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A random sample of 300 parts was taken from a particular grinding machine. Of those parts, 12 were defective. (a) Construct a 99% upper bound for the proportion of all parts produced by this machine that will be defective. 0.0109 (b) Construct a 99% confidence interval for the proportion of all parts produced by this machine that will be defective. [ Select ] (c) It is decided that the interval produced in part (b) is too wide, and thus a new sample must be taken and a new confidence interval calculated. What sample size would be necessary to ensure that the width of the 99% confidence interval will be 0.02 or less

Respuesta :

Answer:

a) Ub = 0,069

b)  CI 99 %  = ( 0,0287 ; 0,069 )

c) n = 2536

Step-by-step explanation:

Confidence Interval (CI) for proportion is

p₀  ±  z(c) * √ ( p₀*q₀ ) /n

Where     p₀  is the sample proportion

n = size of the sample

and z(c)   is z score for the significance level α

As    CI  = 99 %       α = 1%      α = 0,01      α/2 = 0,005

from z-table we find   z(c) = 2,57

p₀ = 12/300      p₀ = 0,04  ;    

n*p₀ = 12  > 10 and  n*q₀  = 300*0,96        n*q₀ = 288 > 10

We can use normal distribution as a valid  aproximation of binomial distibution

Then     z(c) = 2,57

a) upper bound for CI     Ub

Ub = p₀ + z(c) * √ p₀q₀/300

Ub = 0,04  +  2,57 * √ (0,04*0,96)/300

Ub = 0,04 + 0,0113

Ub = 0,069

b) CI at 99%  is

CI  = p₀  ±  z(c) * √ ( p₀*q₀ ) /n

we have already calculated the Ub = 0,069

the Lower bound would be   Lb = 0,04 - 0,0113

Lb = 0,0287

Then     CI 99 %  = ( 0,0287 ; 0,069 )

c) If      CI ≤ 0,02 we can find the minimun n value with the equation:

p₀  ±  z(c) * √ ( p₀*q₀ ) /n = CI   or    p₀  -  z(c) * √ ( p₀*q₀ ) /n = 0,01

For symmetry respect to p₀

z(c) *√ (p₀*q₀) / n  = 0,01

Solving for n

z(c) * √ ( p₀*q₀ ) /n = 0,01

2,57 * √ (0,04*0,96)/n  =  0,01

2,57 * √ 0,0384/n = 0,01

2,57 = 0,01/√ 0,0384/n

Squaring both sides of the equation

6,6049 = (0,01)² *n / 0,0384

0,0384*6,6049 / 0,0001  = n

n = 2536

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