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mariomintchev
Need help with #4
@jim_thompson5910 @Jemurray3
ok so, we have n=100 x bar=80,000 sigma=15,000 and we need mu with a 95% confidence interval
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Okay, so you seem to have it... two standard deviations above and below the mean.
what would be my final answer exactly?
See http://en.wikipedia.org/wiki/Confidence_interval#Practical_example They are *estimating* the mean salary of 80,000 and know (somehow) the population standard deviation. the sample mean (the 80,000) can be considered a random variable, with a normal distribution. It has the same mean (80,000) by a sigma_new= sigma/sqrt(samples)
the sigma on your sample mean is 15,000/sqrt(100) = 1,500 now you can find the interval that contains 95% of the area: ±1.96 s. d.
yeah, i did 80,000/ [15,000/ sqrt(100)] earlier and got 53.333....
"now you can find the interval that contains 95% of the area: ±1.96 s. d." what do you mean?
? I would do 80,000 ± 1.96*1,500 as the interval
yeah, 95 = 1.96 on the table
Oh, I misunderstood the question -- gotta read better. Phi's right about that.
why 1,500? shouldn't it be 15,000?
the idea is you sample 100 people and get a mean of 80,000 if you do that again and again (sampling a random subset of the population, with a sample size of 100) and compute the mean for each, you will get a normal distribution, with a mean of 80,000 and a s.d. of 1500
the *true* (i.e. population mean as opposed to sample mean) will lie within the interval 80,000 ± 1.96*1500 with 95% confidence.
but why 1500? why not 15,000???
If you take a sample of 100 and find their mean, it will be close to the population (i.e. *true* mean). The sample mean will have a s.d. of population sigma/ sqrt(# of samples) I am sure there is a proof of this, but I don't remember it off-hand
ok so the CI is (77,060, 82,940)
i have a question for # 5 as well.... do we take all the numbers given and average them out and then place that mean on a normal distribution curve?
for q4 (and I think question 5 also) you can think of each sample as a random variable with mean u and variance s^2 if you add n samples you get a new random variable with mean n*u and variance n*s^2 and standard deviation sqrt(n)* s now divide each statistic by n. you get n*u/n = u (the same mean as for the population) and a standard deviation of sqrt(n)*s/n = s/sqrt(n)
Q5 is the same type as Q4, except you need to find the sample mean. Then you need to find the sample mean's std. = sigma/ sq rt(15)
ok so i just need to find the mean of all the hours given
yes, that will correspond to the 80,000 in Q4
ok my final answer is (5.3, 10.7) when rounding off
what did you get for the mean and std? what sigma gives 90% confidence?
i did: n=15 sigma=6.3 x bar=8 6.3/sqrt(15) = 1.63 90% CI = 1.645 8 +- (1.645 * 1.63) (5.3, 10.7)
can i bother you for one more question? its the only i have left. ...#11
#11 part a, i got 81.6 and for part b, i got 159.6 but what about the error??
Score = B0 + B1 (Hours) + Error
All you have to do is plug in the numbers given for B0, B1, and the hours, but what about the + error ???
I am wondering if the 159.6 is valid. I would think 100 is the best you can do.
yeah that's what i was thinking, so the error has to be a negative number
but where the heck would we get that number from the information provided?
If you get an answer, let me know. I only know enough of this stuff to be dangerous.
yeah, im trying to get some more help from someone cause i've obviously already tried on my own.
I see no reason that 159 is not a valid score, it doesn't say anything about it being a percentage. The regression neglecting the error is your estimate for an average student -- the error just quantifies how sure or unsure you are of your regression analysis.
so 81.6 and 159.6 seem to be the correct answers?
seems a little too simple... idk
but i understand what you are saying