How bad is the wireless network at JJ? Out of a random sample of 123 JJ students
ID: 3318260 • Letter: H
Question
How bad is the wireless network at JJ? Out of a random sample of 123 JJ students, 75 students said they had a network connection at JJ lost in the past month. Let p be the (unknown) true proportion JJ students who have lost a network connection in the last month. Let phat be the sample proportion
a)Obtain an unbiased point estimate of p.
b) We wish to construct a 96 % classical confidence interval for p. What is the critical value multiplier zstar?
c) Create a 96% classical confidence interval for p?
,
d) How long is the 96% classical confidence interval for p?
f) Based on this data estimate the standard deviation of the sample proportion of JJ students who lost a connection in the last month.
g)Assuming the same value of the sample proportion what is the smallest sample size for which the length of the 96% confidence interval for p is less than or equal to .12?
h) Copy your R script for the above into the text box here.
Explanation / Answer
TRADITIONAL METHOD
a.
given that,
possibile chances (x)=75
sample size(n)=123
success rate ( p )= x/n = 0.6098
I.
sample proportion = 0.6098
standard error = Sqrt ( (0.6098*0.3902) /123) )
= 0.044
II.
margin of error = Z a/2 * (stanadard error)
where,
b.
Za/2 = Z-table value
level of significance, = 0.04
crtical value = from standard normal table, two tailed z /2 =2.054
margin of error = 2.054 * 0.044
= 0.0903
c.
III.
CI = [ p ± margin of error ]
confidence interval = [0.6098 ± 0.0903]
= [ 0.5194 , 0.7001]
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DIRECT METHOD
given that,
possibile chances (x)=75
sample size(n)=123
success rate ( p )= x/n = 0.6098
CI = confidence interval
confidence interval = [ 0.6098 ± 2.054 * Sqrt ( (0.6098*0.3902) /123) ) ]
= [0.6098 - 2.054 * Sqrt ( (0.6098*0.3902) /123) , 0.6098 + 2.054 * Sqrt ( (0.6098*0.3902) /123) ]
= [0.5194 , 0.7001]
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interpretations:
1. We are 96% sure that the interval [ 0.5194 , 0.7001] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 96% of these intervals will contains the true population proportion
f.
standard Deviation ( sd )= sqrt(PQ/n) = sqrt(0.609756097560976*0.3902/123)
=0.044
e.
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.04 is = 2.054
Sample Proportion = 0.6098
ME = 0.12
n = ( 2.054 / 0.12 )^2 * 0.6098*0.3902
= 69.7129 ~ 70
g.
> n = 75 # valid responses count
> k = 123
> pbar = k/n; pbar
[1] 0.6098
Then we estimate the standard error.
> SE = sqrt(pbar(1pbar)/n); SE # standard error
[1] 0.044
Since there are two tails of the normal distribution, the 96% confidence level would imply the 97.5th percentile of the normal distribution at the upper tail. Therefore, z2 is given by qnorm(..96). Hence we multiply it with the standard error estimate SE and compute the margin of error.
> E = qnorm(.96)*SE; E # margin of error
[1] 0.0903
Combining it with the sample proportion, we obtain the confidence interval.
> pbar + c(E, E)
[1] 0.5194 , 0.7001
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