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5. A survey finds that 257 of 867 househol ds owned more than one car twenty yea

ID: 3318527 • Letter: 5

Question

5. A survey finds that 257 of 867 househol ds owned more than one car twenty years ago. Another survey of 382 currently own more than one car. Find a 90t confidence interval for the actual househoids finds that 245 mean increase (over the past two decades) in the percentage of households that own more than one car. 6, suppose we want to find a 90% confidence interval similar to the one in Problem 5 that has a margin of error less than 6t. Find the sample sizes needed to achieve this goal if there have been no preliminary surveys

Explanation / Answer

since the actual percentage increase can be found by finding one tail confidence between the previous rate and to the current rate for who whom owns more than a one car

TRADITIONAL METHOD
given that,
sample one, x1 =257, n1 =867, p1= x1/n1=0.2964
sample two, x2 =245, n2 =382, p2= x2/n2=0.6414
I.
standard error = sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where
p1, p2 = proportion of both sample observation
n1, n2 = sample size
standard error = sqrt( (0.2964*0.7036/867) +(0.6414 * 0.3586/382))
=0.029
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.1
from standard normal table,left tailed z /2 =1.28
margin of error = 1.28 * 0.029
=0.0372
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.2964-0.6414) ±0.0372]
= [ -0.3821 , -0.3078]
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DIRECT METHOD
given that,
sample one, x1 =257, n1 =867, p1= x1/n1=0.2964
sample two, x2 =245, n2 =382, p2= x2/n2=0.6414
CI = (p1-p2) ± sqrt( p1 * (1-p1)/n1 + p2 * (1-p2)/n2 )
where,
p1, p2 = proportion of both sample observation
n1,n2 = size of both group
a = 1 - (confidence Level/100)
Za/2 = Z-table value
CI = confidence interval
CI = [ (0.2964-0.6414) ± 1.28 * 0.029]
= [ -0.3821 , -0.3078 ]
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interpretations:
1) we are 90% sure that the interval [ -0.3821 , -0.3078] contains the difference between
true population proportion P1-P2
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 difference between
true population mean P1-P2

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