Question

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

Calculate np and nq

If we assume the selected sample is a simple random sample, then all of the requirements, in this case, are met to conduct the hypothesis test. True or False

Answer 1

The following information is provided: The sample size is N=291, the number of favorable cases is X=239, and the sample proportion is pˉ= X/N= 239/291=0.8213, and the significance level is α=0.01

np= 291*0.8213= 238.9983>10

nq= 291*0.1787=52.0017>10
(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:p=0.5
Ha:p>0.5
This corresponds to a right-tailed test, for which a z-test for one population proportion needs to be used.

(2) Rejection Region

Based on the information provided, the significance level is α=0.01, and the critical value for a right-tailed test is zc=2.33.

The rejection region for this right-tailed test is R={z:z>2.33}
(3) Test Statistics

The z-statistic is computed as follows:

(3) Test Statistics

The z-statistic is computed as follows:

(4) Decision about the null hypothesis

Since it is observed that z=10.962>zc=2.33, it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value is p=0, and since p=0<0.01, it is concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population proportion p is greater than p0, at the α=0.01 significance level.

If we assume the selected sample is a simple random sample, then all of the requirements, in this case, are met to conduct the hypothesis test. True