Question

7.5 Apartment rents. Refer to Exercise 7.1 (page 360). Do these data give good
reason to believe that the average rent for all advertised one-bedroom apartments
is greater than $650 per month? Make sure to state the hypotheses, find the t
statistic, degrees of freedom, and P-value, and state your conclusion using the
5% significance level.

Reference:

7.1 One-bedroom rental apartment. A large city newspaper contains several
hundred advertisements for one-bedroom apartments. You choose 25 at random
and calculate a mean monthly rent of $703 and a standard deviation of $115.
(a) What is the standard error of the mean?
(b) What are the degrees of freedom for a one-sample t statistic?

Answer 1

Step 1:

Ho:  

Ha:

Null hypothesis states that the average rent is less than or equal to 650

This is a right tailed test a the claim is that the mean value is greater than 650

Step 2: Test statistics

n= 25

sample mean = = 703

sample standard deviation = s = 115

level of significance = = 5% = 0.05

We assume that the data is normally distributed, since the sample size is less than 30 and population standard deviation is not given we will use t statistics

t stat = 2.304

t critical = 1.710882

Since the t stat (2.304) is greater than t critical (1.710882) , i.e. t stat falls in the rejection area we reject the Null hypothesis.

P Value

P value = TDIST (t statistics, df, 2) = TDIST(2.304, 24, 1)= 0.0151

P value = 0.0151

Since the P value is less than the level of significance ( = 0.05), we will reject the null hypothesis. Hence we do not have sufficient evidence to support the claim that the mean is greater than 650

(a) Standard error of mean

Standard error of mean = 23

(b) Degree of freedom = n-1 =25-1 = 24