got a stats question...last time nyc helped me

[doubles1]

nycgirl last time you helped me however i have another problem that I am needing help with...

i will gladly give sum karma

here is the problem:

What sample size is needed to estimate with 95 percent confidence the mean intake of calcium within 20 units of the true mean if the intake is normal with a variance of 1900 units.

thanks for any assistance.

-mark

 

[nycgirl]

Oh, confidence intervals. A bit of weak spot for me. I regression. Absolutely hate the confidence interval.

Give me a minute. I will give it a shot. If I'm not confident about my answer/help I will let you know. Definitely, go with someone else if they beat me to it.

How are you doing?

 

[doubles1]

im doing okay i am taking a test right now with using z and t tables and sample sizes and also hypothesis and alternate hypothesis.

the only thing i am having trouble on is that question and

What sample size would be required to estimate m to within ±3 with 95% confidence if the population standard deviation is 24.5?

and the choices are
769
124
257
91
2321

other than that i have a couple more questions to answer and i should be done with this test...ahhhhhhhhhhhhhhhhhhhhhhhhhh been working on it for about 3 hrs!!!

 

[EnderJE]

WTF? Are you surfing while taking the test? It is an in-home test or an in-class test? If it's in class, you've got the props, maing.

 

[doubles1]

hey its online test...i have 4 hrs to take it ...theres like 35 fuckin problems!!!!


nyc if you have aim or yahoo would you msg me...i will pm you my name!!

 

[doubles1]

hey you exceeded your pm's...

hey my yahoo name is

[email protected]

and aim

lucky5588

hit me up if you dont mind...

 

[nycgirl]

hey its online test...i have 4 hrs to take it ...theres like 35 fuckin problems!!!!


nyc if you have aim or yahoo would you msg me...i will pm you my name!!

I have msn.

Are these the only questions you are having issues with? You may have to guess on these.

 

[Smurfy]

cheater!

 

[doubles1]

umm well if you wouldnt mind msg'ng me i am impressed with your knowledge and wouldnt mind askin u a couple of things.

 

[doubles1]

hey NYC, if your not comfortable msgn me thats fine.

the last prob im working with is this...


Does the sample evidence indicate that the average time an employee stays with a company in their current positions is less than 3 years? A random sample of 50 employees yielded a mean of 2.79 years and s = .76. Use a = .01

 

[nycgirl]

hey NYC, if your not comfortable msgn me thats fine.

the last prob im working with is this...


Does the sample evidence indicate that the average time an employee stays with a company in their current positions is less than 3 years? A random sample of 50 employees yielded a mean of 2.79 years and s = .76. Use a = .01

Good lord. No need to make me look weird. I just sent you a PM.

I will give these a try. But, I think you are going to have to guess. This was never my strong point in stats.

 

[bran987]

it tells you it's a normal distribution so use the Z-table. Look up the Z-score for a 95% confidence interval (i.e. 2.5% on each side, so find the score that gives you 2.5% on each side, I think it's 1.96?)

Instead of converting "X" to a standard normal distribution, reverse the formula and convert the Z-score backwards to give "X" and use the standard deviation in the formula given by the square root of the variance (square root of 1900 = 43.59). So the only variable left to solve for in the equation will be the mean.

X = u + (std. dev. * z-score)

It's the inverse of the standardization rule in other words.

I think.

 

[bran987]

oh wait no hold on

 

[doubles1]

bran,

thanks a lot...i am startin to struggle now been at this for 3+ hrs now.

 

[bran987]

to get the sample size you take n = p (1 - p) * (z alpha/2 / MOE) so we need to solve for N here...

So I think here that would be

 

[bran987]

bran,

thanks a lot...i am startin to struggle now been at this for 3+ hrs now.

what I posted up there in that first post is wrong lemme work on this for another minute haven't done it in awhile. So it's asking for "within 20" so that's the margin of error? it doesn't ask for a certain percentage margin of error right?

 

[doubles1]

hey bran it asks for

What sample size is needed to estimate with 95 percent confidence the mean intake of calcium within 20 units of the true mean if the intake is normal with a variance of 1900 units?



The last question on the test is this..and i am strugglin like a champ...

A random sample of 100 European professional soccer players has an average age of 27 years. The sample standard deviation is 4 years. We would like to decide if there is enough evidence to establish that the average age of European soccer players is more than 26 years.

Answer A B & C Below:

A. Write the null hypothesis.

B. Write the alternative hypothesis.

C. What is the rejection point (given in terms of the value of the test statistic) at a = .05?

 

[bran987]

nycgirl last time you helped me however i have another problem that I am needing help with...

i will gladly give sum karma

here is the problem:

What sample size is needed to estimate with 95 percent confidence the mean intake of calcium within 20 units of the true mean if the intake is normal with a variance of 1900 units.

thanks for any assistance.

-mark

I think the answer to this one is n = ((1.96 * sqrt(1900))/20) ^ 2 = 18.25

n = 18.25

 

[seizer]

sucka, you gotta take stats!


and here I am with 6 hours of bio a head of me....

 

[bran987]

hey bran it asks for

What sample size is needed to estimate with 95 percent confidence the mean intake of calcium within 20 units of the true mean if the intake is normal with a variance of 1900 units?



The last question on the test is this..and i am strugglin like a champ...

A random sample of 100 European professional soccer players has an average age of 27 years. The sample standard deviation is 4 years. We would like to decide if there is enough evidence to establish that the average age of European soccer players is more than 26 years.

Answer A B & C Below:

A. Write the null hypothesis.

B. Write the alternative hypothesis.

C. What is the rejection point (given in terms of the value of the test statistic) at a = .05?

A) Null Hypothesis: u = 26 (or u > 26)
B) Alternate Hypthothesis: u does not = 26


now for C, hold on

 

[bran987]

C) this is a 1-sided test, if alpha = .05,

t(s) = ((27 - 26))/(4/sqrt(100)) = 2.5

t-statistic rejection point = 2.5

 

[doubles1]

bran,

i just pm'ed you lemme know if you got it.