Suppose the average size of a new house built in a certain county in 2006 was 2,

Question

Suppose the average size of a new house built in a certain county in 2006 was 2,275 square feet. A random sample of 25 new homes built in this county was selected in 2010. The average square footage was 2,199, with a sample standard deviation of 227 square feet. a. Using = 0 10, does this sample provide enough evidence to conclude that the average house size of a new home in the county has changed since 2006? b. Use technology to determine the p-value for this test. a. Determine the null and alternative hypotheses. Choose the correct answer below. A. Ho: #2275 and H1:=2.275 O B. Ho: 2 2.275 and H1 : 2.275 OD. Ho : :2275 and H1: #2,275

Explanation / Answer

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sample mean is X¯=2199 and sample standard deviation is s=227, and the sample size is n = 25.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: = 2275

Ha: 2275

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

(2) Rejection Region

The significance level is =0.10, and the critical value for a two-tailed test is tc=1.711.

The rejection region for this two-tailed test is R={t:t>1.711}

(3) Test Statistics

The t-statistic is computed as follows:

t =[ (X¯0) / ( s/n) ] = [ (21992275) / (227/25) ] = 1.674

(4) Decision about the null hypothesis

Since it is observed that t=1.674tc=1.711, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.1071, and since p=0.10710.10, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean is different than 2275, at the 0.10 significance level.

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sample mean is X¯=110 and the known population standard deviation is =27, and the sample size is n =45.

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: =117

Ha: 117

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

(2) Rejection Region

The significance level is =0.05, and the critical value for a two-tailed test zc=1.96.

The rejection region for this two-tailed test is R={z:z>1.96}

(3) Test Statistics

The z-statistic is computed as follows:

z=[ (X¯0) / (/n) ] = [ (110117) / (27/45) ] = 1.74

(4) Decision about the null hypothesis

Since it is observed that z=1.74zc=1.96, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p=0.082, and since p=0.0820.05, it is concluded that the null hypothesis is not rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean is different than 117, at the 0.05 significance level.

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