Question

An article on softball injuries provided a comparison of breakaway bases (designed to reduce injuries) and stationary bases. Consider the accompanying data (which agree with summary values given in the paper).

	Number of Games Played	Number of Games Where a Player Suffered a Sliding Injury
Stationary Bases	1250	92
Breakaway Bases	1250	19

(a)

Is the proportion of games with a player suffering a sliding injury significantly lower for games using breakaway bases? Answer by performing a level 0.01 test. (Use p_Stationary − p_Breakaway. Round your test statistic to two decimal places and your P-value to four decimal places.)

z=

p-value=

(b) which of the following is correct?

(1)We need to assume that the 2500 games were randomly assigned to the two treatments, so that the results are representative of all softball games. It seems likely that treatments would be assigned by league or region for consistency, so it is unlikely that the results are representative of all softball games.

(2)We need to assume that the 2500 games were randomly assigned to the two treatments, so that the results are representative of all softball games. It seems likely that treatments would vary on most softball fields, so it is unlikely that the results are representative of all softball games.

(3)We need to assume that the 2500 games were randomly assigned to the two treatments, so that the results are representative of all softball games. It seems likely that treatments would vary on most softball fields, so it seems reasonable that the results are representative of all softball games.

(4)We need to assume that the 2500 games were randomly assigned to the two treatments, so that the results are representative of all softball games. It seems likely that treatments would be assigned by league or region for consistency, so it seems reasonable that the results are representative of all softball games.

Answer 1

x1                =	92	x2                =	19
p̂1=x1/n1 =	0.0736	p̂2=x2/n2 =	0.0152
n1                       =	1250	n2                       =	1250
estimated prop. diff =p̂1-p̂2    =		0.0584
pooled prop p̂ =(x1+x2)/(n1+n2)=		0.0444
std error Se=√(p̂1*(1-p̂1)*(1/n1+1/n2) =		0.0082
test stat z=(p̂1-p̂2)/Se =		7.09 please try 7.12 if this comes wrong)
P value   =		0.0000

(1)We need to assume that the 2500 games were randomly assigned to the two treatments, so that the results are representative of all softball games. It seems likely that treatments would be assigned by league or region for consistency, so it is unlikely that the results are representative of all softball games.