Question

(b) Suppose a production facility purchases a particular component part in large lots from a supplier. The production manager wants to estimate the proportion of defective parts received from this supplier. She believes the proportion defective is no more than .20 and wants to be within .02 of the true proportion of defective parts with a 90% level of confidence. How large a sample should she take?

(c) A random sample of size 20 is taken, resulting in a sample mean of 16.45 and a sample standard deviation of 3.59. Assume x is normally distributed and use this information and a= 0.5 to test the following hypotheses

(6 marks)

Answer 1

Answer:

sorry I don't know you ask for someone else

Answer 2

Answer:

size of sample should be 957, this value is enough for all P≤0.2

Explanation:-

Given that :

Sample proportion P = 0.20

Margin of error , ME = 0.04

Solution:-

Proportion p has approximately normal distribution with parameters,

μp  = Р.σp =$\sqrt\frac{p.(1-p)}{n}$ thus for p

analisi we, use normal distribution

N$(p,\sqrt \frac{p.(1-p)}{n} )$ so C- confidence interval for

mean M(p) will be P ±z₁* -c/2 .$\sqrt\frac{p,(1-p)}{n}$

where z₁* +c/2 is such level that p (N(0,1)

> Z₁*-c/2 = 1-c/2 therefore margin error will be

E = z₁*-c/2.$\sqrt\frac{p(1-p)}{n}$ From this equation

we define size of sample  n = p.(1-p)(Z₁-c/2)²/E².

we are given with E =0.02 , P = 0.2 we use the worst case

c = 90%= 0.9 and calculate :

Z₁*-c/2 = Z*0.05=q norm (0.95,0.1) = 1.645 where q norm is statistical function from Mathcad therefore we

have n =0.2.(1-0.2).1.645²/0.02² = 956.046≈957

so ,

size of sample should be 957, this value is enough for all P≤0.2