Question

Suppose the average size of a new house built in a certain county in 2014 was 2,275 square feet. A random sample of 25 new homes built in this county was selected in 2018. The average square footage was 2,189​, with a sample standard deviation of 227 square feet. Complete parts a and b.

a. Using α=0.02​, does this sample provide enough evidence to conclude that the average house size of a new home in the county has changed since​ 2014?

Determine the null and alternative hypotheses.

H0​:μ ▼ greater than or equals ≥ not equals ≠ less than or equals ≤ equals =

H1​:μ ▼ equals = not equals ≠ less than < greater than > ​(Type integers or decimals. Do not​ round.)

Determine the appropriate critical value. Select the correct choice below and fill in the answer box within your choice. ​(Round to three decimal places as​ needed.)

A. tα=

B. −tα=

C. tα/2 equals =

Calculate the appropriate test statistic.

t-x= ​(Round to two decimal places as​ needed.)

State the conclusion.

(Reject/Do not reject) H0. There is/is not sufficient evidence to conclude that the average house size of a new home in the county has (stayed the same/decreased/changed/increased) since 2014.

b. Determine the precise​ p-value for this test using Excel.

The​ p-value is . ​(Round to three decimal places as​ needed.)

Answer 1

we want to test that the average house size of a new home in the county has changed since​ 2014 i.e it is different than 2275.

n=25, the sample mean =2189 , sample std deviation = 227

The following null and alternative hypotheses need to be tested:

Ho: μ = 2275

Ha: μ ≠ 2275

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation, will be used.

Rejection Region

Based on the information provided, the significance level is α=0.02

the critical value for a two-tailed test is tc = t alpha /2  = t 0.01

and the critical value for a two-tailed test is tc​=2.492 ( from t table)

The rejection region for this two-tailed test is R={t:∣t∣>2.492}

Test Statistics

The t-statistic is computed as follows:

=​( Xˉ−μ0) / (S /√n) ​​

=   (2189-2275)   / (227 / √25)

=−1.894

Decision about the null hypothesis

Since it is observed that ∣t∣=1.894 ≤ tc​ =2.492,

it is then concluded that the null hypothesis is not rejected.

Answer:- We fail to reject the Null Hypothesis

Do not reject H0. There is not sufficient evidence to conclude that the average house size of a new home in the county has changed since 2014.

P-value = P( |tn-1| > |t| )

               =2* P(t24 > 1.894 )

               = 2 *0.0352

             = 0.0703

P-value = 0.0703

Using Excel :- two tailed test so will use function “T.DIST.2T

In excel = T.DIST.2T(1.894,24)

Using the P-value approach: p=0.0703 ≥ 0.02,

it is concluded that the null hypothesis is not rejected.