Question

Suppose a random sample of 200 Americans is asked to disclose whether they can order a meal in a foreign language. Describe the sampling distribution of ˆp , the proportion of Americans who can order a meal in a foreign language.

Answer 1

Answer:

The distribution of sample proportion Americans who can order a meal in a foreign language is,

$\hat p\sim N(p,\ \sqrt{\frac{p(1-p)}{n}})$

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

 $\mu_{\hat p}=p$

The standard deviation of this sampling distribution of sample proportion is:

$\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}$

The sample size of Americans selected to disclose whether they can order a meal in a foreign language is, n = 200.

The sample selected is quite large.

The Central limit theorem can be applied to approximate the distribution of sample proportion.

The distribution of sample proportion is,

$\hat p\sim N(p,\ \sqrt{\frac{p(1-p)}{n}})$