Question

The Genetics and IVF Institute conducted a clinical trial of the YSORT method designed to increase the probability of conceiving a boy. As this book was being written, 51 babies were born to parents using the YSORT method, and 39 of them were boys. Use the sample data with a 0.01 significance level to test the claim that with this method, the probability of a baby being a boy is greater than 0.5. Does the method appear to work?

Answer 1

Null hypothesis:
`p\leq 0.5`

Alternative hypothesis:
`p > 0.5`

`z=\frac{0.765 -0.5}{\sqrt{(0.5(1-0.5))/(51)}}=3.785`

`p_v =P(z>3.785)=7.68x10^(-5)`

So the p value obtained was a very low value and using the significance level given
`\alpha=0.01` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of boys is significantly higher than 0.5.

Explanation:

1) Data given and notation

n=51 represent the random sample taken

X=39 represent the number of boys

`\hat p=(39)/(51)=0.765` estimated proportion of boys

`p_o=0.5` is the value that we want to test

`\alpha=0.01` represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

`p_v` represent the p value (variable of interest)

2) Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that with this method, the probability of a baby being a boy is greater than 0.5.:

Null hypothesis:
`p\leq 0.5`

Alternative hypothesis:
`p > 0.5`

When we conduct a proportion test we need to use the z statisitc, and the is given by:

`z=\frac{\hat p -p_o}{\sqrt{(p_o (1-p_o))/(n)}}` (1)

The One-Sample Proportion Test is used to assess whether a population proportion
`\hat p` is significantly different from a hypothesized value
`p_o`.

3) Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

`z=\frac{0.765 -0.5}{\sqrt{(0.5(1-0.5))/(51)}}=3.785`

4) Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided
`\alpha=0.01`. The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

`p_v =P(z>3.785)=7.68x10^(-5)`

So the p value obtained was a very low value and using the significance level given
`\alpha=0.01` we have
`p_v<\alpha` so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of boys is significantly higher than 0.5.