Question

Researchers claim that the birth rate in Bonn, Germany is more than the national average. A sample of 1200 Bonn residents produced 12 births in 2018, whereas a sample of 1000 people from all over Germany had 8 births the same year. Test this claim at the 0.05 level of significance.

Answer 1

p1cap = X1/N1 = 12/1200 = 0.01
p1cap = X2/N2 = 8/1000 = 0.008
pcap = (X1 + X2)/(N1 + N2) = (12+8)/(1200+1000) = 0.0091

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 = (0.01-0.008)/sqrt(0.0091*(1-0.0091)*(1/1200 + 1/1000))
z = 0.49

P-value Approach
P-value = 0.3121
As P-value >= 0.05, fail to reject null hypothesis.

The birth rate in Bonn is not more than national birth rate