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. As of this writing, among the babies born to parents using the YSORT method, 172 were boys and 39 were girls. Use the sample data with with 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.

1. Does the YSORT method of gender selection appear to work?

Answer 1

`z=\frac{0.815 -0.5}{\sqrt{(0.5(1-0.5))/(211)}}=9.151`

We can calculate the p value using the fact that we are conducting a right tailed test:

`p_v =P(z>9.151)\approx 0`

Since the p value i approximately 0 we can conclude that we have enough evidence to say that with this method, the probability of a baby being a boy is significantly greater than 0.5

Explanation:

Information given

n=172+39=211 represent the sample size

X=172 represent the number of boys in the sample

`\hat p = (172)/(211)= 0.815` estimated proportion of of boys

`p_o=0.5` is the value to verify

z would represent the statistic

`p_v` represent the p value

Hypothesis to test

The ides is verify if with this method, the probability of a baby being a boy is greater than 0.5, so then the system of hypothesis are.:

Null hypothesis:
`p\leq 0.5`

Alternative hypothesis:
`p > 0.5`

The statitic is given by:

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

Replacing the info we got:

`z=\frac{0.815 -0.5}{\sqrt{(0.5(1-0.5))/(211)}}=9.151`

We can calculate the p value using the fact that we are conducting a right tailed test:

`p_v =P(z>9.151)\approx 0`

Since the p value i approximately 0 we can conclude that we have enough evidence to say that with this method, the probability of a baby being a boy is significantly greater than 0.5