Question

In a sample of 49 roads repaired with asphalt mix E, 25 showed damage within two years. In a sample of 38 roads repaired with asphalt mix F, 11 showed damage within two years.

a. Use your calculator and 0.01 significance to test the claim that the two-year damage rate is higher using mix E than mix F.

b. Use formula/chart to construct a 99% confidence interval estimate for the difference between the two-year damage rates of mix E and mix F.

Answer 1

a)

p1cap = X1/N1 = 25/49 = 0.5102
p1cap = X2/N2 = 11/38 = 0.2895
pcap = (X1 + X2)/(N1 + N2) = (25+11)/(49+38) = 0.4138

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 = 0.01
Critical value of z is 2.33.
Hence reject H0 if z > 2.33

Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.5102-0.2895)/sqrt(0.4138*(1-0.4138)*(1/49 + 1/38))
z = 2.07

P-value Approach
P-value = 0.0192
As P-value >= 0.01, fail to reject null hypothesis.

b)

Here, , n1 = 49 , n2 = 38
p1cap = 0.5102 , p2cap = 0.2895

Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.5102 * (1-0.5102)/49 + 0.2895*(1-0.2895)/38)
SE = 0.1025

For 0.99 CI, z-value = 2.58
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.5102 - 0.2895 - 2.58*0.1025, 0.5102 - 0.2895 + 2.58*0.1025)
CI = (-0.0438 , 0.4852)