Since an instant replay system for tennis was introduced at a major tournament,
ID: 3227507 • Letter: S
Question
Since an instant replay system for tennis was introduced at a major tournament, men challenged 1415 referee calls, with the result that 429 of the calls were overturned. Women challenged 770 referee calls, and 221 of the calls were overturned. Use a 0.01 level to test the claim that men and women have equal success in challenging calls. Complete parts (a) through (c) below. a. Test the claim using a hypothesis test. Consider the first sample to be the sample of male tennis players who challenged referee calls and second sample to be the sample of female tennis players who challenged referee calls. What are the null and alternative hypothesis for the hypothesis test? A. H_0: p_1 = p_2 H_1: p_1 > p_2 B. H_0: p_1 notequalto p_2 H_1: p_1 = p_2 C. H_0: p_1 lessthanorequalto p_2 D. H_0: p_1 greaterthanorequalto p_2 H_1: p_1 notequalto p_2 E. H_0: p_1 = p_2 H_1: p_1Explanation / Answer
Given that,
sample one, x1 =429, n1 =1415, p1= x1/n1=0.303
sample two, x2 =221, n2 =770, p2= x2/n2=0.287
null, Ho: p1 = p2
alternate, H1: p1 != p2
level of significance, = 0.01
from standard normal table, two tailed z /2 =
since our test is two-tailed
reject Ho, if zo < -2.576 OR if zo > 2.576
we use test statistic (z) = (p1-p2)/(p^q^(1/n1+1/n2))
zo =(0.303-0.287)/sqrt((0.297*0.703(1/1415+1/770))
zo =0.79
| zo | =0.79
critical value
the value of |z | at los 0.01% is 2.576
we got |zo| =0.79 & | z | =2.576
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 0.7897 ) = 0.4297
hence value of p0.01 < 0.4297,here we do not reject Ho
ANSWERS
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null, Ho: p1 = p2
alternate, H1: p1 != p2
test statistic: 0.79
critical value: -2.576 , 2.576
decision: do not reject Ho
p-value: 0.4297
Confidence Interval for Diffrence of Proportion
CI = (p1 - p2) ± Z a/2 Sqrt(p1(1-p1)/n1 + p2(1-p2)/n2 )
Proportion 1
No. of chances( X1 )=429
No.Of Observed (n1)=1415
P1= X1/n1=0.303
Proportion 2
No. of chances(X2)=221
No.Of Observed (n2)=770
P2= X2/n2=0.287
C.I = (0.303-0.287) ±Z a/2 * Sqrt( (0.303*0.697/1415) + (0.287*0.713/770) )
=(0.303-0.287) ± 2.58* Sqrt(0)
=0.016-0.053,0.016+0.053
=[-0.036,0.069]