It has been reported that the probability that an individual will develop schizo
ID: 3319250 • Letter: I
Question
It has been reported that the probability that an individual will develop schizophrenia over their lifetime is 0.004. In a random sample of 3000 individuals, it was determined that 17 developed schizophrenia. Is there evidence to support the claim that the true proportion of people who will develop schizophrenia is different from 0.004 at the 0.05 level of significance? (Source: Saha S, Chant D, Welham J, McGrath J. A systematic review of the prevalence of schizophrenia. PLOS Med 2(5): e141.)
State your null and alternative hypothesis:
Describe what you type into your calculator to have it run the numbers.
State your P-value
State your conclusion is a way a non-stat student would understand.
Explanation / Answer
Given that,
possibile chances (x)=17
sample size(n)=3000
success rate ( p )= x/n = 0.0057
success probability,( po )=0.004
failure probability,( qo) = 0.996
null, Ho:p=0.004
alternate, H1: p!=0.004
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=0.00567-0.004/(sqrt(0.003984)/3000)
zo =1.4463
| zo | =1.4463
critical value
the value of |z | at los 0.05% is 1.96
we got |zo| =1.446 & | 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 != 1.44627 ) = 0.1481
hence value of p0.05 < 0.1481,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.004
alternate, H1: p!=0.004
test statistic: 1.4463
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.1481
we don't have evidence to say that there is a way a non-stat student would understand
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