Question

Average price of homes sold in the U.S. in 2012 was $240,000. A sample of 144 homes sold in Chattanooga in 2012 showed an average price of $246,000. It is known that the standard deviation of the population (σ) is $36,000. We are interested in determining whether or not the average price of homes sold in Chattanooga is significantly more than the national average. Use Ti where appropriate.

a.	State the null and alternative hypotheses to be tested.
b.	Compute the test statistic.
c.	The null hypothesis is to be tested at the 5% level of significance. Determine the critical value(s) for this test.
d.	Construct a distribution curve showing the rejection and no-rejection regional with supporting value(s).
e.	Using the critical-value test, what do you conclude?

Answer 1

The provided sample mean is Xˉ=246000

and the known population standard deviation is σ=36000,

and the sample size is n = 144.

a)

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:μ=240,000

Ha:μ>240,000

This corresponds to a right-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

b)

(2) Test Statistics

The z-statistic is computed as follows:

c)

(3) Rejection Region

Based on the information provided, the significance level is α=0.05, and the critical value for a right-tailed test is z_c = 1.64

.

The rejection region for this right-tailed test is

R={z:z>1.64}

d)

rejection region = black region[ Z > 1.64 ]

Non rejection region = white region [ Z < 1.64 ]

e)

(4) Decision about the null hypothesis

Since it is observed that

z=81.92>zc​=1.64,

it is then concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is rejected.

Therefore, there is enough evidence to claim that the population mean μ is greater than 240,000, at the 0.05 significance level.

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