A new cream that advertises that it can reduce wrinkles and improve skin was sub
ID: 3181673 • Letter: A
Question
A new cream that advertises that it can reduce wrinkles and improve skin was subject to a recent study. A sample of 42 women over the age of 50 used the new cream for 6 months. Of those 42 women, 30 of them reported skin improvement(as judged by a dermatologist). Is this evidence that the cream will improve the skin of more than 50% of women over the age of 50? Test using =0.05.
(a) Test statistic: z=
(b) Critical Value: z=
(c) The final conclusion is
A. There is not sufficient evidence to reject the null hypothesis that p=0.5. That is, there is not sufficient evidence to reject that the cream can improve the skin of more than 50% of women over 50.
B. We can reject the null hypothesis that p=0.5 and accept that p>0.5. That is, the cream can improve the skin of more than 50% of women over 50.
Explanation / Answer
Given that,
possibile chances (x)=30
sample size(n)=42
success rate ( p )= x/n = 0.7143
success probability,( po )=0.5
failure probability,( qo) = 0.5
null, Ho:p=0.5
alternate, H1: p>0.5
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.64
since our test is right-tailed
reject Ho, if zo > 1.64
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.71429-0.5/(sqrt(0.25)/42)
zo =2.7775
| zo | =2.7775
critical value
the value of |z | at los 0.05% is 1.64
we got |zo| =2.777 & | z | =1.64
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: right tail - Ha : ( p > 2.77746 ) = 0.00274
hence value of p0.05 > 0.00274,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.5
alternate, H1: p>0.5
test statistic: 2.7775
critical value: 1.64
decision: reject Ho
p-value: 0.00274
no sufficient evidence to reject that the cream can improve the skin of more than 50% of women.
we can't reject
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