Question

A new PC operating system is being developed. The designer claims "more than 90% of all computer programs currently being run on Microsoft's Vista operating system will run successfully on the new PC operating system." A random sample of 100 programs are selected and run. Only 86 of the 100 programs run successfully. The designer decides to calculate a 90% confidence interval from his data for

p= proportion of programs that will run successfully on the new PC operating system

a. Are the conditions met by his data to calculate a 90% confidence interval for p? Explain

b. Find a 90% confidence interval for p.

c. If he continues to sample another 100 programs to run, what is the probability that more than 86% of the selected programs will run successfully if the true p=0.9

Answer 1

a.
TRADITIONAL METHOD
given that,
possible chances (x)=86
sample size(n)=100
success rate ( p )= x/n = 0.86
I.
sample proportion = 0.86
standard error = Sqrt ( (0.86*0.14) /100) )
= 0.035
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.1
from standard normal table, two tailed z α/2 =1.645
margin of error = 1.645 * 0.035
= 0.057
III.
CI = [ p ± margin of error ]
confidence interval = [0.86 ± 0.057]
= [ 0.803 , 0.917]
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DIRECT METHOD
given that,
possible chances (x)=86
sample size(n)=100
success rate ( p )= x/n = 0.86
CI = confidence interval
confidence interval = [ 0.86 ± 1.645 * Sqrt ( (0.86*0.14) /100) ) ]
= [0.86 - 1.645 * Sqrt ( (0.86*0.14) /100) , 0.86 + 1.645 * Sqrt ( (0.86*0.14) /100) ]
= [0.803 , 0.917]
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interpretations:
1. We are 90% sure that the interval [ 0.803 , 0.917] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion
Answer:
a.
yes,
the conditions met by his data to calculate a 90% confidence interval for p
sample proportion = 0.86
b.
90% sure that the interval [ 0.803 , 0.917]
c.
the PDF of normal distribution is = 1/σ * √2π * e ^ -(x-u)^2/ 2σ^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
proportion ( p ) = 0.9
standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.9*0.1/100)
=0.03

the probability that more than 86% of the selected programs will run successfully if the true p=0.9
P(X > 0.86) = (0.86-0.9)/0.03
= -0.04/0.03 = -1.3333
= P ( Z >-1.333) From Standard Normal Table
= 0.9088