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

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

`\hat p\sim N(p,\ \sqrt{(p(1-p))/(n)})`

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{(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{(p(1-p))/(n)})`