Question

An epidemiologist is studying cases of tuberculosis throughout the United States. In 2017, the national average for cases of tuberculosis was 2.8 per 100 people. Indiana and Illinois are not at the national average. Their data is in the table below.

	Indiana	Illinois
Number of TB Cases	15	26
Sample Size	1000	1000

Let p1 be the proportion of TB cases in Indiana and p2 be the proportion of TB cases in Illinois.

Is there enough evidence at α=0.01 to show Illinois had a different proportion of cases of tuberculosis than Indiana based on the 2017 data?

Find p̂ 1,p̂ 2, and p̂ pooled. If needed, round to three decimal places.

Answer 1

p1cap = X1/N1 = 15/1000 = 0.015

p2cap = X2/N2 = 26/1000 = 0.026

pcap = (X1 + X2)/(N1 + N2) = (15+26)/(1000+1000) = 0.021

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p1 = p2
Alternate Hypothesis, Ha: p1 ≠ p2

Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.015-0.026)/sqrt(0.021*(1-0.021)*(1/1000 + 1/1000))
z = -1.72

P-value Approach
P-value = 0.0854
As P-value >= 0.01, fail to reject null hypothesis.