The question that I need answred is 8.86. I included 8.32 for reference! 8.86 Ch

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The question that I need answred is 8.86. I included 8.32 for reference!

8.86 Changes in credit card usage by undergraduates. in Exercise 8.32 (page 503), we looked at data from a survey of 1430 undergraduate students and their credit use. In the sample, 43% said that they had four or more credit cards. A similar study performed four years earlier by the e same organization reported that 32% of ple said that thev had four or more credit cards. at te sample sizes for the two studies are the Find a 95% confidence interval for the change in er by th sam 31 the th e percent of more credit cards. w ergraduates who report having four or the change. Refer to

Explanation / Answer

8.86.

TRADITIONAL METHOD
given that,
sample one, x1 =614.9, n1 =1430, p1= x1/n1=0.43
sample two, x2 =457.6, n2 =1430, p2= x2/n2=0.32
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( (0.43*0.32/1430) +(0.32 * 0.68/1430))
=0.018
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
margin of error = 1.96 * 0.018
=0.035
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.43-0.32) ±0.035]
= [ 0.075 , 0.145]
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DIRECT METHOD
given that,
sample one, x1 =614.9, n1 =1430, p1= x1/n1=0.43
sample two, x2 =457.6, n2 =1430, p2= x2/n2=0.32
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 = [ (0.43-0.32) ± 1.96 * 0.018]
= [ 0.075 , 0.145 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 0.075 , 0.145] 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, 95% of these intervals will contains the difference between
true population mean P1-P2

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