You can try a proportional one-sample z-test for this one since this problem is using proportions.
Here's a few hints to get you started:
Null hypothesis:
Ho: p = .5 -->meaning: population proportion is equal to .5
Alternative hypothesis:
Ha: p > .5 -->meaning: population proportion is greater than .5 (this is a one-tailed test)
Using a formula for a proportional one-sample z-test with your data included, we have:
z = (.57 - .5) -->test value (174/306 is approximately .57) minus population value (.5) divided by
√[(.5)(.5)/306]
Finish the calculation. Check a z-table for .01 level of significance for a one-tailed test. Remember if the null is not rejected, then there is no difference. If the null is rejected, then p > .5 and there is a difference.
I hope this will help get you started.
A clinical trial tests a method designed to increase the probability of conceiving a girl. In a study 306 babies were born, and 174 of them were girls. the sample data with a 0.01 significance level to test the claim that with this method, the
probability of being a girl is greater than 0.5.
1 answer
1 answer