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Need help with #4

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wats number 4??
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ok so, we have n=100 x bar=80,000 sigma=15,000 and we need mu with a 95% confidence interval
Okay, so you seem to have it... two standard deviations above and below the mean.
what would be my final answer exactly?
50k -110k.
  • phi
See 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)
  • phi
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?
  • phi
? 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?
  • phi
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
  • phi
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???
  • phi
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?
  • phi
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)
  • phi
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
  • phi
yes, that will correspond to the 80,000 in Q4
ok my final answer is (5.3, 10.7) when rounding off
  • phi
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)
  • phi
OK, I got the same numbers
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 ???
  • phi
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?
  • phi
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

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