Question 1011624: Please help me
In a study of pregnant women and their ability to correctly predict the sex of their baby, 58 of the pregnant women had 12 years of education or less, and 36.2% of them correctly predicted
the sex of their baby. Use a 0.05 significance level to test the claim that these women have no ability to predict the sex of their baby, and the results are not significantly different from those that would be expected with random guesses.
Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method. Use the normal distribution as an approximation of the binomial distribution.
Identify the null and alternative hypotheses.
The test statistic is z=
The P-value is
Identify the conclusion about the null hypothesis and the final conclusion that addresses the original claim.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Ho: p is <+0.50
Ha:p>0.50. The Ho includes LESS than 0.50 because correctly predicting the sex of the baby should be at least 50%. Given that interpretation, there is no need to do the test on a probability of 36.2%.
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If one expands the criteria that it is not random, that can be done, but the comment of "no ability to predict the sex of their baby" might be better written solely as "not random"
Ho: p=0.50
Ha:p NE 0.50
alpha=0.05
z=(phat-p)/sqrt{p*(1-p)}/n
reject if |z|>1.96
=0.362-0.5/sqrt{.362*.638)/160; 58/160 is closest to 36.2%.
=-0.138/0.0380
= -3.63
Reject Ho; their choices are worse than random.
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