A medical research team studied the number ot head and neck

A medical research team studied the number ot head and neck injunies sustained by hokey players. Of the 319 players who wore a full-face shield, 195 sustained an injury. Of the 323 players who wore a hald-face shield, 204 sustained an injury. At -0.10, can you reject the claim that the proportions of players sustaining head and neck injuries are the same for the two groups? Interpret the decision in the context of the original claim. Do the 8 steps of hypothesis testing, like outlined below and found in the similar examples on blackboard. Use technology(like STATCRUNCH) to generate your results, like P-value. A Identify the population parameters being discussed above. (use words to describe them) B. Write a Hypothesis test using those variables. Write the word \"claim\" next to one statement. c List everything that is given, including the Alpha, and sample size. Are the two samples dependent or independent? D. Decide what test to use, Z-test, T-test, or Chi-squared Test. Explain how you derived to this distribution. E. Decide if this is a Left-tailed Test, Right-Tailed Test, or Two-Tailed Test. Draw it with correct shaded area and test statistics. F Use the Table above and Find the Test Statistic and P- value G. Decide if to Reject Ho or to Fail to Reject Ho. Explain, using alpha

Solution

A)

Sample 1 : n1 = 319 ,x1 =195 , p1 = 195/319 = 0.61

Sample 1 : n1 = 323 ,x1 =204 , p1 = 204/323 = 0.63

B)

Null Hypothesis : p1=p2

Alternate hypothesis : p1 != p2

C)

Sample 1 : n1 = 319 ,x1 =195 , p1 = 195/319 = 0.61

Sample 1 : n1 = 323 ,x1 =204 , p1 = 204/323 = 0.63

alpha = 0.1

These samples are independent.

D)

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.

E)

Its a two-tailed test because we are testing for the equality condition.

F)

First of all proportion p = (195 + 204)/(319+323) = 0.62

z = (0.61 - 0.63)/(0.62*0.38*(1/319 + 1/323)) = -0.52

P-value = 2*0.3015 = 0.6030 Answer

G)

Since 0.6>0.1 , failed to reject null hypothesis.