Need help with #4

- anonymous

Need help with #4

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- e.cociuba

wats number 4??

- anonymous

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- anonymous

^^^

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## More answers

- anonymous

@jim_thompson5910 @Jemurray3

- anonymous

ok so,
we have
n=100
x bar=80,000
sigma=15,000
and we need mu with a 95% confidence interval

- anonymous

|dw:1361314863406:dw|

- anonymous

Okay, so you seem to have it... two standard deviations above and below the mean.

- anonymous

what would be my final answer exactly?

- anonymous

50k -110k.

- phi

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)

- 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.

- anonymous

yeah, i did 80,000/ [15,000/ sqrt(100)] earlier and got 53.333....

- anonymous

"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

- anonymous

yeah, 95 = 1.96 on the table

- anonymous

Oh, I misunderstood the question -- gotta read better. Phi's right about that.

- anonymous

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.

- anonymous

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

- anonymous

ok so the CI is (77,060, 82,940)

- anonymous

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)

- anonymous

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

- anonymous

ok my final answer is (5.3, 10.7) when rounding off

- anonymous

right?

- phi

what did you get for the mean and std?
what sigma gives 90% confidence?

- anonymous

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

- anonymous

shhweet

- anonymous

can i bother you for one more question? its the only i have left. ...#11

- anonymous

#11
part a, i got 81.6
and for part b, i got 159.6
but what about the error??

- anonymous

Score = B0 + B1 (Hours) + Error

- anonymous

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.

- anonymous

yeah that's what i was thinking, so the error has to be a negative number

- anonymous

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.

- anonymous

yeah, im trying to get some more help from someone cause i've obviously already tried on my own.

- anonymous

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.

- anonymous

so 81.6 and 159.6 seem to be the correct answers?

- anonymous

seems a little too simple... idk

- anonymous

but i understand what you are saying

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