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Ten years ago, the average acreage of farms in a certain geographic region was 65 acres. The standard deviation of the population was 7 acres. A recent study consisting of 22 randomly selected farms showed that the average was 63.2 acres per farm. Test the claim, at \(\alpha=0.10,\) that the average has not changed by finding the \(P\) -value for the test. Assume that \(\sigma\) has not changed and the variable is normally distributed.

Short Answer

Expert verified
The p-value is 0.2052, which means we fail to reject the null hypothesis at \(\alpha = 0.10\).

Step by step solution

01

Define the Hypotheses

First, let's define the null and alternative hypotheses for this test. The null hypothesis is that the average acreage of farms has not changed from 65 acres, so \(H_0: \mu = 65\). The alternative hypothesis is that the average acreage has changed, \(H_1: \mu eq 65\). This is a two-tailed test since we are checking for any change, not a specific direction.
02

Determine the Test Statistic

The next step is to calculate the test statistic. We use the formula for the z-test statistic since we know the population standard deviation: \[z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}\] where \(\bar{x} = 63.2\), \(\mu = 65\), \(\sigma = 7\), and \(n = 22\). Substitute these values to calculate the z-score: \[z = \frac{63.2 - 65}{\frac{7}{\sqrt{22}}} \approx -1.267\].
03

Find the P-Value

With the z-score calculated, the next step is to find the corresponding p-value. Since this is a two-tailed test, we must find the area in both tails. Use a z-table or technology to find the p-value for \(z = -1.267\). The p-value obtained is approximately \(2 \times 0.1026 = 0.2052\).
04

Make a Decision

Compare the p-value to the significance level \(\alpha = 0.10\). Since the p-value \(0.2052\) is greater than \(0.10\), we fail to reject the null hypothesis. This means there is not enough evidence to claim that the average acreage of farms has changed.

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Key Concepts

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

Understanding the Z-Test
The z-test is a type of hypothesis test that helps us determine if there is a significant difference between sample and population means. It's commonly used when the population standard deviation is known and the sample size is sufficiently large, typically over 30. However, it can also be used with smaller samples if the data is normally distributed.

In this scenario, the z-test is applied because the population standard deviation is given as 7 acres, and even though our sample size of 22 farms is below 30, the problem indicates that the variable is normally distributed. When conducting a z-test, we calculate a z-score, which tells us how many standard deviations our sample mean is from the population mean.
  • Z-test requirements include knowing the population standard deviation.
  • It is applicable for samples where the distribution is approximately normal.
  • The test checks if the sample mean significantly differs from the population mean.
The formula for calculating the z-score in a z-test is: \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, \( \sigma \) is the population standard deviation, and \( n \) is the sample size. This formula helps in determining the position of the sample mean within the population distribution.
Discovering the P-Value
The p-value is a measure that helps us determine the strength of the evidence against the null hypothesis. It indicates the probability of observing the sample data, or something more extreme, assuming that the null hypothesis is true.

In essence, the p-value quantifies the evidence and helps decide whether to reject the null hypothesis. A lower p-value suggests stronger evidence against the null hypothesis. In hypothesis testing, we compare the p-value to a significance level \( \alpha \) (often 0.05 or 0.10) to make a decision.
  • If the p-value is less than \( \alpha \), we reject the null hypothesis.
  • If the p-value is greater than \( \alpha \), we fail to reject the null hypothesis.
For this exercise, the calculated p-value is approximately 0.2052. This value is more than the chosen \( \alpha = 0.10 \), and thus, we do not have enough evidence to reject the null hypothesis. This means the data does not support a significant change in farm acreage from the previous average of 65 acres.
Exploring the Null Hypothesis
The null hypothesis, denoted as \( H_0 \), is a statement that there is no effect or no difference. It serves as the starting assumption in a hypothesis test. In statistical testing, the null hypothesis typically states that something is equal to a standard or expected value.

For the given problem, the null hypothesis is \( H_0: \mu = 65 \), which means that the average acreage of farms has not changed and remains at 65 acres. Hypothesis tests aim to gather evidence against \( H_0 \) and show that an alternative explanation, the alternative hypothesis, might be true.
  • \( H_0 \) provides a baseline or reference point for testing.
  • It is assumed true unless evidence strongly indicates otherwise.
  • Failing to reject \( H_0 \) does not prove it true, just insufficient evidence to discount it.
In hypothesis testing, do not equate a failure to reject \( H_0 \) with proving that \( H_0 \) is true. It merely indicates that there isn't enough statistical evidence to support a change or effect.
Understanding the Alternative Hypothesis
The alternative hypothesis, represented by \( H_1 \), is a statement that contradicts the null hypothesis. It proposes that there is indeed a significant effect or difference present. In testing, if sufficient evidence suggests that the null hypothesis is unlikely, the alternative hypothesis may be accepted.

In this specific example, the alternative hypothesis is \( H_1: \mu eq 65 \). It postulates that the average acreage of farms has changed from 65 acres, which could mean either an increase or decrease. This makes it a two-tailed test, where potential outcomes on both sides of the mean are considered.
  • \( H_1 \) provides an alternative to \( H_0 \) in hypothesis testing.
  • It suggests that a real change or difference exists.
  • Accepting \( H_1 \) requires solid evidence against \( H_0 \).
Determining whether \( H_1 \) is accepted depends on comparing the calculated p-value with the significance level \( \alpha \). If the null hypothesis is rejected, it suggests that the data supports some difference or change, aligned with \( H_1 \).

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Most popular questions from this chapter

The average farm size in the United States is 444 acres. A random sample of 40 farms in Oregon indicated a mean size of 430 acres, and the population standard deviation is 52 acres. At \(\alpha=0.05,\) can it be concluded that the average farm in Oregon differs from the national mean? Use the \(P\) -value method.

How is the power of a test related to the type II error?

Assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. A statistics professor is used to having a variance in his class grades of no more than \(100 .\) He feels that his current group of students is different, and so he examines a random sample of midterm grades as shown. At \(\alpha=0.05,\) can it be concluded that the variance in grades exceeds \(100 ?\) \(\begin{array}{lllll}92.3 & 89.4 & 76.9 & 65.2 & 49.1 \\ 96.7 & 69.5 & 72.8 & 67.5 & 52.8 \\ 88.5 & 79.2 & 72.9 & 68.7 & 75.8\end{array}\)

Perform each of the following 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. A magazine article reported that \(11 \%\) of adults buy takeout food every day. A fast-food restaurant owner surveyed 200 customers and found that 32 said that they purchased takeout food every day. At \(\alpha=0.02,\) is there evidence to believe the article's claim? Would the claim be rejected at \(\alpha=0.05 ?\)

A survey taken several years ago found that the average time a person spent reading the local daily newspaper was 10.8 minutes. The standard deviation of the population was 3 minutes. To see whether the average time had changed since the newspaper's format was revised, the newspaper editor surveyed 36 individuals. The average time that these 36 randomly selected people spent reading the paper was 12.2 minutes. At \(\alpha=0.02\), is there a change in the average time an individual spends reading the newspaper? Find the \(98 \%\) confidence interval of the mean. Do the results agree? Explain.

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