A publisher reports that67% of their readers own a particular make of car. A mar

Question

A publisher reports that67% of their readers own a particular make of car. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage. A random sample of 150 found that 64% of the readers owned a particular make of car. Is there sufficient evidence at the 0.05 level to support the executive's claim? There is sufficient evidence to support the claim that the percentage of readers who own a particular make of car is different from 67%. There is not sufficient evidence to support the claim that the percentage of readers who own a particular make of car is different from 67%.

Explanation / Answer

Given that,
possibile chances (x)=96
sample size(n)=150
success rate ( p )= x/n = 0.64
success probability,( po )=0.67
failure probability,( qo) = 0.33
null, Ho:p=0.67
alternate, H1: p!=0.67
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.64-0.67/(sqrt(0.2211)/150)
zo =-0.7814
| zo | =0.7814
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =0.781 & | z | =1.96
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -0.7814 ) = 0.43457
hence value of p0.05 < 0.4346,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.67
alternate, H1: p!=0.67
test statistic: -0.7814
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.43457
We don't have evidence to support it is diffrent from the actual reported percentage

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