. A randomized study was performed to determine whether a diet of processed meats causes a higher proportion of incidents of breast cancer. Researchers compared high intake versus low intake diets of processed meats for women with breast cancer and found the following results shown in the two-way table below: Breast CancerNo Breast CancerTotalHigh Intake processed meat93157250Low Intake processed meat67163230Conduct a hypothesis test for this problem, using for the proportion with a diet high in processed meat who got breast cancer and for the proportion of women with a diet low in processed meat who got breast cancerState the hypotheses. State sample statistic. Find the p-value. Write a generic conclusion. Write a conclusion in context. If we compute a randomization statistic and assume the null hypothesis is true, what does that mean in the context of this problem? Where would the randomization distribution be centered for a hypothesis test for this problem? Where would a bootstrap distribution be centered for a confidence interval for this problem?

Suppose, p1 be the population proportion of a diet high in processed meat who got breast cancer. p2 be the population proportion of a diet low in processed meat who got breast cancer. 1. The test is to check the compare the proportion with a diet high in processed meat who got breast cancer and for the proportion of women with a diet low in processed meat who got breast cancer. The null hypothesis of this test is H0: p1=p2 and the alternative hypothesis of this test is Ha: p1≠p2 2. The sample statistics of this test is z=p^1-p^2p^1-p^1n+1m,where, p^1 is the sample proportion of a diet high in processed meat who got breast cancer. p^2 is the sample proportion of a diet low in processed meat who got breast cancer.n=first sample size,m=second sample size,p^=n p^1+m p^2n+m. Given that, n=250, m=230. p^1=93250, p^2=67230,p^=93+67250+230=160480=13. Then the sample statistics value is z=p^1-p^2p^1-p^1n+1m,=93250-67230131-131250+1230=1.8736 Therefore, the sample statistics value is 1.8736.3. The p-value of the test is p=2PZ>z, Z~aNormal(0,1)=2PZ>1.8736=21-PZ≤1.8736=2×1-0.9695=2×0.0305=0.061. Therefore, the p-value of the test is 0.061. Note: Since you have posted a question with multiple sub-parts, we will solve first three sub-parts for you. To get remaining sub-part solved please repost the complete question and mention the sub-parts to be solved.

Answered: A randomized study was performed to…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

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. A randomized study was performed to determine whether a diet of processed meats causes a higher proportion of incidents of breast cancer. Researchers compared high intake versus low intake diets of processed meats for women with breast cancer and found the following results shown in the two-way table below:

	Breast Cancer	No Breast Cancer	Total
High Intake processed meat	93	157	250
Low Intake processed meat	67	163	230

Conduct a hypothesis test for this problem, using for the proportion with a diet high in processed meat who got breast cancer and for the proportion of women with a diet low in processed meat who got breast cancer
State the hypotheses.

State sample statistic.

Find the p-value.

Write a generic conclusion.

Write a conclusion in context.

If we compute a randomization statistic and assume the null hypothesis is true, what does that mean in the context of this problem?

Where would the randomization distribution be centered for a hypothesis test for this problem? Where would a bootstrap distribution be centered for a confidence interval for this problem?

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