Please Help Probability Questions

sos

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Oct 17, 2010
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Can someone please help me figure out if my attempts are right? or if I am using the right method?

Calculate P(X < x); where x = 54.10 and the distribution of X is N(54.1, unknown )


*I am gussing 0.5 but won't I have to assume a normal distributioN?


Calculate xbar where P(Mu > xbar) = 0.025, df = 20, Muxbar = 39.4, sigmaxbar = 28.7


*I tried to use the t-distriubution formula but I am still confused about the difference between xbar, Mu0, muxbar and Mu they seem like the same :S espiecally muxbar (is that the SD of the mean sample but then isn't that xbar :S Can I just plug sigmaxbar into S in the formula t = (xbar - mu)/(s/sqrt n) :S:S


What is the probability that you made a type I error in #11. In a similar test that was powered at 95%, you examined whether the use of advil among women attending your store was different from the general population. You conducted the test with 98% confidence, and found that the use of advil at your store was higher, but similar to and not significantly different than the general population. What was the probability that you were wrong?

Choices
a. 0.050
b. 0.020
c. 0.200
d. None of the other answers


*I thought that it was 0.02 at first but then I started thinking about the 95%, what does that mean? What about alpha error?


Your friend is saying that you should not be using a Z or T test to test your hypothesis because the distribution of advil use in the general population of women is highly skewed. What do you say?
1) we will redo the tests in a way that does not rely on normality
2) It is fine because the Central Limit Theorem states that Sigmaxbar = Sigma0 / sqrt (N)


***I narrowed it down to these two but I can't figure out whether it's 1 or 2. How do I reason this out? I know that with CLT that statement is right but does that mean it's correct for this question?


in one study on 20 men. the upper confidence limit of a 90% confidence interval is 62.3 grams, and the upper confidence limit of an 80% confidence interval is 54.8 grams. What would the lower confidence limit of a more appropriate 95% confidence limit be?


*I got a negative value for the lower confidence limit of -9.2 but it doesn't make sense. I solved for xbar by substituting the number I got from simga thru the CI formula

thank you so so much
 
sos said:
Calculate P(X < x); where x = 54.10 and the distribution of X is N(54.1, unknown )[/b]

*I am gussing 0.5 but won't I have to assume a normal distributioN?
?? You do not have to assume anything. You are GIVEN the Normal Distribution. N(54.1,unknown) MEANS a Normal Distribution with Mean = 54.1 and an unknown standard deviation.
 
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