Question

In: Statistics and Probability

Suppose the incidence rate of myocardial infarction (MI) was 5 per 1000 among 45- to 54-year-old...

Suppose the incidence rate of myocardial infarction (MI)
was 5 per 1000 among 45- to 54-year-old men in 2000.
To look at changes in incidence over time, 5000 men in this
age group were followed for 1 year starting in 2010. Fifteen
new cases of MI were found.

7.12 Using the critical-value method with α = .05, test the
hypothesis that incidence rates of MI changed from 2000
to 2010.


7.13 Report a p-value to correspond to your answer to
Problem 7.12.


Suppose that 25% of patients with MI in 2000 died within
24 hours. This proportion is called the 24-hour case-fatality
rate.


7.14 Of the 15 new MI cases in the preceding study,
5 died within 24 hours. Test whether the 24-hour case fatality
rate changed from 2000 to 2010.

7.15 Suppose we eventually plan to accumulate 50 MI
cases during the period 2010–2015. Assume that the
24-hour case-fatality rate is truly 20% during this period.
How much power would such a study have in distinguishing
between case-fatality rates in 2000 and 2010–2015

Solutions

Expert Solution

Test and CI for One Proportion (without normality assumption)

Test of p = 0.005 vs p not = 0.005


Exact
Sample X N Sample p 95% CI P-Value
1 15 5000 0.003000 (0.001680, 0.004943) 0.036


Test and CI for One Proportion (using normality assumption)

Test of p = 0.005 vs p not = 0.005


Sample X N Sample p 95% CI Z-Value P-Value
1 15 5000 0.003000 (0.001484, 0.004516) -2.01 0.045

Since both situations, p-value<0.05 so we reject null hypothesis at 5% level of significance and conclude that incidence rates of MI changed from 2000 to 2010 significantly.

7.14:

Let p=true proportion of patient with MI died within 24 hours.

Test and CI for One Proportion (without normality assumption)

Test of p = 0.25 vs p not = 0.25


Exact
Sample X N Sample p 95% CI P-Value
1 5 15 0.333333 (0.118241, 0.616196) 0.550

Test and CI for One Proportion (using normality assumption)

Test of p = 0.25 vs p not = 0.25


Sample X N Sample p 95% CI Z-Value P-Value
1 5 15 0.333333 (0.094774, 0.571893) 0.75 0.456

Using the normal approximation.
From above we see that both p-values>0.05 so we fail to reject null hypothesis at 5% level of significance and conclude that the  24-hour case fatality rate is not changed significantly from 2000 to 2010.



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