Question

A study was conducted to determine the proportion of people who dream in black and white instead of color. Among 302 people over the age of​ 55, 79 dream in black and​ white, and among 307 people under the age of​ 25, 17 dream in black and white. Use a 0.01 significance level to test the claim that the proportion of people over 55 who dream in black and white is greater than the proportion for those under 25.

a) Identify the test statistic

Identify the P-Value

b) Test the claim by constructing an appropriate confidence interval

Answer 1

a)

p1cap = X1/N1 = 79/302 = 0.2616
p1cap = X2/N2 = 17/307 = 0.0554
pcap = (X1 + X2)/(N1 + N2) = (79+17)/(302+307) = 0.1576

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 = (0.2616-0.0554)/sqrt(0.1576*(1-0.1576)*(1/302 + 1/307))
z = 6.9826

P-value Approach
P-value = 0
As P-value < 0.01, reject the null hypothesis.

b)

Here, , n1 = 302 , n2 = 307
p1cap = 0.2616 , p2cap = 0.0554

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.2616 * (1-0.2616)/302 + 0.0554*(1-0.0554)/307)
SE = 0.0285

For 0.99 CI, z-value = 2.58
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.2616 - 0.0554 - 2.58*0.0285, 0.2616 - 0.0554 + 2.58*0.0285)
CI = (0.1327 , 0.2797)