Problem #5 : (a) Suppose that we identify 165 women 50 to 54 years of age who ha

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Problem #5 : (a) Suppose that we identify 165 women 50 to 54 years of age who have both a mother and a sister with a history of breast cancer. 18 of these women themselves have developed breast cancer at some time in their lives. If we assume that the proportion of breast cancer cases in women whose mothers have had breast cancer is 8%. does having a sister with the disease increase the risk? Find the p-value (b) At the 2% significance level, what is the conclusion of the above hypothesis test? Problem #5(a): p-value (correct to 4 decimals) (A) We conclude that having a sister with the disease increases the rislk since the p-value is greater than or equal to.01 (B) We conclude that having a sister with the disease increases the risk since the p-value is greater than or equal to 0.02 (C) We cannot conclude that having a sister with the disease increases the risk since the p-value is less than 0.02 (D) We cannot conclude that having a sister with the disease increases the risk since the p-value is greater or equal to .01 (E) We cannot conclude that having a sister with the disease increases the risk since the p-value is less than .01 (F) We conclude that having a sister with the disease increases the risk since the p-value is less than 0.02 (G) We cannot conclude that having a sister with the disease increases the risk since the p-value is greater or equal to 0.02 (H) We conclude that having a sister with the disease increases the rislk since the p-value is less than .01 Problem #5(b): Select conclusion Just Save Submit Problem #5 for Grading

Explanation / Answer

Given that,
possibile chances (x)=18
sample size(n)=165
success rate ( p )= x/n = 0.1091
success probability,( po )=0.08
failure probability,( qo) = 0.92
null, Ho:p=0.08  
alternate, H1: p>0.08
level of significance, = 0.02
from standard normal table,right tailed z /2 =2.05
since our test is right-tailed
reject Ho, if zo > 2.05
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.10909-0.08/(sqrt(0.0736)/165)
zo =1.3774
| zo | =1.3774
critical value
the value of |z | at los 0.02% is 2.05
we got |zo| =1.377 & | z | =2.05
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: right tail - Ha : ( p > 1.3774 ) = 0.08419
hence value of p0.02 < 0.08419,here we do not reject Ho
ANSWERS
---------------
null, Ho:p=0.08
alternate, H1: p>0.08
test statistic: 1.3774
critical value: 2.05
decision: do not reject Ho
p-value: 0.08419

we cannot conclude that having a sister with the disease increases the
risk since the p value greatet than equal to 0.02

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