Question

According to statistics reported on CNBC, a surprising number of motor vehicles are not covered by insurance. Simple results, consistent with the CNBC report, showed 49 of 216 vehicles were not covered by insurance. (See exercise 36 on page 372 of your textbook for a similar problem.) 

Question 2A Based on the data, can you say the proportion of vehicles not covered by insurance is different from 17% at α=0.1?

Question 2B Construct a 98% confidence interval for the true proportion of vehicles not covered by insurance.

Answer 1

2A.
Given that,
possibile chances (x)=49
sample size(n)=216
success rate ( p )= x/n = 0.227
success probability,( po )=0.17
failure probability,( qo) = 0.83
null, Ho:p=0.17  
alternate, H1: p>0.17
level of significance, alpha = 0.1
from standard normal table,right tailed z alpha/2 =1.282
since our test is right-tailed
reject Ho, if zo > 1.282
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.22685-0.17/(sqrt(0.1411)/216)
zo =2.224
| zo | =2.224
critical value
the value of |z alpha| at los 0.1% is 1.282
we got |zo| =2.224 & | z alpha | =1.282
make decision
hence value of | zo | > | z alpha| and here we reject Ho
p-value: right tail - Ha : ( p > 2.22437 ) = 0.01306
hence value of p0.1 > 0.01306,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.17
alternate, H1: p>0.17
test statistic: 2.224
critical value: 1.282
decision: reject Ho
p-value: 0.01306
Given that,
possibile chances (x)=49
sample size(n)=216
success rate ( p )= x/n = 0.227
success probability,( po )=0.17
failure probability,( qo) = 0.83
null, Ho:p=0.17  
alternate, H1: p!=0.17
level of significance, alpha = 0.1
from standard normal table, two tailed z alpha/2 =1.645
since our test is two-tailed
reject Ho, if zo < -1.645 OR if zo > 1.645
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.22685-0.17/(sqrt(0.1411)/216)
zo =2.224
| zo | =2.224
critical value
the value of |z alpha| at los 0.1% is 1.645
we got |zo| =2.224 & | z alpha | =1.645
make decision
hence value of | zo | > | z alpha| and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 2.22437 ) = 0.02612
hence value of p0.1 > 0.0261,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.17
alternate, H1: p!=0.17
test statistic: 2.224
critical value: -1.645 , 1.645
decision: reject Ho
p-value: 0.02612
we have enough evidence to support the claim

2B.
TRADITIONAL METHOD
given that,
possibile chances (x)=49
sample size(n)=216
success rate ( p )= x/n = 0.2269
I.
sample proportion = 0.2269
standard error = Sqrt ( (0.2269*0.7731) /216) )
= 0.0285
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, α = 0.02
from standard normal table, two tailed z α/2 =2.326
margin of error = 2.326 * 0.0285
= 0.0663
III.
CI = [ p ± margin of error ]
confidence interval = [0.2269 ± 0.0663]
= [ 0.1606 , 0.2931]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
possibile chances (x)=49
sample size(n)=216
success rate ( p )= x/n = 0.2269
CI = confidence interval
confidence interval = [ 0.2269 ± 2.326 * Sqrt ( (0.2269*0.7731) /216) ) ]
= [0.2269 - 2.326 * Sqrt ( (0.2269*0.7731) /216) , 0.2269 + 2.326 * Sqrt ( (0.2269*0.7731) /216) ]
= [0.1606 , 0.2931]
-----------------------------------------------------------------------------------------------
interpretations:
1. We are 98% sure that the interval [ 0.1606 , 0.2931] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 98% of these intervals will contains the true population proportion