Question

2 Sociologists and sports psychologists have noticed that in many sports, the athletes who make it to the professional level in their respective sport were the older participants in the sport as a youth. That is, the athlete’s birthdate is early in the “sport year”. This phenomenon has been noticed around the world in various sports such as soccer (Europe), hockey (Canada), and baseball (USA). In youth baseball leagues, the “sport year” begins on August 1, so children born in August are the older participants. In a random sample of 230 major league players born since 1975, 35 were born in the month of August.

A. Construct the 98% confidence interval for the true proportion of professional baseball players with an August birth month.

B. National demographic statistics show that 9% of all births in the general population occur in August. Is there a higher percentage of professional baseball players born in August than in the general population?

C. If the researchers want to cut the margin of error in half for part A, how many athletes should they survey?

Answer 1

Solution-A;

sample proportion for the true proportion of professional baseball players with an August birth month.

=p^=x/n=35/230= 0.1521739

z critical for 98%==NORM.S.INV(0.99)=2.326347874

98% confidence interval for p is

p^-z*sqrt(p^*(1-p^)/n,p^+z*sqrt(p^*(1-p^)/n

0.1521739-2.326347874*sqrt(0.1521739*(1-0.1521739)/230),0.1521739+2.326347874*sqrt(0.1521739*(1-0.1521739)/230)

0.09707611, 0.2072717

0.09707611<p<0.2072717

we are  98% confident that the true proportion of professional baseball players with an August birth month lies in between 0.09707611 and 0.2072717

98 % confidence interval for p is

(0.09707611 ,0.2072717)

Solution-b"

Ho:p=0.09

Ha:p>0.09

since this true popualtion proportion does not lie in the

98% confidence interval range( 0.09707611 ,0.2072717)

Reject null hypothesis

Accept alternative hypothesis

There is suffcient statistical evidence at 2% level of signifcance to conclude that

higher percentage of professional baseball players born in August than in the general population

Solution-c:

margin of error=2.326347874*sqrt(0.1521739*(1-0.1521739)/230)/2=0.0275489

sample proportion=0.1521739

sample size

n=(Z/E)^2*p^*(1-p^)

=(2.326347874/0.0275489)^2*0.1521739*(1-0.1521739)

=919.9998

Required sample size,n=920