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mariomintchevBest ResponseYou've already chosen the best response.1
@jim_thompson5910 @Jemurray3
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.1
ok so, we have n=100 x bar=80,000 sigma=15,000 and we need mu with a 95% confidence interval
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.1
dw:1361314863406:dw
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.0
Okay, so you seem to have it... two standard deviations above and below the mean.
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.1
what would be my final answer exactly?
 one year ago

phiBest ResponseYou've already chosen the best response.2
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)
 one year ago

phiBest ResponseYou've already chosen the best response.2
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.
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.1
yeah, i did 80,000/ [15,000/ sqrt(100)] earlier and got 53.333....
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.1
"now you can find the interval that contains 95% of the area: ±1.96 s. d." what do you mean?
 one year ago

phiBest ResponseYou've already chosen the best response.2
? I would do 80,000 ± 1.96*1,500 as the interval
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.1
yeah, 95 = 1.96 on the table
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.0
Oh, I misunderstood the question  gotta read better. Phi's right about that.
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.1
why 1,500? shouldn't it be 15,000?
 one year ago

phiBest ResponseYou've already chosen the best response.2
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
 one year ago

phiBest ResponseYou've already chosen the best response.2
the *true* (i.e. population mean as opposed to sample mean) will lie within the interval 80,000 ± 1.96*1500 with 95% confidence.
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.1
but why 1500? why not 15,000???
 one year ago

phiBest ResponseYou've already chosen the best response.2
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 offhand
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.1
ok so the CI is (77,060, 82,940)
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.1
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?
 one year ago

phiBest ResponseYou've already chosen the best response.2
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)
 one year ago

phiBest ResponseYou've already chosen the best response.2
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)
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.1
ok so i just need to find the mean of all the hours given
 one year ago

phiBest ResponseYou've already chosen the best response.2
yes, that will correspond to the 80,000 in Q4
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.1
ok my final answer is (5.3, 10.7) when rounding off
 one year ago

phiBest ResponseYou've already chosen the best response.2
what did you get for the mean and std? what sigma gives 90% confidence?
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.1
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)
 one year ago

phiBest ResponseYou've already chosen the best response.2
OK, I got the same numbers
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.1
can i bother you for one more question? its the only i have left. ...#11
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.1
#11 part a, i got 81.6 and for part b, i got 159.6 but what about the error??
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.1
Score = B0 + B1 (Hours) + Error
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.1
All you have to do is plug in the numbers given for B0, B1, and the hours, but what about the + error ???
 one year ago

phiBest ResponseYou've already chosen the best response.2
I am wondering if the 159.6 is valid. I would think 100 is the best you can do.
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.1
yeah that's what i was thinking, so the error has to be a negative number
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.1
but where the heck would we get that number from the information provided?
 one year ago

phiBest ResponseYou've already chosen the best response.2
If you get an answer, let me know. I only know enough of this stuff to be dangerous.
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.1
yeah, im trying to get some more help from someone cause i've obviously already tried on my own.
 one year ago

Jemurray3Best ResponseYou've already chosen the best response.0
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.
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.1
so 81.6 and 159.6 seem to be the correct answers?
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.1
seems a little too simple... idk
 one year ago

mariomintchevBest ResponseYou've already chosen the best response.1
but i understand what you are saying
 one year ago
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