3. A researcher collected data on a sample of 610 patients. The variable of inte

Question

3. A researcher collected data on a sample of 610 patients. The variable of interest was the percentage of subjects with impaired fasting glucose (IFG). IFG refers to a metabolic stage intermediate between normal glucose homeostasis and diabetes. In the study, 48 patients were classified in the IFG stage. An article cites population estimates for IFG among all patients in that area as 6.3 percent. Is there sufficient evidence to indicate that our sample comes from a population with a prevalence of IFG different from 6.3 percent? Use a level of significance of 0.01 to conduct the test, use the test statistic method, and report the p-value.

Explanation / Answer

Given that,
possibile chances (x)=48
sample size(n)=610
success rate ( p )= x/n = 0.0787
success probability,( po )=0.063
failure probability,( qo) = 0.937
null, Ho:p=0.063
alternate, H1: p!=0.063
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.58
since our test is two-tailed
reject Ho, if zo < -2.58 OR if zo > 2.58
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.07869-0.063/(sqrt(0.059031)/610)
zo =1.5948
| zo | =1.5948
critical value
the value of |z | at los 0.01% is 2.58
we got |zo| =1.595 & | z | =2.58
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.5948 ) = 0.11076
hence value of p0.01 < 0.1108,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.063
alternate, H1: p!=0.063
test statistic: 1.5948
critical value: -2.58 , 2.58
decision: do not reject Ho
p-value: 0.11076

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