4. An entrepreneur wants to provide pet care while her clients are on vacation.
ID: 3325131 • Letter: 4
Question
4. An entrepreneur wants to provide pet care while her clients are on vacation. As part of her market research, she interviews 80 people, chosen randomly from among the pet owners in her community, to ask whether they travel frequently. Twenty-eight respondents say that they travel frequently, while the other 52 say they do not travel frequently. (a) (+2) Construct a two-sided 95% confidence interval for the fraction of pet owners that travel frequently (b) (+3) Suppose that among the population of people without pets, 45% travel frequently. Is there sufficient evidence to conclude that pet owners travel less frequently than people without pets? Perform the hypothesis test at the = .05 level. What is the p-value associated with this test?Explanation / Answer
PART A.
TRADITIONAL METHOD
given that,
possibile chances (x)=28
sample size(n)=80
success rate ( p )= x/n = 0.35
I.
sample proportion = 0.35
standard error = Sqrt ( (0.35*0.65) /80) )
= 0.0533
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 * 0.0533
= 0.1045
III.
CI = [ p ± margin of error ]
confidence interval = [0.35 ± 0.1045]
= [ 0.2455 , 0.4545]
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DIRECT METHOD
given that,
possibile chances (x)=28
sample size(n)=80
success rate ( p )= x/n = 0.35
CI = confidence interval
confidence interval = [ 0.35 ± 1.96 * Sqrt ( (0.35*0.65) /80) ) ]
= [0.35 - 1.96 * Sqrt ( (0.35*0.65) /80) , 0.35 + 1.96 * Sqrt ( (0.35*0.65) /80) ]
= [0.2455 , 0.4545]
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interpretations:
1. We are 95% sure that the interval [ 0.2455 , 0.4545] contains the true population proportion
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 true population proportion
PART B.
Given that,
possibile chances (x)=28
sample size(n)=80
success rate ( p )= x/n = 0.35
success probability,( po )=0.45
failure probability,( qo) = 0.55
null, Ho:p=0.45
alternate, owners travel less frequencey than people without pets H1: p<0.45
level of significance, = 0.05
from standard normal table,left tailed z /2 =1.64
since our test is left-tailed
reject Ho, if zo < -1.64
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.35-0.45/(sqrt(0.2475)/80)
zo =-1.7979
| zo | =1.7979
critical value
the value of |z | at los 0.05% is 1.64
we got |zo| =1.798 & | z | =1.64
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: left tail - Ha : ( p < -1.79787 ) = 0.0361
hence value of p0.05 > 0.0361,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.45
alternate, H1: p<0.45
test statistic: -1.7979
critical value: -1.64
decision: reject Ho
p-value: 0.0361
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