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

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

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

Jemurray3
 one year ago
Best ResponseYou've already chosen the best response.0Okay, so you seem to have it... two standard deviations above and below the mean.

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

phi
 one year ago
Best ResponseYou've already chosen the best response.2See 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)

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

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

mariomintchev
 one year ago
Best 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?

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

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

Jemurray3
 one year ago
Best ResponseYou've already chosen the best response.0Oh, I misunderstood the question  gotta read better. Phi's right about that.

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

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

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

phi
 one year ago
Best ResponseYou've already chosen the best response.2If 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

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

mariomintchev
 one year ago
Best ResponseYou've already chosen the best response.1i 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
 one year ago
Best ResponseYou've already chosen the best response.2for 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
 one year ago
Best ResponseYou've already chosen the best response.2Q5 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)

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

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

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

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

mariomintchev
 one year ago
Best ResponseYou've already chosen the best response.1i 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
 one year ago
Best ResponseYou've already chosen the best response.2OK, I got the same numbers

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

mariomintchev
 one year ago
Best 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??

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

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

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

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

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

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

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

Jemurray3
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

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

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

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