The Genetics & IVF Institute conducted a clinical trial of the YSORT method designed to increase the probability of conceiving a boy. As of this writing, 291 babies were born to parents using the YSORT method, and 239 of them were boys. Use a 0.01 significance level to test the claim that the YSORT method is effective in increasing the likelihood that a baby will be a boy. What is the value of the test statistic needed to test this claim?

Answer: The test statistic needed to test this claim= 10.92Step-by-step explanation:We know that the probability of giving birth to a boy : p= 0.5i..e The population proportion of giving birth to a boy =  0.5As per given , we haveNull hypothesis : [tex]H_0: p\leq0.5[/tex]Alternative hypothesis :  [tex]H_a: p>0.5[/tex]Since [tex]H_a[/tex] is right-tailed , so the hypothesis test is a right-tailed z-test.Also, it is given that , the sample size : n= 291Sample proportion: [tex]\hat{p}=\dfrac{239}{291}\approx0.82[/tex]Test statistic : [tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex] , where n is sample size ,  [tex]\hat{p}[/tex] is sample proportion and p is the population proportion.[tex]\Rightarrow\ z=\dfrac{0.82-0.5}{\sqrt{\dfrac{0.5(1-0.5)}{291}}}\approx10.92[/tex]i.e. the test statistic needed to test this claim= 10.92Critical value ( one-tailed) for  0.01 significance level = [tex]z_{0.01}=2.326[/tex]Decision : Since Test statistic value (10.92)> Critical value (2.326), so we reject the null hypothesis .[When test statistic value is greater than the critical value , then we reject the null hypothesis.]Thus , we concluded that we have enough evidence at 0.01 significance level to support the claim that the YSORT method is effective in increasing the likelihood that a baby will be a boy.