Question

Test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, and then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim. Among

2173

passenger cars in a particular region,

235

had only rear license plates. Among

335335

commercial trucks,

5151

had only rear license plates. A reasonable hypothesis is that commercial trucks owners violate laws requiring front license plates at a higher rate than owners of passenger cars. Use a

0.100.10

significance level to test that hypothesis.

a. Test the claim using a hypothesis test.

b. Test the claim by constructing an appropriate confidence interval.

b. Identify the confidence interval limits for the appropriate confidence interval. Let population 1 correspond to the

passenger carspassenger cars

and population 2 correspond to the

commercial truckscommercial trucks.

Let a success be a vehicle that only has a rear license plate.

nothingless than<p 1p1minus−p 2p2less than<nothing

(Round to four decimal places as needed.)

Because the confidence interval limits

▼

do not contain

contain

0, there

▼

is not

is

a significant difference between the two proportions. There

▼

is

is not

sufficient evidence to support the claim that commercial trucks owners violate laws requiring front license plates at a higher rate than owners of passenger cars.

Answer 1

Given that,
sample one, x1 =235, n1 =2173, p1= x1/n1=0.108
sample two, x2 =51, n2 =335, p2= x2/n2=0.152
finding a p^ value for proportion p^=(x1 + x2 ) / (n1+n2)
p^=0.114
q^ Value For Proportion= 1-p^=0.886
null, Ho: p1 = p2
alternate, H1: p1 < p2
level of significance, α = 0.1
from standard normal table,left tailed z α/2 =1.282
since our test is left-tailed
reject Ho, if zo < -1.282
we use test statistic (z) = (p1-p2)/√(p^q^(1/n1+1/n2))
zo =(0.108-0.152)/sqrt((0.114*0.886(1/2173+1/335))
zo =-2.363
| zo | =2.363
critical value
the value of |z α| at los 0.1% is 1.282
we got |zo| =2.363 & | z α | =1.282
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: left tail - Ha : ( p < -2.3634 ) = 0.00905
hence value of p0.1 > 0.00905,here we reject Ho
ANSWERS
---------------
null, Ho: p1 = p2
alternate, H1: p1 < p2
test statistic: -2.363
critical value: -1.282
decision: reject Ho
p-value: 0.00905
we have enough evidence to support the claim that commercial trucks owners violate laws requiring front license plates at a higher rate than owners of passenger cars

b.
TRADITIONAL METHOD
given that,
sample one, x1 =235, n1 =2173, p1= x1/n1=0.1081
sample two, x2 =51, n2 =335, p2= x2/n2=0.1522
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( (0.1081*0.8919/2173) +(0.1522 * 0.8478/335))
=0.0207
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, α = 0.1
from standard normal table,left tailed z α/2 =1.28
margin of error = 1.28 * 0.0207
=0.0265
III.
CI = (p1-p2) ± margin of error
confidence interval = [ (0.1081-0.1522) ±0.0265]
= [ -0.0706 , -0.0176]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample one, x1 =235, n1 =2173, p1= x1/n1=0.1081
sample two, x2 =51, n2 =335, p2= x2/n2=0.1522
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 = [ (0.1081-0.1522) ± 1.28 * 0.0207]
= [ -0.0706 , -0.0176 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ -0.0706 , -0.0176] 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