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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.

Short Answer

Expert verified
No, there is not enough evidence to conclude a difference.

Step by step solution

01

Define the Null and Alternative Hypotheses

The null hypothesis is that the average farm size in Oregon is equal to the national average, meaning \( H_0: \mu = 444 \). The alternative hypothesis is that the average farm size in Oregon is different from the national average, \( H_a: \mu eq 444 \).
02

Gather the Given Information

From the problem, we know \( \bar{x} = 430 \) acres (sample mean), \( \sigma = 52 \) acres (population standard deviation), \( n = 40 \) (sample size), and significance level \( \alpha = 0.05 \).
03

Compute the Test Statistic

The test statistic is calculated using the formula for the Z-test: \( Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \). Substituting the given values, we have \( Z = \frac{430 - 444}{52/\sqrt{40}} \).
04

Calculate the Z-value

Now calculate the Z-value: \[ Z = \frac{430 - 444}{52/\sqrt{40}} = \frac{-14}{8.223} = -1.703 \].
05

Find the P-value

Since this is a two-tailed test, we find the P-value by finding the probability for \( Z \leq -1.703 \) and then multiplying by 2. Using a Z-table or calculator, \( P(Z \leq -1.703) \approx 0.044 \), therefore the P-value is \( 2 \times 0.044 = 0.088 \).
06

Compare the P-value with Significance Level

The P-value (0.088) is greater than the significance level (0.05). This means we do not reject the null hypothesis.
07

Conclusion

Based on the P-value method, there is not enough evidence to conclude that the average farm size in Oregon is different from the national mean of 444 acres.

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Key Concepts

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

Understanding the Z-Test
In hypothesis testing, the Z-test is an essential statistical method used to determine whether the means of two groups are different. This test assumes data follows a normal distribution and the population standard deviation is known.
The Z-test formula is:
  • \[ Z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} \]
where:
  • \(\bar{x}\) = sample mean,
  • \(\mu\) = population mean,
  • \(\sigma\) = population standard deviation,
  • \(n\) = sample size.
The Z-test provides the Z-value, which indicates how many standard deviations away the sample mean is from the population mean.
This value helps in assessing the probability that the observed result could have occurred under the null hypothesis.
Explaining the Null Hypothesis
The null hypothesis, denoted as \( H_0 \), is a key concept in hypothesis testing, representing the default or status quo assumption that there is no difference or effect. In the exercise, \( H_0 \) states that the average farm size in Oregon is equal to the national average, i.e., \( \mu = 444 \) acres.
The purpose of testing the null hypothesis is to determine whether there is enough statistical evidence to reject it. In simpler terms, we want to see if the data we collected (the sample) convincingly indicates that the average is not what we initially assumed based on the entire population.
Accepting or rejecting this hypothesis depends on the results of the test statistic and the comparison of the P-value to a predetermined significance level, often referred to as \( \alpha \), like 0.05 used in the exercise.
Defining the P-value
The P-value is a crucial measure in statistics, indicating the strength of the evidence against the null hypothesis. It expresses the probability that the observed data would be at least as extreme as the actual observed outcomes if the null hypothesis were true.
A low P-value (typically less than or equal to \( \alpha \)) suggests that the observed data is unlikely under the null hypothesis and, therefore, the null hypothesis can be rejected. Conversely, a high P-value indicates there is not enough evidence to reject \( H_0 \).
In the exercise example, the computed P-value is 0.088. Since it is greater than \( \alpha = 0.05 \), we do not have sufficient evidence to reject the null interpretation that Oregon's average farm size equals the national average.
Exploring the Two-tailed Test
A two-tailed test is used when we are interested in deviations in either direction from the expected value. This means we are looking for differences that could be either higher or lower than the hypothesized population mean.
In the exercise, the null hypothesis asserts that the mean is exactly 444 acres, and the alternative hypothesis allows for the mean to either exceed or fall short of this value, represented as \( \mu eq 444 \).
To perform a two-tailed test, we calculate the probability in both tails of the distribution. The P-value essentially captures the extreme values on both sides, reflecting the complete probability of obtaining a result as extreme as, or more extreme than, the observed data. This results in doubling the P-value from one tail to account for both potential directions of deviation.

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Most popular questions from this chapter

Workers with a formal arrangement with their employer to be paid for time worked at home worked an average of 19 hours per week. A random sample of 15 mortgage brokers indicated that they worked a mean of 21.3 hours per week at home with a standard deviation of 6.5 hours. At \(\alpha=0.05,\) is there sufficient evidence to conclude a difference? Construct a \(95 \%\) confidence interval for the true mean number of paid working hours at home. Compare the results of your confidence interval to the conclusion of your hypothesis test and discuss the implications.

In hypothesis testing, why can't the hypothesis be proved true?

A special cable has a breaking strength of 800 pounds. The standard deviation of the population is 12 pounds. A researcher selects a random sample of 20 cables and finds that the average breaking strength is 793 pounds. Can he reject the claim that the breaking strength is 800 pounds? Find the \(P\) -value. Should the null hypothesis be rejected at \(\alpha=0.01 ?\) Assume that the variable is normally distributed.

Assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. A manufacturing process produces machine parts with measurements the standard deviation of which must be no more than \(0.52 \mathrm{~mm}\). A random sample of 20 parts in a given lot revealed a standard deviation in measurement of \(0.568 \mathrm{~mm}\). Is there sufficient evidence at \(\alpha=0.05\) to conclude that the standard deviation of the parts is outside the required guidelines?

Assume that the variables are normally or approximately normally distributed. Use the traditional method of hypothesis testing unless otherwise specified. The number of carbohydrates found in a random sample of fast-food entrees is listed. Is there sufficient evidence to conclude that the variance differs from \(100 ?\) Use the 0.05 level of significance. \(\begin{array}{lllll}53 & 46 & 39 & 39 & 30 \\ 47 & 38 & 73 & 43 & 41\end{array}\)

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