Question

The Genetics and IVF Institute conducted a clinical trial of the YSORT method designed to increase the probability that a baby is a boy. Among the babies born to parents in this trial, 172 were boys and 39 were girls. Use the sample data with a 0.01 significance level to test the claim that with the YSORT method, the probability of a baby being a boy is greater than 0.5.

Answer 1

One-Proportion Z test

The following information is provided: The sample size is N = 211, the number of favorable cases is X = 172 and the sample proportion is pˉ​=X/N​=172/211​=0.8152, and the significance level is α=0.01 (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. (2a) Critical Value Based on the information provided, the significance level is α=0.01, therefore the critical value for this Right-tailed test is Zc​=2.3263. This can be found by either using excel or the Z distribution table. (2b) Rejection Region The rejection region for this Right-tailed test is Z>2.3263 (3) Test Statistics The z-statistic is computed as follows: (4) The p-value The p-value is the probability of obtaining sample results as extreme or more extreme than the sample results obtained, under the assumption that the null hypothesis is true. In this case, the p-value is p =P(Z>9.1561)=0 (5) The Decision about the null hypothesis (a) Using traditional method Since it is observed that Z=9.1561 > Zc​=2.3263, it is then concluded that the null hypothesis is rejected. (b) Using p-value method 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. (6) 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 0.5, at the 0.01 significance level.