Answer :
To determine if there's evidence to prove that Chicago has a higher average price for 3-bedroom, 2-bath homes than Denver, we'll be conducting a hypothesis test. Here’s how you can approach the problem step-by-step:
1. Set Up Hypotheses:
- Null Hypothesis, [tex]\( H_0 \)[/tex]: The average price in Chicago is equal to or less than the average price in Denver, [tex]\(\mu_C \leq \mu_D\)[/tex].
- Alternative Hypothesis, [tex]\( H_a \)[/tex]: The average price in Chicago is greater than the average price in Denver, [tex]\(\mu_C > \mu_D\)[/tex].
2. Significance Level:
- We are given a significance level, [tex]\(\alpha = 0.05\)[/tex].
3. Determine Decision Rule:
- Since this is a one-tailed test and we are testing if Chicago’s average price is higher, we will reject [tex]\( H_0 \)[/tex] if the calculated z-value is greater than 1.64 (the critical value for a one-tailed test at [tex]\(\alpha = 0.05\)[/tex]).
4. Calculate the Test Statistic (z-score):
We use the formula for the z-score for the difference between two means:
[tex]\[
z = \frac{\overline{X}_C - \overline{X}_D}{\sqrt{\frac{\sigma_C^2}{n_C} + \frac{\sigma_D^2}{n_D}}}
\][/tex]
where:
- [tex]\(\overline{X}_C = 148,000\)[/tex] and [tex]\(\overline{X}_D = 142,500\)[/tex] are the sample means,
- [tex]\(\sigma_C = 12,000\)[/tex] and [tex]\(\sigma_D = 10,000\)[/tex] are the standard deviations,
- [tex]\(n_C = 20\)[/tex] and [tex]\(n_D = 18\)[/tex] are the sample sizes.
Plug in the values to compute the z-score:
- Calculate the standard error:
[tex]\[
SE = \sqrt{\frac{12,000^2}{20} + \frac{10,000^2}{18}} = 3,557.03
\][/tex]
- Calculate the z-score:
[tex]\[
z = \frac{148,000 - 142,500}{3,557.03} = 1.54
\][/tex]
5. Make a Decision:
- Since the calculated z-value [tex]\(1.54\)[/tex] is less than [tex]\(1.64\)[/tex], we do not reject the null hypothesis [tex]\(H_0\)[/tex].
Thus, based on this test, there is insufficient evidence at the 0.05 significance level to conclude that Chicago has a higher average home price than Denver.
1. Set Up Hypotheses:
- Null Hypothesis, [tex]\( H_0 \)[/tex]: The average price in Chicago is equal to or less than the average price in Denver, [tex]\(\mu_C \leq \mu_D\)[/tex].
- Alternative Hypothesis, [tex]\( H_a \)[/tex]: The average price in Chicago is greater than the average price in Denver, [tex]\(\mu_C > \mu_D\)[/tex].
2. Significance Level:
- We are given a significance level, [tex]\(\alpha = 0.05\)[/tex].
3. Determine Decision Rule:
- Since this is a one-tailed test and we are testing if Chicago’s average price is higher, we will reject [tex]\( H_0 \)[/tex] if the calculated z-value is greater than 1.64 (the critical value for a one-tailed test at [tex]\(\alpha = 0.05\)[/tex]).
4. Calculate the Test Statistic (z-score):
We use the formula for the z-score for the difference between two means:
[tex]\[
z = \frac{\overline{X}_C - \overline{X}_D}{\sqrt{\frac{\sigma_C^2}{n_C} + \frac{\sigma_D^2}{n_D}}}
\][/tex]
where:
- [tex]\(\overline{X}_C = 148,000\)[/tex] and [tex]\(\overline{X}_D = 142,500\)[/tex] are the sample means,
- [tex]\(\sigma_C = 12,000\)[/tex] and [tex]\(\sigma_D = 10,000\)[/tex] are the standard deviations,
- [tex]\(n_C = 20\)[/tex] and [tex]\(n_D = 18\)[/tex] are the sample sizes.
Plug in the values to compute the z-score:
- Calculate the standard error:
[tex]\[
SE = \sqrt{\frac{12,000^2}{20} + \frac{10,000^2}{18}} = 3,557.03
\][/tex]
- Calculate the z-score:
[tex]\[
z = \frac{148,000 - 142,500}{3,557.03} = 1.54
\][/tex]
5. Make a Decision:
- Since the calculated z-value [tex]\(1.54\)[/tex] is less than [tex]\(1.64\)[/tex], we do not reject the null hypothesis [tex]\(H_0\)[/tex].
Thus, based on this test, there is insufficient evidence at the 0.05 significance level to conclude that Chicago has a higher average home price than Denver.
