Question

In a random sample of​ males, it was found that 26 write with their left hands and 216 do not. In a random sample of​ females, it was found that 64 write with their left hands and 455 do not. Use a 0.01 significance level to test the claim that the rate of​ left-handedness among males is less than that among females. Complete parts​ (a) through​ (c) below.

a. Test the claim using a hypothesis test.

b. Identify the test statistic

c. P-value

d. Confidence interval

Answer 1

a)
p1cap = X1/N1 = 26/242 = 0.1074
p1cap = X2/N2 = 64/519 = 0.1233
pcap = (X1 + X2)/(N1 + N2) = (26+64)/(242+519) = 0.1183

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

b)
Test statistic
z = (p1cap - p2cap)/sqrt(pcap * (1-pcap) * (1/N1 + 1/N2))
z = (0.1074-0.1233)/sqrt(0.1183*(1-0.1183)*(1/242 + 1/519))
z = -0.63

c)
P-value Approach
P-value = 0.2643
As P-value >= 0.01, fail to reject null hypothesis.

There is not sufficient evidence to conclude that the rate of​left-handedness among males is less than that among females.

d)
Standard Error, sigma(p1cap - p2cap),
SE = sqrt(p1cap * (1-p1cap)/n1 + p2cap * (1-p2cap)/n2)
SE = sqrt(0.1074 * (1-0.1074)/242 + 0.1233*(1-0.1233)/519)
SE = 0.0246

For 0.99 CI, z-value = 2.58
Confidence Interval,
CI = (p1cap - p2cap - z*SE, p1cap - p2cap + z*SE)
CI = (0.1074 - 0.1233 - 2.58*0.0246, 0.1074 - 0.1233 + 2.58*0.0246)
CI = (-0.0794 , 0.0476)