The average price of homes sold in the U.S. in 2012 was $240,000. A sample of 14

Q: The average price of homes sold in the U.S. in 2012 was $240,000. A sample of 14

Solution:- The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below: State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis. Null hypothesis: >= 240,000, average price of homes sold in Chattanooga is significantly more than the national average. Alternative hypothesis:

ID: 3227495 • Letter: T

Question

The 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 (a) 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. a. State the null and alternative hypotheses to be tested. b. Compute the test statistic. c. The null hypothesis is to be tested at 95% confidence. Determine the critical value(s) for this test. d. What do you conclude?

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Solution:-

The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: >= 240,000, average price of homes sold in Chattanooga is significantly more than the national average.
Alternative hypothesis: < 240,000, average price of homes sold in Chattanooga is not significantly more than the national average.

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. (as we are given 95% confidence). The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n) = 36,000 / sqrt(144) = 36,000 / 12 = 3000
DF = n - 1 = 144 - 1 = 143
t = (x - ) / SE = (246,000 - 240,000)/3000 = 2

where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.

Here is the logic of the analysis: Given the alternative hypothesis ( < 240,000), we want to know whether the observed sample mean is small enough to cause us to reject the null hypothesis.

The observed sample mean produced a t statistic test statistic of 2. We use the t Distribution Calculator to find P(t < 2) = 0.023697.

Interpret results. Since the P-value (0.023697) is smaller than the significance level (0.05), we can reject the null hypothesis. Thus accepting the alternate hypothesis.

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The average price of homes sold in the U.S. in 2012 was $240,000. A sample of 14

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