Question

Problem #1: (a) Suppose that we identify 161 women 50 to 54 years of age who have both a mother and a sister with a history of breast cancer. 16 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 #1(a): 0.1480 p-value (correct to 4 decimals) (A) We cannot conclude that having a sister with the disease increases the risk since the p-value is less than .01 (B) We cannot conclude that having a sister with the disease increases the risk since the p-value is greater 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 conclude that having a sister with the disease increases the risk since the p-value is less than .01 (E) We cannot conclude that having a sister with the disease increases the risk since the p-value is greater or equal to .01 (F) We conclude that having a sister with the disease increases the risk since the p-value is greater than or equal to .01 (G) We conclude that having a sister with the disease increases the risk since the p-value is greater than or equal to 0.02 (H) We conclude that having a sister with the disease increases the risk since the p-value is less than 0.02 Problem #1(b): D conclusion Just Save Submit Problem #1 for Grading Attempt #3 1(a) Problem #1 Attempt #1 Your Answer: 1(a) 0.1492 1(b) H Your Mark: 1(a) 0/2X 1(b) 0/2x Attempt #2 1(a) 0.1480 1(b) D 1(a) 0/2X 1(b) 0/2X 1(b) | 1(a)

Answer 1

a)

Below are the null and alternative Hypothesis,
Null Hypothesis: p = 0.08
Alternative Hypothesis: p > 0.08

Rejection Region
This is right tailed test, for α = 0.02
Critical value of z is 2.05.
Hence reject H0 if z > 2.05

Test statistic,
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.0994 - 0.08)/sqrt(0.08*(1-0.08)/161)
z = 0.91

P-value Approach
P-value = 0.1814
As P-value >= 0.02, fail to reject null hypothesis.

b)

Option B)