Breast cancer is strongly influenced by a woman's reproductive history. In particular, the longer the length of time from the age at menarche (the age when menstruation begins) to the age at first childbirth, thegreater the risk is for breast cancer.A projection was made based on a mathematical model that the 30-year risk of a woman in the general U.S. population developing breast cancer from age 40 to age 70 is 7%. Suppose a special subgroup of fivehundred six 40-year-old women without breast cancer was studied whose age at menarche was 17 (compared with an average age at menarche of 13 in the general population) and age at first birth was 20(compared with an average age at first birth of 25 in the general population). These women were followed for development of breast cancer between ages 40 and 70. The study found that 17 of the womendeveloped breast cancer between age 40 and age 70.You can use SALT to answer parts of this question. Please note that the Inferential Statistics page does not use the continuity-corrected version of the test statistic.(a) Test the hypothesis that the underlying rate of breast cancer is the same or different in this group as in the general population. (Use a = 0.05.)Suppose 100 million women in the U.S. population have not developed breast cancer by the age of 40.What is your best estimate of the number of breast-cancer cases that would be prevented from age 40 to 70 if all women in the U.S. population reached menarche at age 17 and gave birth to their first childat age 20? (Round your answer to the nearest integer.)casesProvide a 95% CI for the number of breast-cancer cases prevented. (Enter your answer using interval notation. Round your numerical values to the nearest integer.)

a) From the given information, The percentage of women in the general U.S population developing breast cancer from age 40 to age 70 is 7%. Total number of women is 506. The study found that that 17 of the women developed breast cancer between age 40 and age 70. Let X=17 and n=506. p^=XN =17506 =0.0336 Null Hypothesis: H0:p=7%=0.07 Alternative Hypothesis: H1:p≠0.07 Test statistic: z=p^-pp1-pn =0.0336-0.070.071-0.07506 =-3.2091 p value: p=2Pz≤test statistic since tail of the test is two tailed =2Pz≤-3.2091 =2*0.0007, from the excel function, =NORM.DIST(-3.2091,0,1,TRUE) =0.0014 Decision Rule: If p-value ≤ α, then reject the null hypothesis. If p-value > α, then do not reject the null hypothesis. Conclusion: Let us consider the level of significance is α=0.05. Here, the p-value is lesser than the level of significance. From the rejection rule, reject the null hypothesis. The percentage of women in the general U.S population developing breast cancer from age 40 to age 70 is not 7%.Given that 100 million women in the U.S population have not developed breast cancer by the age of 40. Suppose a special subgroup of five hundred six 40 year old women without breast cancer was studied whose age at menarche was 17 and age at first birth was 20. From the study, the sample proportion of developing breast cancer is 0.0336. The sample proportion of not developing breast cancer is 1-0.0336=0.9664 The best estimate of the number of breast-cancer cases that would be prevented from age 40 to 70 if all women in the U.S population menarche at age 17 and gave birth to their first child at age 20 is, 100 million*0.9664=96.64 million ≈97 million cases. 95% confidence interval for the proportion of breast-cancer cases prevented is, p^±zcriticalp^1-p^n=0.9664±1.960.96641-0.9664506, 95% z critical value=1.96 =0.9664±0.0157 =0.9507, 0.9821 95% confidence interval for the number of breast-cancer cases prevented: 0.9507*100 million, 0.9821*100 million=95.07 million, 98.21 million ≈95 million, 98 million