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The average size of a home in 2008 fell to 2343 square feet, according to the National Association of Home Builders and reported in USA Today (January 11, 2009). The homebuilders of a northeastern city feel that the average size of homes continues to grow each year. To test their claim, a random sample of 45 new homes was selected and revealed an average of 2490 square feet. Assuming that the population standard deviation is approximately 450 square feet, is there evidence that the average size is larger in the northeast compared to the national 2008 figure? Use a 0.05 level of significance.

Short Answer

Expert verified
The answer depends on the calculated Z score from Step 2. If Z > 1.645, we reject the null hypothesis and conclude that the average home size in the northeastern city is larger than the national average in 2008. If Z <= 1.645, we do not reject the null hypothesis and conclude that there is not enough evidence to suggest the average home size in the northeastern city is larger.

Step by step solution

01

State the Hypotheses

The null hypothesis (\(H_0\)) is that the average size of homes in the northeastern city is not larger than the national average in 2008, or in other words, the mean homes size is less than or equal to 2343 square feet. The alternative hypothesis (\(H_a\)) is that the mean home size in the northeastern city is larger than the national average i.e, greater than 2343 square feet.
02

Compute the Test Statistic

In a one-sample z-test, the test statistic is given by \(Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\), where \(\overline{x}\) is the sample mean (2490), \(\mu\) is the population mean (2343), \(\sigma\) is the standard deviation (450), and \(n\) is the sample size (45). Plugging in these values and calculating will give the Z-score.
03

Determine the Critical Value and Make a Decision

For a significance level of 0.05 in a one-tailed test, the critical value of Z is 1.645. If the computed Z is greater than 1.645, then the null hypothesis is rejected in favor of the alternative hypothesis. Otherwise, there is insufficient evidence to reject the null hypothesis.
04

Interpret the Result

If the null hypothesis is rejected, we conclude there is evidence to suggest that the average home size in the northeastern city is larger than the national average in 2008. If not rejected, we conclude there is insufficient evidence to suggest the same.

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Key Concepts

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

Null and Alternative Hypothesis
Understanding the null (\( H_0 \)) and alternative hypotheses (\( H_a \) is crucial in hypothesis testing. The null hypothesis typically represents a statement of no effect or no difference; it's what we assume to be true before collecting any data. In our home size example, the null hypothesis posits that the average size of homes in the northeastern city isn't larger than the national average of 2343 square feet.

The alternative hypothesis, on the other hand, is a statement that indicates what we want to test for; it's a claim that there is a difference, and in this exercise, it's the hypothesis that the mean home size in the northeastern city is greater than 2343 square feet. It's important to precisely define these hypotheses because they determine the structure and direction of the test.
One-Sample z-Test
A one-sample z-test is a statistical method used when comparing the sample mean (\( \bar{x} \) to a known population mean (\( \bar\mu \) with a known population standard deviation. The formula for the z-test statistic is given by \( Z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \), where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, \( \sigma \) is the population standard deviation, and \( n \) is the sample size.

In this scenario, the calculated z-score will indicate where the sample mean falls in relation to the population mean, considering the standard deviation. A high z-score suggests the sample mean is much larger than the population mean, which is critical for deciding whether to reject the null hypothesis.
Statistical Significance
The concept of statistical significance is pivotal in hypothesis testing. It’s a measure of whether the observed results are likely due to chance or if they reflect an actual trend. Statistical significance is denoted by the p-value, which is the probability of observing the results, or something more extreme, if the null hypothesis were true.

To evaluate significance, we compare the p-value to a pre-specified significance level, often denoted by \( \alpha \)—in our case, 0.05. If the p-value is less than \( \alpha \) we declare the results statistically significant and reject the null hypothesis. If the p-value is greater, we don't have enough evidence to dismiss the null hypothesis. This doesn't confirm the null hypothesis, but rather indicates our sample data isn't sufficient to support the alternative hypothesis.
Critical Value
The critical value in hypothesis testing is a threshold to which test statistics are compared to determine whether to reject the null hypothesis. This value defines the boundary of the rejection region for the null hypothesis. In a one-sample z-test, the critical value corresponds to a point on the z-distribution. For a given significance level \( \alpha \) the critical value marks the start of the tail in which we would find results that are unlikely under the null hypothesis.

For example, with a 0.05 significance level in a right-tailed test, our critical z-value is around 1.645. This means if our calculated z-score is greater than 1.645, the sample data falls into the tail end of the distribution and is considered unexpected if the null hypothesis is accurate, leading us to reject \( H_0 \).

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Most popular questions from this chapter

There are only two possible decisions as a result of a hypothesis test. a. State the two possible decisions. b. Describe the conditions that will lead to each of the two decisions identified in part a.

Consider the hypothesis test where the hypotheses are \(H_{o}: \mu=26.4\) and \(H_{a}: \mu<26.4 .\) A sample of size 64 is randomly selected and yields a sample mean of 23.6 a. If it is known that \(\sigma=12,\) how many standard errors below \(\mu=26.4\) is the sample mean, \(\bar{x}=23.6 ?\) b. \(\quad\) If \(\alpha=0.05,\) would you reject \(H_{o} ?\) Explain.

At a very large firm, the clerk-typists were sampled to see whether salaries differed among departments for workers in similar categories. In a sample of 50 of the firm's accounting clerks, the average annual salary was \(\$ 16,010 .\) The firm's personnel office insists that the average salary paid to all clerk-typists in the firm is \(\$ 15,650\) and that the standard deviation is \(\$ 1800 .\) At the 0.05 level of significance, can we conclude that the accounting clerks receive, on average, a different salary from that of the clerk-typists? a. Solve using the \(p\) -value approach. b. Solve using the classical approach.

A manufacturer wishes to test the hypothesis that "by changing the formula of its toothpaste, it will give its users improved protection." The null hypothesis represents the idea that "the change will not improve the protection," and the alternative hypothesis is "the change will improve the protection." Describe the meaning of the two possible types of errors that can occur in the decision when the test of the hypothesis is conducted.

Calculate the \(p\) -value, given \(H_{a}: \mu \neq 245\) and \(z \star=1.1\)

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