Can someone help please
1 answer:
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Answer:
d. t distribution with df = 80
Step-by-step explanation:
Assuming this problem:
Consider independent simple random samples that are taken to test the difference between the means of two populations. The variances of the populations are unknown, but are assumed to be equal. The sample sizes of each population are n1 = 37 and n2 = 45. The appropriate distribution to use is the:
a. t distribution with df = 82.
b. t distribution with df = 81.
c. t distribution with df = 41.
d. t distribution with df = 80
Solution to the problem
When we have two independent samples from two normal distributions with equal variances we are assuming that
And the statistic is given by this formula:
Where t follows a t distribution with degrees of freedom and the pooled variance
is given by this formula:
This last one is an unbiased estimator of the common variance
So on this case the degrees of freedom are given by:
And the best answer is:
d. t distribution with df = 80
We see that in these 3 terms, 6 is a common factor. So, let's factor out a 6:
We can set this equal to 0 and factor by using the quadratic formula which is:
So, let's do just that:
Note that the 6 goes away if you divide both sides by 6. In this case, a = 1, b = -1, and c = -2. Let's plug that into the quadratic equation:
So, we can write this as:
Notice that the 6 comes back because it was only temporarily mad. And that the roots have opposite signs in the parentheses because to find the roots, you need to set each parentheses equal to 0 and solve for n. With that in mind, your final answer is 6(n-1)(n+2). Hope I could help you!