Question

An organization believes that the number of prospective home buyers who want their next house to be larger, smaller, or the same size as their current house is uniformly distributed (all proportions the same). To test this claim, you randomly select 900 prospective home buyers and ask them what size they want their next house to be. The results are shown in the table. Using a Chi-square Goodness of Fit test, can you reject the claim that the distribution is uniform? Use a=0.05. For credit, you must show a filled out standardized test statistic formula. You may refer back to problem 15 for the steps required for a test of hypothesis. Response Larger Same Size Smaller Frequency 330 245 325

Answer 1

Using a Chi-square Goodness of Fit test and a significance level of 0.05, the test statistic was calculated to be 14.25, which exceeds the critical value of 5.99 for 2 degrees of freedom. Thus, we reject the claim that the distribution is uniform.

Step-by-step explanation:

To determine whether the claim that the distribution of the size preference for prospective home buyer's next house is uniform, we can conduct a Chi-square Goodness of Fit test. The null hypothesis (H0) states that the number of prospective home buyers preferring a larger, same size, or smaller next house is equally distributed. With a total sample size of 900 and three categories, the expected frequency for each category is 900/3 = 300. We apply the formula for the chi-square test statistic:

χ² = ∑ [(Observed - Expected)² / Expected]

For our data:

• Larger: (330 - 300)² / 300 = 3.0
• Same Size: (245 - 300)² / 300 = 9.17
• Smaller: (325 - 300)² / 300 = 2.08

The sum of these values gives us the chi-square test statistic:

χ² = 3.0 + 9.17 + 2.08 = 14.25

With a significance level (α) of 0.05 and degrees of freedom (df) equal to the number of categories minus one, which is 3 - 1 = 2, we compare our calculated chi-square statistic to the critical value from the chi-square distribution table. If the calculated value exceeds the critical value, we reject the null hypothesis.

Since each observed value and the expected value are greater than 5, we meet the test requirement. Consulting the chi-square distribution table for df = 2 and α = 0.05, we find that the critical value is 5.99. Because our calculated test statistic of 14.25 > 5.99, we reject the null hypothesis. Therefore, we have sufficient evidence to conclude that the distribution is not uniform.

Answer 2

Using a Chi-square Goodness of Fit test, we calculated a chi-square statistic of 15.17. Since this value exceeds the critical value at the 0.05 significance level for 2 degrees of freedom, we reject the null hypothesis, indicating that the distribution of homebuyers' preferences is not uniform.

Step-by-step explanation:

To test the claim that the distribution of the sizes of houses that prospective homebuyers want (larger, same, or smaller) is uniform, we can use a Chi-square Goodness of Fit test. The null hypothesis (H0) states that the observed frequencies for the categories will fit a uniform distribution.

Under the null hypothesis, we would expect each category (larger, same, smaller) to have the same number of prospective home buyers. With a sample size of 900, this means we would expect 300 prospective buyers for each category. We can now calculate the chi-square test statistic using the formula:

χ² = ∑ ((O - E)² / E)

where O is the observed frequency and E is the expected frequency. Plugging the values into the formula:

χ² = ((330 - 300)² / 300) + ((245 - 300)² / 300) + ((325 - 300)² / 300)

χ² = (900 + 3025 + 625) / 300

χ² = 4550 / 300

χ² = 15.17

To determine whether to reject the null hypothesis, we compare the calculated chi-square statistic to the critical value from the chi-square distribution table with degrees of freedom df = k - 1, where k is the number of categories.

Since we have 3 categories, df = 3 - 1 = 2. At α = 0.05, the critical value for df = 2 is 5.991.

Since our test statistic of 15.17 is greater than 5.991, we reject the null hypothesis and conclude that the distribution is not uniform.