Is population variance of the two data equal?

Question

A professor has two classes in psychology: a morning class of 9 students, and an afternoon class of 12 students. On final exams at the same time for all students, the classes received the grades shown in the table below. Can one conclude at .05 significance level, that the average performance of morning class is worse than average performance of afternoon class? Morning class: 73,87,79,75,82,66,95,75,70 Afternoon class:86,81,84,88,90,85,84,92,83,91,53,84

Answer 1

I had gave you some hints on this post yesterday, why did you delete it and post the question again?? Your title mentions population variances, but the question asks about the average performance between the 2 groups. I told you to use a t-test, and that the population variance cannot be found based on these 2 samples.

You can do this in a graphing calculator using a 2-SampleTtest which will yield the same results that I am about get using R. I'm assuming your more familiar with the graphing calculator, but I want to see if you can do that by yourself. You can just create 2 lists and use them in the 2-SampleTtest, its very straightforward

I will demonstrate how to do this problem in R for you and other viewers who may find it useful.. Before proceeding with the t-test, we can evaluate the sample variances (not population) of the two groups.

 Morning = c(73,87,79,75,82,66,95,75,70)
 Afternoon =c(86,81,84,88,90,85,84,92,83,91,53,84)
 var.test(Morning,Afternoon)

F test to compare two variances
F = 0.7716, num df = 8, denom df = 11, p-value = 0.7293
alternative hypothesis: true ratio of variances is not equal to 1
 95 percent confidence interval:
  0.2106011 3.2742363
  sample estimates:
 ratio of variances 
     0.7716045

All you need to know from this output is that p-value = .7293, thus the two variances are equal, since the p-value is much larger than .05

Alternatively you can even compare the F-value of to a F-critical value The syntax is $qf(p, df.num, df.den)$

qf(0.95, 8, 11)
= 2.948

Since $.7716 < 2.948$, the two variances are equal.

Now onto the actual t-test:

 t.test(Morning, Afternoon, var.equal = TRUE, paired = FALSE, alternative = "less")

Where var.equal is True because I just evaluated that the sample variances are equal, the alternative is less because you want to test if morning < afternoon and paired is false because the data is 2 independent samples, they are not related(paired).

 Two Sample t-test
 t = -1.2709, df = 19, p-value = 0.1095
 alternative hypothesis: true difference in means is less than 0
 95 percent confidence interval:
 -Inf 1.952838
 sample estimates:
 mean of x mean of y 
 78.00000  83.41667

Thus, since the p-value = .1095 and is greater than .05, we can conclude that the average performance of morning class is the same as the average performance of afternoon class. Or we can say that we have insufficient evidence to conclude that the morning class performs worse than the afternoon class.

As I stated at the beginning, you can achieve this task using a 2-SampleTtest, in either the graphing calculator or other statistical software. You should leave this question up (and not delete it) because I supplied a lot of useful information to you and other viewers.

I hope this helped.