Question

A sample of size 100 is selected from a population with p = .40.

Answer 1

Answer:

(a) 0.40

(b) 0.049

(c) $\bar p\sim N(0.40,0.049^{2})$

(d) Explained below

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:

 $E(\bar p)=p$

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

 $SE(\bar p)=\sqrt{\frac{p(1-p)}{n}}$

Given:

n = 100

p = 0.40

As n = 100 > 30 the Central limit theorem is applicable.

(a)

Compute the expected value of $\bar p$ as follows:

$E(\bar p)=p=0.40$

The expected value of $\bar p$ is 0.40.

(b)

Compute the standard error of $\bar p$ as follows:

$SE(\bar p)=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.40(1-0.40)}{100}}=0.049$

The standard error of $\bar p$ is 0.049.

(c)

The sampling distribution of $\bar p$ is:

$\bar p\sim N(0.40,0.049^{2})$

(d)

The sampling distribution of p show that as the sample size is increasing the distribution is approximated by the normal distribution.