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Biomass Exercise 7.63 reported that the biomass for tropical woodlands, thought to be about 35 kilograms per square meter \(\left(\mathrm{kg} / \mathrm{m}^{2}\right),\) may in fact be too high and that tropical biomass values vary regionally - from about 5 to \(55 \mathrm{~kg} / \mathrm{m}^{2} .20\) Suppose you measure the tropical biomass in 400 randomly selected square- meter plots and obtain \(\bar{x}=31.75\) and \(s=10.5 .\) Do the data present sufficient evidence to indicate that scientists are overestimating the mean biomass for tropical woodlands and that the mean is in fact lower than estimated? a. State the null and alternative hypotheses to be tested. b. Locate the rejection region for the test with \(\alpha=.01\). c. Conduct the test and state your conclusions.

Step by step solution

01

Null hypothesis (H0)

The null hypothesis states that the mean biomass for tropical woodlands is equal to the assumed value of 35 kg/m². Mathematically, this can be written as \(H_0: \mu = 35\).

02

Alternative hypothesis (H1)

The alternative hypothesis states that the mean biomass for tropical woodlands is less than the assumed value of 35 kg/m². Mathematically, this can be written as \(H_1: \mu < 35\). b. Determine the rejection region for the test with \(\alpha=.01\).

03

Rejection region calculation

We are given a significance level of \(\alpha = 0.01\). Since this is a left-tailed test, we will find the z-score that corresponds to the \(0.01\) probability in the left tail of the standard normal distribution. This can be found using a z-table or a calculator. The z-score is approximately -2.33. Any test statistic smaller than -2.33 will lead us to reject the null hypothesis in favor of the alternative hypothesis. c. Conduct the test and state the conclusion.

04

Test statistic calculation

To conduct the test, we will calculate the test statistic using the given sample mean \(\bar{x}=31.75\), sample standard deviation \(s=10.5\), and sample size \(n=400\). The test statistic is given by \(z = \frac{\bar{x}-\mu_0}{\frac{s}{\sqrt{n}}} = \frac{31.75-35}{\frac{10.5}{\sqrt{400}}} \approx -6.38\).

05

Conclusion

Since the calculated test statistic, \(z = -6.38\), is smaller than the critical value, \(-2.33\), we reject the null hypothesis in favor of the alternative hypothesis. Therefore, we have sufficient evidence to conclude that the mean biomass for tropical woodlands is indeed lower than the estimated value of 35 kg/m².

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Key Concepts

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

Null Hypothesis

The null hypothesis is an essential part of hypothesis testing. It represents a default or starting assumption in a statistical test. The null hypothesis is typically denoted as \(H_0\). In our exercise, it states that the mean biomass of tropical woodlands is equal to the hypothesized value of 35 kg/m². This hypothesis assumes no effect or no difference. We begin with this assumption and look for evidence to challenge it.

• The null hypothesis serves as a baseline, suggesting no change or no impact.
• It is usually tested with the aim of it's rejection.
• Mathematically, it is expressed as \(H_0: \mu = 35\) in this scenario.

It is crucial to understand that rejecting the null hypothesis implies finding sufficient evidence for an alternative scenario. But failing to reject it doesn't necessarily prove it true, just that there's not enough evidence against it.

Alternative Hypothesis

The alternative hypothesis is a key component of hypothesis testing, denoted as \(H_1\). It is what researchers aim to support through their data analysis. In our particular exercise, the alternative hypothesis suggests the mean biomass is less than 35 kg/m². This hypothesis indicates an expected effect or difference from the null hypothesis.

• The alternative hypothesis proposes a specific change or effect, contrary to the null.
• It is a statement that researchers hope to support with evidence.
• Here, it is written as \(H_1: \mu < 35\).

The alternative hypothesis can be one-sided, as in this example, where the mean is hypothesized to only be lower. Alternatively, it can be two-sided which suggests the mean could differ in either direction.

Significance Level

The significance level, symbolized as \(\alpha\), sets the threshold for determining if a test result is statistically significant. In our example, the significance level is \(\alpha = 0.01\), meaning there is a 1% risk of rejecting the null hypothesis if it is actually true.

• It's a predefined probability used as a cut-off point for determining statistical significance.
• A lower \(\alpha\) reduces the chance of a Type I error — incorrectly rejecting a true null hypothesis.
• Choosing a significance level depends on the context and desired level of certainty.

Thus, this level reflects how much confidence one requires to claim the alternative hypothesis is supported by evidence. The more critical the consequences of error, the smaller \(\alpha\) is chosen.

Rejection Region

The rejection region is an integral part of deciding whether to reject the null hypothesis. It is determined based on the significance level and the distribution used in the particular statistical test. For our exercise, with \(\alpha = 0.01\), this is a one-tailed test pointing to the left.

• The rejection region is defined by critical values beyond which the null hypothesis is rejected.
• In this exercise, the critical value is approximately \(-2.33\).
• Any calculated test statistic falling within this region leads to rejection of \(H_0\).

This area represents the probability, under the null, of observing a result as extreme as, or more than, our test result. In our example, the calculated test statistic \(z = -6.38\) falls in this region, indicating significant evidence against the null hypothesis.