Question

HYPOTHESIS TESTING FOR TWO PROPORTIONS

A nutritionist claims that the proportion of females who consume too much saturated fat is lower than the proportion of males who consume too much saturated fat. In interviews with 513 randomly selected, she determined that 300 consume too much saturated fat. In interviews with 564 randomly selected males, she determined that 391 consume too much saturated fat. Determine whether a lower proportion of females than males consume too much saturated fat alpha = 0.05 level of significance and state the conclusion

1. State the null and alternative hypothesis
2. Give a p-value
3. Give a conclusion for the hypothesis

Answer 1

Given that,
sample one, x1 =300, n1 =513, p1= x1/n1=0.585
sample two, x2 =391, n2 =564, p2= x2/n2=0.693
null, Ho: p1 = p2
alternate, H1: p1 < p2
level of significance, α = 0.05
from standard normal table,left tailed z α/2 =1.645
since our test is left-tailed
reject Ho, if zo < -1.645
we use test statistic (z) = (p1-p2)/√(p^q^(1/n1+1/n2))
zo =(0.585-0.693)/sqrt((0.642*0.358(1/513+1/564))
zo =-3.707
| zo | =3.707
critical value
the value of |z α| at los 0.05% is 1.645
we got |zo| =3.707 & | z α | =1.645
make decision
hence value of | zo | > | z α| and here we reject Ho
p-value: left tail - Ha : ( p < -3.7074 ) = 0.0001
hence value of p0.05 > 0.0001,here we reject Ho
ANSWERS
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null, Ho: p1 = p2
alternate, H1: p1 < p2
test statistic: -3.707
critical value: -1.645
decision: reject Ho
p-value: 0.0001
we have enough evidence to support the claim that whether a lower proportion of females than males consume too much saturated fat