Question

Most married couples have two or three personality preferences in common. A random sample of 375 married couples and found that 132 had three preferences in common. Another random sample of 571 couples showed that 237 had two personality preferences in common. Let ?1 be the population proportion of all married couples who have three personality preferences in common. Let ?2 be the population proportion of all married couples who have two personality preferences in common.

a) Can a normal distribution be used to approximate the ?̂ − ?̂ distribution? Explain.

b) Find a 99% confidence interval for ?1 − ?2. Show all work that leads to your answer

c) Examine the confidence interval in part (b) and explain what it means in the context of this problem. Does the confidence interval contain all positive, all negative, or both positive and negative numbers? What does this tell you about the proportion of married couples with three personality preferences in common compared with the proportion of couples with two preferences in common (at the 99% confidence level)?

d) Suppose a marriage counselor believes that the proportion of married couples with three personality preferences in common is significantly different from the known value of 35%. A survey randomly given to her current and former clients found 130 of 432 couples who shared three personality preferences. Does this indicate that the proportion is significantly different (either higher or lower)? Use a 1% level of significance to test her claim and show all work.

Answer 1

a.
proportion is normally distributed
b.
TRADITIONAL METHOD
given that,
sample one, x1 =132, n1 =375, p1= x1/n1=0.352
sample two, x2 =237, n2 =571, p2= x2/n2=0.4151
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( (0.352*0.648/375) +(0.4151 * 0.5849/571))
=0.0321
II.
margin of error = Z a/2 * (standard error)
where,
Za/2 = Z-table value
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.58
margin of error = 2.58 * 0.0321
=0.0829
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.352-0.4151) ±0.0829]
= [ -0.146 , 0.0199]
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DIRECT METHOD
given that,
sample one, x1 =132, n1 =375, p1= x1/n1=0.352
sample two, x2 =237, n2 =571, p2= x2/n2=0.4151
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 = [ (0.352-0.4151) ± 2.58 * 0.0321]
= [ -0.146 , 0.0199 ]
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interpretations:
1) we are 99% sure that the interval [ -0.146 , 0.0199] 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, 99% of these intervals will contains the difference between
true population mean P1-P2
c.
proportion of married couples with three personality preferences in common compared with the proportion of couples with two common preferences.
positive and negative numbers.
d.
Given that,
possibile chances (x)=130
sample size(n)=432
success rate ( p )= x/n = 0.3009
success probability,( po )=0.35
failure probability,( qo) = 0.65
null, Ho:p=0.35
alternate, H1: p!=0.35
level of significance, α = 0.01
from standard normal table, two tailed z α/2 =2.58
since our test is two-tailed
reject Ho, if zo < -2.58 OR if zo > 2.58
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.30093-0.35/(sqrt(0.2275)/432)
zo =-2.1385
| zo | =2.1385
critical value
the value of |z α| at los 0.01% is 2.58
we got |zo| =2.138 & | z α | =2.58
make decision
hence value of |zo | < | z α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -2.13847 ) = 0.03248
hence value of p0.01 < 0.0325,here we do not reject Ho
ANSWERS
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null, Ho:p=0.35
alternate, H1: p!=0.35
test statistic: -2.1385
critical value: -2.58 , 2.58
decision: do not reject Ho
p-value: 0.03248
we do not have enough evidence to support the claim that proportion of married couples with three personality preferences in common is significantly
different from the known value is 35%