Question

The mean area of homes in a certain city built in 2009 was 2438 square feet. Assume that a simple random sample of 11 homes in the same city built in 2010 had a mean area of 2,293 square feet, with a standard deviation of 225 square feet. An insurance company wants to know if the mean area of homes built in 2010 is less than that of homes built in 2009. Compute the P-value of the test. Write down your P-value. You will need it for the next question.

Answer 1

Answer:

The p-value of the test is 0.029.

Step-by-step explanation:

The mean area of homes in a certain city built in 2009 was 2438 square feet.

An insurance company wants to know if the mean area of homes built in 2010 is less than that of homes built in 2009.

This means that at the null hypothesis we test if the mean is the same as in 2009, that is:

$H_0: \mu = 2438$

At the alternate hypothesis, we test that if it is less, than is;

$H_a: \mu < 2438$

The test statistic is:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

2438 is tested at the null hypothesis:

This means that $\mu = 2438$

Assume that a simple random sample of 11 homes in the same city built in 2010 had a mean area of 2,293 square feet, with a standard deviation of 225 square feet.

This means that $n = 11, X = 2293, s = 225$

Value of the test statistic:

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{2293 - 2438}{\frac{225}{\sqrt{11}}}$

$t = -2.14$

Compute the P-value of the test.

Probability of finding a mean less than a value(in this case, 2293) is a one-tailed test.

The pvalue is found for a one-tailed test, with t = -2.14 and 11 - 1 = 10 degrees of freedom. With the help of a calculator, the pvalue of the test is of 0.029

The p-value of the test is 0.029.