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Statistics : Hypothesis Testing, Variance, Confidence Intervals and Levels of Significance

The MBA department is concerned that dual degree students may be receiving lower grades than the regular MBA students. Two independent random samples have been selected. 100 observations from population 1 (dual degree students) and 100 from population 2 (MBA students). The sample means obtained are X1(bar)=83 and X2(bar)=85. It is known from previous studies that the population variances are 4.0 and 5 respectively. Using a level of significance of .05, is there evidence that the dual degree students are receiving lower grades? Fully explain your answer.

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Hello!
This question can be solved using the regular hypothesis testing procedure, but we must first understand some transformations that need to be applied to the data you provide.

First, let's see some definitions of the variables:

X1 = Grade of a student from group 1
X2 = Grade of a student from group 2
N1 = 100 = Sample size of the sample on group 1
N2 = 100 = Sample size of the sample on group 2
X1(bar) = 83 = Mean of the grades of the 100 students of group 1
X2(bar) = 85 = Mean of the grades of the 100 students of group 2
(s1)^2 = 4 = Variance of group 1
(s2)^2 = 5 = Variance of group 1
u1 = Population mean of group 1 [usually, notation is with the greek letter "mu"]
u2 = Population mean of group 1

[The ^2 means "squared" -- recall that the variance is usually noted with the greek letter "sigma" squared. Here we're using 's' squared].
Basically, the problem here is to determine whether X1(bar) is "significantly" smaller than X2(bar). This is equivalent to determining whether the term

X1(bar) - X2(bar)

(which in this case gives 83 - 85 = -2) is "significantly" smaller than zero, and that's the approach we're going to use here. Now, in order to find whether this is smaller than zero, we must first find the distribution of the term X1(bar) - X2(bar). Intuitively, if we found this term to have a very high variance, we shouldn't be surprised to get a difference of -2 in the sample means (as is the case in question) even if the population means were equal (ie, even if on average, dual degree students perform just as well as regular MBA students). On the other hand, if we found it to have a very low variance, then maybe the -2 difference would be big enough to reject the idea that the population means are equal. We're going to find that right now.

First of all we need to find the distribution function of X1(bar) and X2(bar). Since we know nothing about the distribution of X1 or X2, this would seem a difficult task. However, it's pretty simple, because we'll make use of the Central Limit ...

Solution Summary

The solution is comprised of a discussion on hypothesis testing, variance, confidence intervals and levels of significance.

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