The national tobacco use statistics indicates that 23.1% of male population smok
ID: 3218348 • Letter: T
Question
The national tobacco use statistics indicates that 23.1% of male population smoke cigarettes. In a sample of n- 825 of IUPUI male students the proportion who smoked was p^- = 0.204. Does the sample provide significant evidence that the proportion of male IUPUI students who smoke is different from the national rate? State the null and alternative hypotheses and compute the test statistic. Use the 5% level of significance to determine the critical value. TS = 2.07 DNR H_0. Conclude that proportion IUPUI male students who smoke is not different from the national rate. B 2.07 Reject H_0 Conclude that proportion IUPUI male students who smoke is different from the national rate. C. 1.84 DNR H_0. Conclude that proportion IUPUI male students who smoke is not different from the national rate. D. 1.84 Reject H_0. Conclude that proportion IUPUI male students who smoke is different from the national rate.Explanation / Answer
Given that,
possibile chances (x)=168.3
sample size(n)=825
success rate ( p )= x/n = 0.204
success probability,( po )=0.231
failure probability,( qo) = 0.769
null, Ho:p=0.231
alternate, H1: p!=0.231
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=0.204-0.231/(sqrt(0.177639)/825)
zo =-1.84
| zo | =1.84
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =1.84 & | 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 != -1.84002 ) = 0.06577
hence value of p0.05 < 0.0658,here we do not reject Ho
ANSWERS
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null, Ho:p=0.231
alternate, H1: p!=0.231
test statistic: -1.84
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.06577