Question

An SRS of n=57 in Community Y finds 17 smokers (??̂= 0.2982). The national average for smoking prevalence is 0.25. We want to test whether the proportion of smokers in Community Y is significantly difference than the national average at 95% confidence level. Assume that we can use a z-test

.Calculate a z statistics.

a. 0.84

b. 0.50

c. 0.24

Answer 1

Given that,
possibile chances (x)=17
sample size(n)=57
success rate ( p )= x/n = 0.3
success probability,( po )=0.25
failure probability,( qo) = 0.75
null, Ho:p=0.25  
alternate, H1: p!=0.25
level of significance, alpha = 0.05
from standard normal table, two tailed z alpha/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=0.29825-0.25/(sqrt(0.1875)/57)
zo =0.84
| zo | =0.84
critical value
the value of |z alpha| at los 0.05% is 1.96
we got |zo| =0.841 & | z alpha | =1.96
make decision
hence value of |zo | < | z alpha | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.84119 ) = 0.40024
hence value of p0.05 < 0.4002,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.25
alternate, H1: p!=0.25
test statistic: 0.84
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.40024
we do not have enough evidence to support the claim that The national average for smoking prevalence is 0.25
option:a