Question

the proportion of twins born in a town is ( p = 0.12 ). suppose we randomly select 100 women from this town who give birth in the next year. which of the following is the mean of the sampling distribution of ( hat{p} )?
( mu_{hat{p}}=p = 0.12 )
( mu_{hat{p}}=np = 100(0.12)=12 )
( mu_{hat{p}}=1 - p = 1 - 0.12 = 0.88 )
( mu_{hat{p}}=n(1 - p)=100(1 - 0.12)=88 )

Explanation:

Step1: Recall the formula for the mean of the sampling distribution of \(\hat{p}\)

The mean of the sampling distribution of the sample proportion \(\hat{p}\), denoted as \(\mu_{\hat{p}}\), is equal to the population proportion \(p\). Mathematically, \(\mu_{\hat{p}}=p\).

Step2: Substitute the given value of \(p\)

We are given that \(p = 0.12\). So, \(\mu_{\hat{p}}=0.12\).

Let's check the other options:

• For the option \(\mu_{\hat{p}}=np\), \(np\) is the mean of a binomial distribution \(X\) (where \(X\) is the number of successes, not the sample proportion \(\hat{p}\)). Here \(X\) would be the number of women having twins. But we are interested in the mean of \(\hat{p}=\frac{X}{n}\).
• For the option \(\mu_{\hat{p}}=1 - p\), this formula is not related to the mean of the sampling distribution of \(\hat{p}\).
• For the option \(\mu_{\hat{p}}=n(1 - p)\), \(n(1 - p)\) is the mean of the number of failures in a binomial distribution, not relevant to the mean of \(\hat{p}\).

Answer:

\(\mu_{\hat{p}}=p = 0.12\) (the first option)