Confidence Intervals

[CanadaAotS]

I really need help here lol.

I have an assignment due this friday, the whole thing is on confidence intervals, and I have no clue what I'm doing -_-'

 

OK!

 

So, what I do know is that a confidence interval is used to test how sure you are that a predicted SRS of size n has a certain mean.

 

How would I do an example question like

We know some statistic has standard deviation of 5.00 (nice and easy lol)

 

If we wanted to know how big of an SRS is needed to predict the mean within a standard error of 2.00 with .95 confidence, how would I do that?

 

I know if I had even a simple question like this explained to me, I'd be able to tackle any of the harder questions.

 

This is how far I can get:

 

[math]M - z * \frac{\theta}{\sqrt{n}} \leq \mu \leq M + z * \frac{\theta}{\sqrt{n}}[/math]

Where [math]M[/math] is the sample mean, [math]\theta[/math] is the standard deviation, [math]\mu[/math] is the mean and [math]n[/math] is the SRS size.

 

For the above example, I get

[math]M - z * \frac{5}{\sqrt{n}} \leq \mu \leq M + z * \frac{5}{\sqrt{n}}[/math]

I believe P(-z<Z<z) = .95, which would make z = 1.96

[math]M - 1.96 * \frac{5}{\sqrt{n}} \leq \mu \leq M + 1.96 * \frac{5}{\sqrt{n}}[/math]

 

And thats as far as it goes. I'm stuck with that damn [math]M[/math] lol. I know I'm supposed to be solving for n...

 

Thanks in advance!

~CanadaAotS

[Mokele]

Standard error is just standard deviation over the square root of sample size. If you want one of 2, with a standard deviation of 5, just plug it in: 2 = 5 / (n)^1/2

 

The rest is just for show, and gives you the 95% confidence interval you'd get from that standard error.

[CanadaAotS]

Mokele...

 

you have no idea how much you just helped me :p

 

I can now do an entire assignment (due tomorrow btw)

[CanadaAotS]

Lets see if I have this right

 

"... time per week studying for a first year student follows a normal distribution with a standard deviation of 70.1 minutes."

 

What sample size is required to estimate the true mean weekly study time to within 24 minutes with 96% confidence?

 

This is an actual question on the assignment.

 

So...

 

[math]24 = \frac{71}{\sqrt{n}}[/math]

Would be the equation, right? Solved its:

[math]\sqrt{n} = \frac{71}{24}[/math]

[math]n = (\frac{71}{24})^2[/math]

 

which is equivalent to 8.751736111...

Is this what I'm looking for? lmao.

I'm still confused

 

EDIT: so does this mean that an SRS of size 8.751736111... would accurately predict the mean within 24 minutes at a .96 confidence?

[CanadaAotS]

Wait so

[math]-z*\theta_{error} \leq \mu \leq z*\theta_{error}[/math]

[math]P(-z < Z < z) = .96[/math]

[math]P(Z < z) - P(Z < -z) = .96[/math]

[math]P(Z < z) = .98, P(Z < -z) = .02[/math]

[math]z = 2.05[/math]

[math]-2.05*8.7517 \leq \mu \leq 2.05*8.7517[/math]

 

OH! I GET IT NOW!

 

[math]-24 \leq \mu_x \leq +24[/math]

So, if I divde away 2.05 from 24 (the z value to get .96 confidence)

[math]-11.7073 \leq \mu_x \leq 11.7073 [/math]

 

I plunk that into the equation you gave me...

[math]-11.7073 = \frac{71}{\sqrt{n}}[/math]

Solve

[math]n = (\frac{71}{-11.7073})^2[/math]

Giving me [math]n = 36.779[/math]

 

THEREFORE I need an SRS of size 36.779 (app) to predict [math]\mu[/math] within .96 confidence.

 

YAY!

I think its funny when I learn the material the day before the assignment is due

(I'm the worst procrastinator in the world btw)