Question

I believe the proportions of individuals who own an Nintendo Switch is greater than those that own an Xbox One. A poll showed that out of 6,000 people, 3,560 out of 6,000 individuals owned a Nintendo Switch and 2,440 out of 6,000 individuals owned an Xbox One.

Test this data at a 90% confidence interval.

Make sure that you use appropriate terminology and specify whether you are using the classical method or the p-value method.

Answer 1

p1cap = X1/N1 = 3560/6000 = 0.5933
p1cap = X2/N2 = 2440/6000 = 0.4067
pcap = (X1 + X2)/(N1 + N2) = (3560+2440)/(6000+6000) = 0.5

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

Rejection Region
This is right tailed test, for α = 0.1
Critical value of z is 1.28.
Hence reject H0 if z > 1.28

Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.5933-0.4067)/sqrt(0.5*(1-0.5)*(1/6000 + 1/6000))
z = 20.441

P-value Approach
P-value = 0
As P-value < 0.1, reject the null hypothesis.