A medical research team studied the number of head and neck injuries sustained by hockey players. Of the 200 players who wore a full-faced shield, 40 sustained an injury. Of the 250 players who wore a half-faced shield, 60 sustained an injury. At á = 0.05, can you reject the claim that the proportion of the players sustaining head and neck injuries are the same for the two groups?

Question

A medical research team studied the number of head and neck injuries sustained by hockey players. Of the 200 players who wore a full-faced shield, 40 sustained an injury. Of the 250 players who wore a half-faced shield, 60 sustained  an injury. At á = 0.05, can you reject the claim that the proportion of the players sustaining head and neck injuries are the same for the two groups?

MathGuru · Answer

Use a formula for a binomial proportion two-sample z-test for this one since you have two groups.

Formula:
z = (p1 - p2)/√[pq(1/n1 + 1/n2)]
...note: 'n' represents the sample sizes, 'p' is (x1 + x2)/(n1 + n2), and 'q' is 1-p.

I'll get you started:
p = (40 + 60)/(200 + 250) = ? -->once you have the fraction, convert to a decimal (decimals are easier to use in the formula)
p1 = 40/200
p2 = 60/250
Convert all fractions to decimals. Plug those decimal values into the formula and find z. Once you have this value, compare to the critical or cutoff value you find in a z-table at 0.05 level of significance for a two-tailed test. If the test statistic exceeds the critical value in the table, reject the null and conclude a difference between the two groups. If the test statistic does not exceed the critical value from the table, you cannot reject the null and you cannot conclude a difference.

I hope this will help get you started.