Question

In an opinion survey, a random sample of 1000 adults from the U.S.A. and 1000 adults from Germany were asked whether they supported the death penalty. 560 American adults and 590 German adults indicated that they supported the death penalty. Researcher wants to know if there is sufficient evidence to conclude that the proportion of adults who support the deal penally in the US is different than that in Germany. Use a confidence interval (at confidence interval of 98%) to answer. (Use USA-Germany)

1. What is the correct calculated test statistic?

2. What is the calculated 98% Confidence Interval?

Answer 1

1)
p1cap = X1/N1 = 560/1000 = 0.56
p1cap = X2/N2 = 590/1000 = 0.59
pcap = (X1 + X2)/(N1 + N2) = (560+590)/(1000+1000) = 0.575

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 = (0.56-0.59)/sqrt(0.575*(1-0.575)*(1/1000 + 1/1000))
z = -1.36

2)
Here, , n1 = 1000 , n2 = 1000
p1cap = 0.56 , p2cap = 0.59

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.56 * (1-0.56)/1000 + 0.59*(1-0.59)/1000)
SE = 0.0221

For 0.98 CI, z-value = 2.33
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.56 - 0.59 - 2.33*0.0221, 0.56 - 0.59 + 2.33*0.0221)
CI = (-0.0815 , 0.0215)