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A new method for measuring phosphorus levels in soil is described in the article "A Rapid Method to Determine Total Phosphorus in Soils" (Soil Sci. Amer. J., 1988: 1301-1304). Suppose a sample of 11 soil specimens, each with a true phosphorus content of \(548 \mathrm{mg} / \mathrm{kg}\), is analyzed using the new method. The resulting sample mean and standard deviation for phosphorus level are 587 and 10 , respectively. a. Is there evidence that the mean phosphorus level reported by the new method differs significantly from the true value of \(548 \mathrm{mg} / \mathrm{kg}\) ? Use \(\alpha=.05\). b. What assumptions must you make for the test in part (a) to be appropriate?

Short Answer

Expert verified
Yes, the mean differs significantly with strong evidence. Assumptions include randomness, normality, and known sample standard deviation.

Step by step solution

01

Define the Hypotheses

First, we determine the null and alternative hypotheses.- **Null Hypothesis (H_0):** The mean phosphorus level reported by the new method is equal to the true mean, \( \mu = 548 \, \mathrm{mg/kg} \).- **Alternative Hypothesis (H_a):** The mean phosphorus level reported by the new method is not equal to the true mean, \( \mu eq 548 \, \mathrm{mg/kg} \). This is a two-tailed test.
02

Choose the Significance Level

The problem states that we should use a significance level of \( \alpha = 0.05 \).This is the probability of rejecting the null hypothesis when it is actually true.
03

Calculate the Test Statistic

The test statistic for a two-tailed t-test is calculated using the formula: \[t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \]where \( \bar{x} = 587 \, \mathrm{mg/kg} \) (sample mean), \( \mu = 548 \, \mathrm{mg/kg} \) (true mean), \( s = 10 \) (standard deviation), and \( n = 11 \) (sample size).Substitute the values:\[t = \frac{587 - 548}{10/\sqrt{11}} = \frac{39}{3.015} \approx 12.93\]
04

Determine Critical Values and Decision Rule

For a two-tailed test at \( \alpha = 0.05 \) and \( n - 1 = 10 \) degrees of freedom, use a t-table to find the critical t-values.The critical t-values for \( \alpha = 0.025 \) in each tail (since it's two-tailed) are approximately \( \pm 2.228 \).**Decision Rule:** Reject the null hypothesis if the calculated \( t \)-value is less than \( -2.228 \)or greater than \( 2.228 \).
05

Make a Conclusion

The calculated \( t \approx 12.93 \)is much greater than \( 2.228 \).According to our decision rule, this means we reject the null hypothesis.**Conclusion:** There is strong evidence to suggest that the mean phosphorus level reported by the new method differs significantly from the true mean.
06

State the Assumptions

For the t-test to be valid, we must assume:1. The sample of soil specimens is random.2. The distribution of phosphorus levels in the population is approximately normal or the sample size is large enough for the Central Limit Theorem to apply (however, with \( n = 11 \), normality is more critical).3. The standard deviation of the population is unknown but approximated by the sample standard deviation.

