Q4 The Ericsson method is one of se... [FREE SOLUTION]

Chapter 8: Q4 (page 371)

The Ericsson method is one of several methods claimed to increase the likelihood of a baby girl. In a clinical trial, results could be analysed with a formal hypothesis test with the alternative hypothesis of p>0.5, which corresponds to the claim that the method increases the likelihood of having a girl, so that the proportion of girls is greater than 0.5. If you have an interest in establishing the success of the method, which of the following P-values would you prefer: 0.999, 0.5, 0.95, 0.05, 0.01, and 0.001? Why?

Short Answer

Expert verified

The following p-values should be considered for the method to increase the likelihood of a baby girl:

0.05
0.01
0.001

Since the p-values mentioned above indicate the rejection of the null hypothesis and the claim of $p > 0.05$ holds true, the method successfully increases the likelihood of a baby girl.

Step by step solution

Given information

The Ericsson method is used to increase the likelihood of a baby girl being born.

The alternative hypothesis of $p > 0.5$ is considered to conduct a hypothesis test to establish the success of the Ericsson method.

Interpretation of p-value

The p-value in hypothesis testing suggests that the decision to reject the null hypothesis is due to chance and cannot be considered true in reality.

A smaller p-value will indicate that the existence of the claim under the alternative hypothesis due to random chance is negligible and can be considered true in reality.

Here, the alternative hypothesis is given as $p > 0.5$ , where p is the proportion of baby girls born using the Ericsson method.

P-values such as 0.999, 0.95, and 0.5 indicate a high probability (99.9%, 95%, and 50%) that the proportion of baby girls born greater than 0.5 is due to chance and is not true in reality.

P-values such as 0.05, 0.01, and 0.001 indicate that there are 5%, 1%, and 0.1% probabilities that the proportion of baby girls born greater than0.5 is due to chance, respectively. Since the probability is very low, these p-values will establish the success of the method.

You are interested in the success of the method and, thus, are interested in supporting the alternative hypothesis of $p > 0.5$ , which then corresponds with rejecting the null hypothesis $p = 0.5 .$

If the p-value is less than the significance level $伪$ , then the null hypothesis is rejected.Since you want to reject the null hypothesis, this is the easiest when the p-value is as small as possible and the smallest given p-value is