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Chapter 9: Problem 26

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

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? $$ \begin{array}{cccc|cccc}{} & {\text { City } \mathbf{A}} & {} & {} & {} & {\text { City } \mathbf{B}} & {} & {} \\ \hline 113 & {22} & {14} & {8} & {82} & {11} & {5} & {15} \\ {25} & {23} & {23} & {30} & {295} & {50} & {12} & {9} \\ {44} & {11} & {19} & {7} & {12} & {68} & {81} & {2} \\ {31} & {19} & {5} & {2} & {} & {20} & {16} & {4} & {5}\end{array} $$

Classify each as independent or dependent samples. a. Heights of identical twins b. Test scores of the same students in English and psychology c. The effectiveness of two different brands of aspirin on two different groups of people d. Effects of a drug on reaction time of two different groups of people, measured by a before-and-after test e. The effectiveness of two different diets on two different groups of individuals

For Exercises 2 through \(12,\) 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. Pulse Rates of Identical Twins A researcher wanted to compare the pulse rates of identical twins to see whether there was any difference. Eight sets of twins were randomly selected. The rates are given in the table as number of beats per minute. At \(\alpha=0.01,\) is there a significant difference in the average pulse rates of twins? Use the \(P\) -value method. Find the \(99 \%\) confidence interval for the difference of the two. $$ \begin{array}{l|cccccccc}{\text { Twin } \mathbf{A}} & {87} & {92} & {78} & {83} & {88} & {90} & {84} & {93} \\ \hline \text { Twin B } & {83} & {95} & {79} & {83} & {86} & {93} & {80} & {86}\end{array} $$

If there is a significant difference between \(p_{1}\) and \(p_{2}\) and between \(p_{2}\) and \(p_{3},\) can you conclude that there is a significant difference between \(p_{1}\) and \(p_{3} ?\)

Teacher Salaries A researcher claims that the mean of the salaries of elementary school teachers is greater than the mean of the salaries of secondary school teachers in a large school district. The mean of the salaries of a random sample of 26 elementary school teachers is dollar 48,256, and the sample standard deviation is dollar 3,912.40 . The mean of the salaries of a random sample of 24 secondary school teachers is dollar 45,633 . The sample standard deviation is dollar 5533 . At \(\alpha=0.05,\) can it be concluded that the mean of the salaries of the elementary school teachers is greater than the mean of the salaries of the secondary school teachers? Use the \(P\) -value method.
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