Question

Self-reported injuries among left- and right-handed people were compared in a survey of 1896 college students in British Columbia, Canada. Of the 180 left-handed students, 93 reported at least one injury. In the same period, 619 of the 1716 right-handed students reported at least one injury. Conduct a significance test to determine if the proportions of left-handed injured students is equal to the proportion of right-handed injured students. State clearly your conjecture, null and alternative hypotheses, test statistic, p-value, test decision, and conclusion for this significance test. Use an α = 0.05 level of significance for this test.

Answer 1

p1cap = X1/N1 = 93/180 = 0.5167
p1cap = X2/N2 = 619/1716 = 0.3607
pcap = (X1 + X2)/(N1 + N2) = (93+619)/(180+1716) = 0.3755

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 = (0.5167-0.3607)/sqrt(0.3755*(1-0.3755)*(1/180 + 1/1716))
z = 4.11

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

There is sufficient evidence to conclude that the proportions of left-handed injured students is equal to the proportion of right-handed injured students.