Question

Since an instant replay system for tennis was introduced at a major​ tournament, men challenged 13861386 referee​ calls, with the result that 427427 of the calls were overturned. Women challenged 763763 referee​ calls, and 224224 of the calls were overturned. Use a 0.010.01 significance 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 the second sample to be the sample of female tennis players who challenged referee calls. What are the null and alternative hypotheses for the hypothesis​ test? A. Upper H 0H0​: p 1p1equals=p 2p2 Upper H 1H1​: p 1p1less than

p 2p2 E. Upper H 0H0​: p 1p1greater than or equals≥p 2p2 Upper H 1H1​: p 1p1not equals≠p 2p2 F. Upper H 0H0​: p 1p1not equals≠p 2p2 Upper H 1H1​: p 1p1equals=p 2p2 Identify the test statistic. zequals=nothing ​(Round to two decimal places as​ needed.) Identify the​ P-value. ​P-valueequals=nothing ​(Round to three decimal places as​ needed.) What is the conclusion based on the hypothesis​ test? The​ P-value is ▼ greater than less than the significance level of alphaαequals=0.010.01​, so ▼ fail to reject reject the null hypothesis. There ▼ is not sufficient is sufficient evidence to warrant rejection of the claim that women and men have equal success in challenging calls. b. Test the claim by constructing an appropriate confidence interval. The 9999​% confidence interval is nothingless than

Answer 1

Ho:   p1 - p2 =   0
Ha:   p1 - p2 ╪   0

sample #1   ----->   male          
first sample size,     n1=   1386          
number of successes, sample 1 =     x1=   427          
proportion success of sample 1 , p̂1=   x1/n1=   0.3081          
                  
sample #2   ----->   female          
second sample size,     n2 =    763          
number of successes, sample 2 =     x2 =    224          
proportion success of sample 1 , p̂ 2=   x2/n2 =    0.2936          
                  
difference in sample proportions, p̂1 - p̂2 =     0.3081   -   0.2936   =   0.0145
                  
pooled proportion , p =   (x1+x2)/(n1+n2)=   0.3029          
                  
std error ,SE =    =SQRT(p*(1-p)*(1/n1+ 1/n2)=   0.0207          
Z-statistic = (p̂1 - p̂2)/SE = (   0.015   /   0.0207   ) =   0.70

p-value =        0.4839 [excel formula =2*NORMSDIST(z)]      
decision :    p-value>α,Don't reject null hypothesis

fail to reject the null hypothesis

There ▼ is not sufficient evidence to warrant rejection of the claim that women and men have equal success in challenging calls

b)

level of significance, α =   0.01              
Z critical value =   Z α/2 =    2.576   [excel function: =normsinv(α/2)      
                  
Std error , SE =    SQRT(p̂1 * (1 - p̂1)/n1 + p̂2 * (1-p̂2)/n2) =     0.0206          
margin of error , E = Z*SE =    2.576   *   0.0206   =   0.0531
                  
confidence interval is                   
lower limit = (p̂1 - p̂2) - E =    0.015   -   0.0531   =   -0.0386
upper limit = (p̂1 - p̂2) + E =    0.015   +   0.0531   =   0.0676
                  
so, confidence interval is (   -0.0386   < p1 - p2 <   0.0676   )