27. A random sample of n observations is taken from a population... SEE ATTACHMENT© BrainMass Inc. brainmass.com October 24, 2018, 6:06 pm ad1c9bdddf
a) since u^ = C1 X1 + C2 X2 + ... + Cn Xn
then the expected value of u^ is:
E(u^)= E( C1 X1 + C2 X2 + ... + Cn Xn )
= E(C1 X1) + E(C2 X2) + ... + E(Cn Xn )
= C1 E(X1) + C2 E(X2) + ... + Cn E(Xn )
also, because E(Xi) = u for all i = 1, 2, ... n
Then E(u^)= C1* u + C2*u + ... + Cn*u
= (C1 + C2 + ... + Cn) u
so for u^ to be unbiased, we need E(u^) = u, or
(C1 + C2 + ... + Cn) u = u
therefore, (C1 + C2 + ... + Cn) = 1 is the necessary and sufficient condition for u^ to be ...
When we want to test two samples to determine if it is likely that the population means (estimated by the sample means) are different, we typically use a t-test. If the samples are large, we can also use a z-test. (Note that the formulas for computing s, t and/or z in the case of a two-sample test are different than the formulas for computing the same values in a one-sample test. Use Excel data analysis to conduct tests comparing two sample means.)
Using ANOVA (short for Analysis of Variance), however, we can test 3 or more sample means to determine if at least one of the sample means comes from a population with a mean that is significantly different from all of the others in the test. We actually do this by estimating a combined population variance two different ways and comparing the two estimates (the ratio of these two variance estimates follows the so-called "F distribution").
Why do we need a new test method to compare the means of 3 or more populations? Why can't we just use a series of z-tests or t-tests to compare all of the possible pairs of population means to see if one (or more) is different?
Most of the testing is to determine one or two things:
1. Is there a statistically significant difference between two or more population means? (based on comparison of 2 or more sample means)
2. Is there a statistically significant relationship between two or more variables? We can use regression analysis or chi-square tests to answer this second question.)View Full Posting Details