Question

According to a study conducted in one​ city, 25​% of adults in the city have credit card debts of more than​ $2000. A simple random sample of n equals 150 adults is obtained from the city. Describe the sampling distribution of ModifyingAbove p with caret​, the sample proportion of adults who have credit card debts of more than​ $2000. Round to three decimal places when necessary.

Answer 1

Answer:

The sampling distribution of $\hat p$ is N (0.25, 0.0354²).

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:

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

Let p = proportion of adults in the city having credit card debts of more than​ $2000.

It is provided that the proportion of adults in the city having credit card debts of more than​ $2000 is, p = 0.25.

A random sample of size n = 150 is selected from the city.

Since n = 150 > 30 the Central limit theorem can be used to approximate the distribution of p by the Normal distribution.

The mean is:

$\mu_{\hat p}=p=0.25$

The standard deviation is:

$\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.25(1-0.25)}{150}}=0.0354$

Thus, the sampling distribution of $\hat p$ is N (0.25, 0.0354²).