Question

According to the National Association of Realtors, it took an average of three weeks to sell a home in 2017. Suppose data for the sale of 39 randomly selected homes sold in Greene County, Ohio, in 2017 showed a sample mean of 3.6 weeks with a sample standard deviation of 2 weeks. Conduct a hypothesis test to determine whether the number of weeks until a house sold in Greene County differed from the national average in 2017. Useα = 0.05for the level of significance, and state your conclusion.(a)State the null and alternative hypothesis. (Enter != for ≠ as needed.)H0:_____________Ha:__________(b)Find the value of the test statistic. (Round your answer to three decimal places.)Find the p-value. (Round your answer to four decimal places.)p-value =___________.

Answer 1

Answer:

1.8732, 0.068

Step-by-step explanation:

Given that according to the National Association of Realtors, it took an average of three weeks to sell a home in 2017.

Sample size n = 39

a) $\bar x = 3.6\\s = 2$

Std error of sample = $\frac{2}{\sqrt{39} } \\=0.3203$

$H_0: \bar x = 3\\H_a: \bar x \neq 3$

(Two tailed test at 5% significance level)

Mean difference = 0.6

Since population std deviation is not known, t test to be used.

b) Test statistic t = mean diff/std error =$\frac{0.6}{0.3203 } \\=1.8732$

df = 38

c) p value = 0.068

Answer 2

From the information given, we have that:

a)

$H_0: \mu = 3$

$H_a: \mu \neq 3$

b)

The value of the test statistic is t = 1.873.

c)

The p-value of the test is of 0.0688.

Item a:

At the null hypothesis, it is tested if the mean is the same as the national average of 3, that is:

$H_0: \mu = 3$

At the alternative hypothesis, it is tested if the mean is different of the national average of 3, that is:

$H_a: \mu \neq 3$

Item b:

We have the standard deviation for the sample, thus, the t-distribution is used. The test statistic is given by:

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

The parameters are:

• $\overline{x}$ is the sample mean.
• $\mu$ is the value tested at the null hypothesis.
• s is the standard deviation of the sample.
• n is the sample size.

In this problem, the parameters are:

$\overline{x} = 3.6, \mu = 3, s = 2, n = 39$

Thus, the value of the test statistic is:

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

$t = \frac{3.6 - 3}{\frac{2}{\sqrt{39}}}$

$t = 1.873$

The value of the test statistic is t = 1.873.

Item c:

We have a two-tailed test(test if the mean is different from a value), with t = 1.873 and 39 - 1 = 38 df.

Using a t-distribution calculator, the p-value is of 0.0688.

The p-value of the test is of 0.0688.

A similar problem is given at https://brainly.com/question/16194574