Question

Questions 23–33: Heart Attack Survival, revisited Some people who are having a heart attack do not experience chest pain, although most do. A study of people admitted to emergency rooms with heart attacks compared the death rates of people who had chest pains with those of people who did not have chest pains (Brieger et al. 2004). Of the 1,763 people who had heart attacks without chest pain, 229 died, while of the 19,118 people who had heart attacks with chest pain, 822 died. Assume that these are a random sample of all those having heart attacks coming to emergency rooms. To answer the following questions you may find it useful to build the two-way table for this study (that is, fill in the following). Chest Pain No Chest Pain Total Died Survived Total 23. What’s the marginal percentage of people who did not have chest pains?

24. What’s the conditional percentage of people who died given they had chest pains?

25. What is the sample proportion of those that died in the group that did not have chest pains?

26. Find the 95% confidence interval for the plus 4 sample proportion of those that died in the group that did not have chest pains (to two decimal places). You may assume that the critical value is 2 for this case.

27. It is typically believed that about 5% of those that have heart attacks die from them. You want to conduct an hypothesis test at the 95% to see if the data from the group that did not have chest pains is consistent with this value. What is the value of your test statistic to one decimal place?

28. Based on your test statistic from the last part, what conclusion do you draw? Now we want to formally compare the proportion of individuals dying between the two groups – those without chest pains and those with. We will step through a few calculations. We use a 99% confidence level for all calculations.

29. To start, are the requirements met for the large sample size proportion calculations in this case?

30. We think that the group without chest pains may be less likely to be identified and treated quickly and so may be at greater risk of death. What are the appropriate null and alternative hypotheses to represent this?

31. What is the appropriate test statistic corresponding to the hypothesis you stated in the last question?

32. Calculate the value of test statistic for these data to two decimal places.

33. At the 99% confidence level, what do you conclude?

Answer 1

first we will make a 2x2 contingency table

	Chest pain	No chest pain	total
died	822	229	1051
survived	18296	1534	19830
total	19118	1763	20881

23)marginal percentage for people who did not have chest pain= (1763/20881)*100=8.443082~8%

24) no.of people died given that they had chest pain=822

percentage of people died given that they had chest pain =(822/19118)*100=4.299613~4%

25)proportion of people who died given that they did not have chest pains=229/1763

26)here we have to find the confidence interval for the population proportion

the confidence interval for population proportion p is given by at 95% confidence interval

then here the level of confidence is 95% then

then

then the CI is given by

=(0.11,0.12)

27)here we have to test hypothesis of one proportion

here it is said that about 5% of those that have heart attacks die from them meaning the proportion is 0.05(even if we manipulate it by 5%*total no. of people died =5*1051/100=52.55

now the proportion is given by 52.55/1051=0.05)

then

H0:p=0.05 vs H1:p≠0.05

here p0=0.05

then the test statistic is given by

where = is the sample proportion =1763/20881=0.08

=30

now z score at 0.05 is 1.645<30

hence we reject our null hypothesis stating that both are not consistent

28)here we have to do the hypothesis for two proportions

the test statistic is given by

where =sample proportion of the thosedying without chest pain=229/1763=0.12

=sample proportion of those dying with chest pain=822/19118=0.04

  =229+822/20881=0.05

=80

z score at 0.01 is 2.33<80 hence we reject the null hypothesis stating that there is a significant difference in the two proportions

29)

identically independently distributed

   in probability as

30,31,32,33)we will do the chi square test of independence