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Sale Prices for Houses The average sales price of new one-family houses in the Midwest is dollar 250,000 and in the South is dollar 253,400. A random sample of 40 houses in each region was examined with the following results. At the 0.05 level of significance, can it be concluded that the difference in mean sales price for the two regions is greater than dollar 3400 ? $$ \begin{array}{lcc}{} & {\text { South }} & {\text { Midwest }} \\ \hline \text { Sample size } & {40} & {40} \\ {\text { Sample mean }} & {\$ 261,500} & {\$ 248,200} \\ {\text { Population standard deviation }} & {\$ 10,500} & {\$ 12,000}\end{array} $$

Short Answer

Expert verified
Yes, the test statistic exceeds the critical value, so we conclude the difference is greater than $3400.

Step by step solution

01

Define the Hypotheses

The null hypothesis is that the difference in population means for the two regions is equal to \(3,400\): \( H_0: \mu_{\text{South}} - \mu_{\text{Midwest}} = 3,400 \). The alternative hypothesis we want to test is that the difference is greater than \(3,400\): \( H_A: \mu_{\text{South}} - \mu_{\text{Midwest}} > 3,400 \).
02

Calculate the Standard Error

The standard error for the difference in two independent means is calculated by: \( SE = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \), where \( \sigma_1 = 10,500 \), \( \sigma_2 = 12,000 \), and \( n_1 = n_2 = 40 \). Therefore, \( SE = \sqrt{\frac{10,500^2}{40} + \frac{12,000^2}{40}} \).
03

Calculate the Test Statistic

The test statistic is calculated using the formula: \( z = \frac{(\bar{x}_1 - \bar{x}_2) - 3,400}{SE} \), where \( \bar{x}_1 = 261,500 \) and \( \bar{x}_2 = 248,200 \). First, calculate the difference in sample means: \( \Delta x = 261,500 - 248,200 = 13,300 \). Now, substitute \( SE \) from Step 2 and \( \Delta x \) into the formula to find \( z \).
04

Determine the Critical Value

With a significance level of \( \alpha = 0.05 \) for a one-tailed test (since we are testing \( > 3,400 \)), find the critical z-value from the standard normal distribution table. The critical z-value for \( \alpha = 0.05 \) in a one-tailed test is approximately 1.645.
05

Make a Decision

Compare the calculated test statistic from Step 3 to the critical value from Step 4. If the test statistic is greater than the critical value, we reject \( H_0 \). Otherwise, we do not reject \( H_0 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Error
The standard error is a crucial concept when dealing with hypothesis testing for differences between two independent means.
It's a measure of how much we expect the sample means to vary if many samples were drawn from the population. In this exercise, we used the formula for standard error of the difference between two means:
  • Calculate the variance for each population by squaring their standard deviations.
  • Divide each by the sample size, 40 in both cases.
  • Finally, sum these two results and take the square root to obtain the standard error.
This calculation helps us understand and quantify sampling variability, essentially setting the groundwork for further analysis.
Test Statistic
The test statistic is a number calculated from your data that is used in hypothesis testing.
It helps determine whether to reject or not reject the null hypothesis.For this process, the test statistic formula used is:\[z = \frac{(\bar{x}_1 - \bar{x}_2) - 3,400}{SE}\]Here, \((\bar{x}_1 - \bar{x}_2)\) refers to the difference between the sample means of the two regions, which is 13,300.
After calculating the standard error, plug it into the formula above.The test statistic tells us how far our observed difference is from 3,400, which we hypothesize as the true difference under the null hypothesis. With this statistic, we can proceed to compare it against the critical z-value to make a decision about the hypotheses.
Z-Value
The Z-value, also known as the Z-score, is a measure used in hypothesis testing to indicate how many standard deviations an element is from the mean.
In the context of this exercise, it's used to test the significance of the difference in house sale prices between two regions.To find the z-value:
  • We first need a predetermined level of significance, \( \alpha = 0.05 \) in this case.
  • Using the standard normal distribution table, determine the critical z-value for a one-tailed test, which is about 1.645.
This critical value represents the cutoff point beyond which we would reject the null hypothesis.
Comparing our calculated test statistic to this z-value helps us see if our observed difference is statistically significant.
Population Mean Difference
The idea of population mean difference is central to this hypothesis test.
We aim to determine if the real difference in population means (from the two regions) exceeds the hypothesized value of \\(3,400.Population means are the average values you would find if you could collect data from every possible member of the populations of interest.
In this exercise, while we have sample data, our goal is to make inferences about the overall population means.We utilize the test statistic to assess if our observed sample mean difference of \\)13,300 implies that the population mean difference is indeed greater than \$3,400.
If our hypothesis testing procedure shows the difference is statistically significant, it suggests that this observed difference is not merely due to random sampling variability.

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Most popular questions from this chapter

Home Prices A real estate agent compares the selling prices of randomly selected homes in two municipalities in southwestern Pennsylvania to see if there is a difference. The results of the study are shown. Is there enough evidence to reject the claim that the average cost of a home in both locations is the same? Use \(\alpha=0.01 .\) $$ \begin{array}{cc}{\text { Scott }} & {\text { Ligonier }} \\ \hline \overline{X_{1}=\$ 93,430^{*}} & {\bar{X}_{2}=\$ 98,043^{*}} \\\ {\sigma_{1}=\$ 5602} & {\sigma_{2}=\$ 4731} \\ {n_{1}=35} & {n_{2}=40}\end{array} $$

Explain the difference between testing a single mean and testing the difference between two means.

For these exercises, perform each of these steps. Assume that all variables are normally or approximately normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified and assume the variances are unequal. Tax-Exempt Properties A tax collector wishes to see if the mean values of the tax-exempt properties are different for two cities. The values of the tax- exempt properties for the two random samples are shown. The data are given in millions of dollars. At \(\alpha=0.05,\) is there enough evidence to support the tax collector's claim that the means are different? $$ \begin{array}{cccc|cccc}{} & {\text { City } \mathbf{A}} & {} & {} & {} & {\text { City } \mathbf{B}} & {} & {} \\ \hline 113 & {22} & {14} & {8} & {82} & {11} & {5} & {15} \\ {25} & {23} & {23} & {30} & {295} & {50} & {12} & {9} \\ {44} & {11} & {19} & {7} & {12} & {68} & {81} & {2} \\ {31} & {19} & {5} & {2} & {} & {20} & {16} & {4} & {5}\end{array} $$

For Exercises 7 through \(27,\) perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Commuters A recent random survey of 100 individuals in Michigan found that 80 drove to work alone. A similar survey of 120 commuters in New York found that 62 drivers drove alone to work. Find the \(95 \%\) confidence interval for the difference in proportions.

What are the characteristics of the \(F\) distribution?

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