Question

Reduced Folate intake is a leading risk factor for birth defects. Researchers performed a dietary assessment of 35 adult women (older than 18 years old) whose household income was below the poverty level. Their mean intake of folate was 388 mcg per day with a standard deviation of 37 mcg/day. The USDA recommends 400 mcg\day of folate intake for women older than 18 years.

B.1) Is there significant evidence that the intake of women whose income is below the poverty level differs from the recommended amount? Justify your answer.

B.2) What is the sample size you would need to have 90% power to answer question 1?

Answer 1

1.
Given that,
population mean(u)=400
sample mean, x =388
standard deviation, s =37
number (n)=35
null, Ho: μ=400
alternate, H1: μ!=400
level of significance, alpha = 0.05
from standard normal table, two tailed t alpha/2 =2.032
since our test is two-tailed
reject Ho, if to < -2.032 OR if to > 2.032
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =388-400/(37/sqrt(35))
to =-1.919
| to | =1.919
critical value
the value of |t alpha| with n-1 = 34 d.f is 2.032
we got |to| =1.919 & | t alpha | =2.032
make decision
hence value of |to | < | t alpha | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -1.9187 ) = 0.0634
hence value of p0.05 < 0.0634,here we do not reject Ho
ANSWERS
---------------
null, Ho: μ=400
alternate, H1: μ!=400
test statistic: -1.919
critical value: -2.032 , 2.032
decision: do not reject Ho
p-value: 0.0634
we do not have there is significant evidence that
the intake of women whose income is below the poverty level differs from the recommended amount
so that there is possibility that type2 error because fail to reject the null hypothesis.

2.
power = 1- type2 error
type2 error = 1-0.9
type2 error =0.1
n= ((σ(Zalpha +Zbeta))/(U-Uo))^2
n =((37*(Z0.05+Z0.10)/(400-388))^2
n= ((37*(1.645+1.28))/(400-388))^2
n = 81.337
n= 81
sample size =81