# Homework Help: Confidence Level for Proportion

1. Mar 13, 2016

### masterchiefo

1. The problem statement, all variables and given/known data

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.

A) 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.

B) The Company would estimate at 3.0 % near the proportion of axle shafts located outside the standards , 19 times out of 20. How many additional axle shafts should you consider it ? You must take into account the results obtained in A) .

2. Relevant equations
(ME)/(sqrt((P*(1-P))/n)*sqrt(1-(n/N)) = Zalpha/2
normcdf(-Zalpha/2,Zalpha/2)

3. The attempt at a solution
A)

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

(ME)/(sqrt((P*(1-P))/n)*sqrt(1-(n/N)) = Zalpha/2
(0.05)/(sqrt((0.2*(1-0.2))/n)*sqrt(1-(150/1800)) = Zalpha/2
Zalpha/2 = 1.59901

normcdf(-Zalpha/2,Zalpha/2)
normcdf(-1.59901,1.59901) = 0.89 or 89%

This is correct answer for A) checked in book.

Now for part B)
ME = 3% = 0.03
19 times out of 20= 95% = 1-alpha

[p+invnorm(0.025)*(sqrt((P*(1-P))/n)*sqrt(1-(n/N)) , p+invnorm(1-0.025)*(sqrt((P*(1-P))/n)*sqrt(1-(n/N))]
[0.2+invnorm(0.025)*(sqrt((0.5*(1-0.5))/150)*sqrt(1-(150/1900)) , 0.2+invnorm(1-0.025)*(sqrt((0.5*(1-0.5))/150)*sqrt(1-(150/1800))]

[0.1233,0.2766]

solve(invnorm(1-0.025)*(sqrt((0.2766*(1-0.2766))/n)*sqrt(1-(n/1800))=0.03) for n
n = 579.225

I am doing something wrong for the part B).
Please point me to the right direction, maybe my P that I have is not correct.

2. Mar 14, 2016

### masterchiefo

anyone can help me :(

3. Mar 14, 2016

### Ray Vickson

I cannot understand what part B is is asking, because I cannot understand what it is saying. Are you asking for the sample size that would allow the firm to estimate (with probability >= 95%) the percentage of items outside the standard to an accuracy of within ± 3 percentage points--that is, an interval (p - 3/100, p + 3/100)--or are you asking for an estimate accurate to within a 3% error--that is, an interval (0.97*p, 1.03*p)? These are two very different concepts.