Question

According to a recent​ publication, the mean price of new mobile homes is ​$63 comma 800. Assume a standard deviation of ​$7900. Let x overbar denote the mean price of a sample of new mobile homes. a. For samples of size 25​, find the mean and standard deviation of x overbar. Interpret your results in words. b. Repeat part​ (a) with nequals50. a. For ▼ the mean and standard deviation of ▼ the prices of the mobile homes all possible sample mean prices are ​$ nothing and ​$ nothing​, respectively. ​(Round to the nearest cent as​ needed.) b. For ▼ samples of 50 mobile homes, the 50 mobile homes sampled, the mean and standard deviation of ▼ the prices of the mobile homes all possible sample mean prices are ​$ nothing and ​$ nothing​, respectively. ​(Round to the nearest cent as​ needed.)

Answer 1

Answer:

a. For n=25, the mean and standard deviation of the prices of the mobile homes all possible sample mean prices are ​$63,800 and ​$1,580​, respectively.

b. For n=50, the mean and standard deviation of the prices of the mobile homes all possible sample mean prices are ​$63,800 and ​$1,117​, respectively.

Step-by-step explanation:

In this case, for each sample size, we have a sampling distribution (a distribution for the population of sample means), with the following parameters:

$\mu_s=\mu=63,800\\\\\sigma_s=\sigma/\sqrt{n}=7,900/\sqrt{n}$

For n=25 we have:

$\mu_s=\mu=63,800\\\\\sigma_s=\sigma/\sqrt{n}=7,900/\sqrt{25}=7,900/5=1,580$

The spread of the sampling distribution is always smaller than the population spread of the individuals. The spread is smaller as the sample size increase.

This has the implication that is expected to have more precision in the estimation of the population mean when we use bigger samples than smaller ones.

If n=50, we have:

$\mu_s=\mu=63,800\\\\\sigma_s=\sigma/\sqrt{n}=7,900/\sqrt{50}=7,900/7.07=1,117$