Question

A particular school district is concerned about the percentage of high schoolers who vape. In a random sample of 1478 high schoolers taken last year, 61 of them indicated on the anonymous survey that they vape. This year, the survey was given again and out of 1482 high schoolers, 181 of them indicated on the survey that they vape. Is there evidence that the vaping rate is now higher?

a) Test at the 5% level

b) Compute a 90% confidence interval for the difference in proportions.

Answer 1

a)
p1cap = X1/N1 = 61/1478 = 0.0413
p1cap = X2/N2 = 181/1482 = 0.1221
pcap = (X1 + X2)/(N1 + N2) = (61+181)/(1478+1482) = 0.0818

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 = (0.0413-0.1221)/sqrt(0.0818*(1-0.0818)*(1/1478 + 1/1482))
z = -8.02

P-value Approach
P-value = 0
As P-value < 0.05, reject the null hypothesis.

yes, now the proportion is high

b)
Here, , n1 = 1478 , n2 = 1482
p1cap = 0.0413 , p2cap = 0.1221

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.0413 * (1-0.0413)/1478 + 0.1221*(1-0.1221)/1482)
SE = 0.01

For 0.9 CI, z-value = 1.64
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.0413 - 0.1221 - 1.64*0.01, 0.0413 - 0.1221 + 1.64*0.01)
CI = (-0.0972 , -0.0644)