How bad is the wireless network at JJ? Out of a random sample of 123 J1 students
ID: 3318255 • Letter: H
Question
How bad is the wireless network at JJ? Out of a random sample of 123 J1 students, 75 students said they had a network connection at J1 lost in the past month. Let p be the (unknown) true proportion 3 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 forp. 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 11 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?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
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
d.
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
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