.
the alternative hypothesis is that it does.
the null hypothesis would be that the mean proportion of students passing her course is .8
the alternate hypothesis is that that the mean proportion of students passing her course is < .8
this is a one tailed study since the only alternative you are looking at is less than, rather than just not equal.
alpha is .01 and will all be on the low side of the distribution curve.
a sample of 25 students was taken.
11 students from this sample passed the course.
11 out of 25 is a proportion of .44
that looks real small, and is more then likely statistically significant.
either this was a particularly bad class or having the lecture materials automatically actually was a detriment to the ability of the students to pass the course.
a quick study was undertaken.
population p is .8
sample p is 11/25 = .44
sample q is 1 - .44 = .56
sample error is sqrt(p*q/n) = sqrt(.44*.56/25) = .0993
z-score is (x-m)/s = (.44-.8)/.0993 = -3.63
sample alpha is 1.4175 * 10^-4 = .00014175
test alpha is .01
.00014175 is well below .01 so there's no doubt that a significantly less percentage of students passed the course after the lecture material was available rather than before based on this sample.
intuitively you think the class would do better, therefore the results of the test contradicted intuition.
it could have been an anomaly.
more tests would be required with bigger sample sizes to validate the results, but based on this test alone, you would have to accept the alternate hypothesis that the number of students who passed the course was indeed lower, and the results couldn't be explained by random variation in different samples of this size of students who passed the course.
in other words, the results of the test were statistically significant.
|
|
|
