Estimation - Student-t distribution - Confidence Level

[masterchiefo]

Homework Statement

[/B]
A company produces batches of 1800 axle shafts . These are tested to determine the proportion of those with too rough surface according to the standards set in the industry.

The quality control department of a sample of 150 axle shafts in a lot and concluded that there are between 15 % and 25 % of the axle shafts in this set that are outside the norms . Calculate the confidence level associated with this estimate.

Homework Equations

t_n-1,sigma/2=(ME)/((s/sqrt(n))* (sqrt(1-(n/N)))
(1-sigma)=tcdf(-t_n-1,sigma/2,+t_n-1,sigma/2,n-1)

The Attempt at a Solution

Known variable:
N = 1800
n=150
[15%,25%] ---> average= (15+25)/2 =20% or 0.20
ME = (25-15)/2 = 5% or 0.05

Unknown variable:
(1-sigma)
Average of N
Standard Deviation of nI am stuck here, I cannot continue without knowing SD.

 

[RUber]

This is a proportion. So, standard deviation of a proportion is
##\sqrt{p(1-p)}.##

 

[masterchiefo]

This is a proportion. So, standard deviation of a proportion is
##\sqrt{p(1-p)}.##

Hello,

I don't understand, what would be p ?

 

[masterchiefo]

This is a proportion. So, standard deviation of a proportion is
##\sqrt{p(1-p)}.##

do I still use Student -t formula ?

 

[masterchiefo]

I did that:
p=(15%+25%)/2 =0.2
s= √p(1−p) = √0.2(1−0.2) = 0.4
t=(ME)/((s/sqrt(n))* (sqrt(1-(n/N)))
t = 0.05/((0.4/sqrt(150))* (sqrt(1-(150/1800)))
(1-sigma)=tcdf(-t,+t,n-1)
(1-sigma) = 1

It does not work :(
anyone please help me.

 

[RUber]

I did that:
p=(15%+25%)/2 =0.2
s= √p(1−p) = √0.2(1−0.2) = 0.4
t=(ME)/((s/sqrt(n))* (sqrt(1-(n/N)))
t = 0.05/((0.4/sqrt(150))* (sqrt(1-(150/1800)))
(1-sigma)=tcdf(-t,+t,n-1)

I am not sure what you last line is doing.
Normally, you would take your calculate t standard error of the proportion...in this case:
t=(ME)/((s/sqrt(n))* (sqrt(1-(n/N))) = .031,
And compare that to your range ( +/- 5% ) .
This gives you an idea of what your ##t_{\frac{\alpha}{2}, df}## should be equal to.

 

[Ray Vickson]

Homework Statement

[/B]
A company produces batches of 1800 axle shafts . These are tested to determine the proportion of those with too rough surface according to the standards set in the industry.

The quality control department of a sample of 150 axle shafts in a lot and concluded that there are between 15 % and 25 % of the axle shafts in this set that are outside the norms . Calculate the confidence level associated with this estimate.

Homework Equations

t_n-1,sigma/2=(ME)/((s/sqrt(n))* (sqrt(1-(n/N)))
(1-sigma)=tcdf(-t_n-1,sigma/2,+t_n-1,sigma/2,n-1)

The Attempt at a Solution

Known variable:
N = 1800
n=150
[15%,25%] ---> average= (15+25)/2 =20% or 0.20
ME = (25-15)/2 = 5% or 0.05

Unknown variable:
(1-sigma)
Average of N
Standard Deviation of nI am stuck here, I cannot continue without knowing SD.

Google "confidence interval for proportion".