This statistics problem is insanely hard.

[blinky8225]

Originally posted by: KayGee
You can't use the normal distribution if the population standard deviation (sigma) is unknown. The general rule is to use the t-distribution if the sample size is less than 30 or sigma is unknown. Also, this is a two-tailed test.

You use the t distribution if you are testing the mean of a distribution with unknown variance. You use the Chi-squared distribution when testing the variance.

 

[Suspicious-Teach8788]

Originally posted by: blinky8225

Originally posted by: dbk

Originally posted by: blinky8225
First, you should be doing a two sided test.
H0: u == $30
H1: u =/= $30

Let your test statistic Z = (16)^(1/2) (X - 3000)/551, where X is average of the cost of the dates.

a) if you are going to test at level of significance 0.05, you will reject H0 if Z > 1.96 or Z < -1.96. This is not the case. Accept H0.
b) if you are going to test at level of significance 0.10, you will reject H0 if Z > 1.645 or Z < -1.645. Again this is not the case. Accept H0.
c) if you want to use a confidence interval of 95%, you find that the interval for the mean u is from 28.4701 to 33.8699. 30 falls into this interval. Again, accept H0.
d) if you want to use a confidence interval of 90%, you find that the interval for the mean u is from 28.9040 to 33.4360. 30 falls into this interval. Again, accept H0.
e) it turns out that A & C are the same test. B & D are also the same test.

wow..nicely done brah

Thanks. If only I had questions like this on my statistics test last week, I might have actually passed.

<--Math major at Duke

LOL good thing I read your post. I drew a bell curve with the standard deviation and mean shown on there with some calculations before I said "Maybe I should read this thread to see if it's been done already." Whew.

But a tip for the OP, always frame your H0 and H1 very nicely. Once you do so, it's almost always very easy to set it up. With statistics like this, you're always either doing a t-test, Z-test, or Chi-squared test, and maybe an F-test, but once you nail the first 2, it should be very mechanical and repetitive for the others.

Also, do not be confused by the distribution of the individual measurements and the distribution of averages. Standard error of the mean was what the problem asked for, and you were using the standard deviation of the individual measurements!! This is a huge problem and even I make this error from time to time.

N Mean StDev SE Mean 95% CI T P
16 31.17 5.51 1.38 (28.23, 34.11) 0.85 0.409

 

[KayGee]

Originally posted by: blinky8225

Originally posted by: KayGee
You can't use the normal distribution if the population standard deviation (sigma) is unknown. The general rule is to use the t-distribution if the sample size is less than 30 or sigma is unknown. Also, this is a two-tailed test.

You use the t distribution if you are testing the mean of a distribution with unknown variance. You use the Chi-squared distribution when testing the variance.

He's not testing variances, he's testing means, and sigma is unknown and the sample size is less than 30, which means he needs to find a critical t-value and not z as you and others have suggested. The hypotheses for his problem are :

H0: µ = $30
H1: µ ? $30

 

[Q]

I didn't read this thread cept for first post -- If you need help on it I can do it

 

[Fenixgoon]

Originally posted by: rgwalt
Sorry, I took real math classes, like calculus and differential equations.

😀

J/K, stats is a tough subject. I never wanted any part of it.

me too. after 4 years of engineering, i finally get boltzmann statistics/distributions 😛

 

[Leros]

Originally posted by: Fenixgoon

Originally posted by: rgwalt
Sorry, I took real math classes, like calculus and differential equations.

😀

J/K, stats is a tough subject. I never wanted any part of it.

me too. after 4 years of engineering, i finally get boltzmann statistics/distributions 😛

I haven't done much statistics. They made us take probability and random processes though.