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Key Concepts

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

t-test
The t-test is a powerful statistical tool used to compare the mean of a sample to a known value or the means of two samples. It helps determine whether any observed differences are statistically significant or due to random chance.
In a t-test, we calculate the t-statistic using the formula: \[t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \]where:
  • \(\bar{x}\) is the sample mean.
  • \(\mu\) is the population mean or the mean being tested against.
  • \(s\) is the sample standard deviation.
  • \(n\) is the sample size.
For the problem addressed in the original exercise, the calculated t-statistic is 12.93, indicating a significant deviation from the population mean. Therefore, the new phosphorus measurement method yields results that differ from the known true mean.
Significance Level
The significance level (\(\alpha\)) is a threshold set before conducting a statistical test. It guides us in deciding whether to reject the null hypothesis, which generally implies no effect or no difference.
In hypothesis testing, the significance level is the probability of rejecting the null hypothesis when it is true—a Type I error. Typically, the significance level is set as 0.05 (5%), but it could be more or less depending on the context of the study.
  • If the p-value from the test is less than or equal to the significance level, we reject the null hypothesis.
  • If the p-value is greater, we fail to reject the null hypothesis.
In our example, a significance level of 0.05 was chosen. Since our computed t-value is beyond the critical values of \(\pm 2.228\), the decision was to reject the null hypothesis, suggesting a significant difference.
Null Hypothesis
The null hypothesis \((H_0)\) is a statement we aim to test in hypothesis testing, often representing no effect or no change. It acts as a starting point for the test. The goal is to see if there is enough evidence to reject this hypothesis in favor of an alternative.
For the phosphorus study, the null hypothesis was that the mean phosphorus level measured by the new method equals the true mean of 548 mg/kg.
The hypothesis testing process begins by assuming that the null hypothesis is true. After calculating the test statistic, if the evidence against it is strong enough (i.e., if the test statistic falls into the rejection region based on the chosen significance level), we reject the null hypothesis. In this case, the test showed strong evidence against the null hypothesis.
Central Limit Theorem
The Central Limit Theorem (CLT) is a key concept in statistics that states that the distribution of the sample mean will tend to be normal, regardless of the population's distribution, provided the sample size is sufficiently large. This is crucial when dealing with small sample sizes or unknown population distributions.
The CLT also allows us to make inferences about populations from sample data. It enables the use of statistical tests like the t-test, especially when the sample size is small, under the assumption of normality or when the population distribution is unknown. However, for small samples (like \(n = 11\) in our exercise), it is important to verify or assume the normal distribution of data for the results to be valid.
Within our soil phosphorus measurement context, the CLT helps justify using the t-test by assuming the sample mean distribution approaches normality given the underlying assumptions.

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Most popular questions from this chapter

Suppose the population distribution is normal with known \(\sigma\). Let \(\gamma\) be such that \(0<\gamma<\alpha\). For testing \(H_{0}: \mu=\mu_{0}\) versus \(H_{\mathrm{a}}: \mu \neq \mu_{0}\), consider the test that rejects \(H_{0}\) if either \(z \geq z_{\gamma}\) or \(z \leq-z_{\alpha-\gamma}\), where the test statistic is \(Z=\left(\bar{X}-\mu_{0}\right) /\) \((\sigma / \sqrt{n})\) a. Show that \(P\) (type I error) \(=\alpha\). b. Derive an expression for \(\beta\left(\mu^{\prime}\right)\). [Hint: Express the test in the form "reject \(H_{0}\) if either \(\bar{x} \geq c_{1}\) or \(\leq c_{2}\)."] c. Let \(\Delta>0\). For what values of \(\gamma\) (relative to \(\alpha\) ) will \(\beta\left(\mu_{0}+\Delta\right)<\beta\left(\mu_{0}-\Delta\right) ?\)

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from an exponential distribution with parameter \(\lambda\). Then it can be shown that \(2 \lambda \sum X_{i}\) has a chi-squared distribution with \(\nu=2 n\) (by first showing that \(2 \lambda X_{i}\) has a chi-squared distribution with \(\nu=2\) ). a. Use this fact to obtain a test statistic and rejection region that together specify a level \(\alpha\) test for \(H_{0}: \mu=\mu_{0}\) versus each of the three commonly encountered alternatives. [Hint: \(E\left(X_{i}\right)=\mu=1 / \lambda\), so \(\mu=\mu_{0}\) is equivalent to \(\lambda=1 / \mu_{0}\). b. Suppose that ten identical components, each having exponentially distributed time until failure, are tested. The resulting failure times are $$ \begin{array}{llllllllll} 95 & 16 & 11 & 3 & 42 & 71 & 225 & 64 & 87 & 123 \end{array} $$ Use the test procedure of part (a) to decide whether the data strongly suggests that the true average lifetime is less than the previously claimed value of 75 .

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