Question

A publisher reports that 63 % of their readers own a personal computer. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage. A random sample of 170 found that 59 % of the readers owned a personal computer. Make the decision to reject or fail to reject the null hypothesis at the 0.05 level. Answer 8 Points Keypad Keyboard Shortcuts Reject Null Hypothesis Fail to Reject Null Hypothesis Prev N

Answer 1

Given that,
possibile chances (x)=100.3
sample size(n)=170
success rate ( p )= x/n = 0.59
success probability,( po )=0.63
failure probability,( qo) = 0.37
null, Ho:p=0.63
alternate, H1: p!=0.63
level of significance, α = 0.05
from standard normal table, two tailed z α/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=0.59-0.63/(sqrt(0.2331)/170)
zo =-1.08
| zo | =1.08
critical value
the value of |z α| at los 0.05% is 1.96
we got |zo| =1.08 & | z α | =1.96
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 != -1.08022 ) = 0.28004
hence value of p0.05 < 0.28,here we do not reject Ho
ANSWERS
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null, Ho:p=0.63
alternate, H1: p!=0.63
test statistic: -1.08
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.28004
we donot have enough evidence to support the claim that the percentage is actually different from the reported percentage.