91Ó°ÊÓ

A random sample of soil specimens was obtained, and the amount of organic matter \((\%)\) in the soil was determined for each specimen, resulting in the accompanying data (from "Engineering Properties of Soil," Soil Science, 1998: 93-102). \(\begin{array}{llllllll}1.10 & 5.09 & 0.97 & 1.59 & 4.60 & 0.32 & 0.55 & 1.45 \\\ 0.14 & 4.47 & 1.20 & 3.50 & 5.02 & 4.67 & 5.22 & 2.69 \\ 3.98 & 3.17 & 3.03 & 2.21 & 0.69 & 4.47 & 3.31 & 1.17 \\ 0.76 & 1.17 & 1.57 & 2.62 & 1.66 & 2.05 & & \end{array}\) The values of the sample mean, sample standard deviation, and (estimated) standard error of the mean are \(2.481,1.616\), and \(.295\), respectively. Does this data suggest that the true average percentage of organic matter in such soil is something other than \(3 \%\) ? Carry out a test of the appropriate hypotheses at significance level .10. Would your conclusion be different if \(\alpha=.05\) had been used? [Note: A normal probability plot of the data shows an acceptable pattern in light of the reasonably large sample size.]

Step by step solution

01

Define the Hypotheses

To determine if the true average percentage of organic matter is different from 3%, we set up the null and alternative hypotheses. The null hypothesis, \(H_0\), is that the population mean \(\mu = 3\%\). The alternative hypothesis, \(H_a\), is that \(\mu eq 3\%\).

02

Determine the Test Statistic

We use the sample mean, \(\bar{x} = 2.481\), the hypothesized population mean \(\mu_0 = 3\), the sample standard deviation \(s = 1.616\), and the sample size \(n = 30\) to calculate the test statistic. The test statistic for the t-test is computed using: \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} = \frac{2.481 - 3}{1.616 / \sqrt{30}} \approx -1.759. \]

03

Determine the Critical Value and Decision Rule

For a two-tailed test at \(\alpha = 0.10\) with \(df = n - 1 = 29\), we find the critical t-value from the t-distribution table. The critical values are approximately \(\pm 1.699\). If the test statistic is less than \(-1.699\) or greater than \(1.699\), we reject the null hypothesis.

04

Compare Test Statistic to Critical Values

Our calculated t-statistic is \(-1.759\). Since \(-1.759 < -1.699\), it falls in the rejection region. Thus, we reject the null hypothesis at \(\alpha = 0.10\). This suggests there is evidence the true mean is different from 3%.

05

Consider the Impact of a Different Significance Level

For \(\alpha = 0.05\), the critical t-values for two tails with \(df = 29\) are approximately \(\pm 2.045\). Our test statistic of \(-1.759\) does not fall outside these values, so we fail to reject the null hypothesis at \(\alpha = 0.05\).

06

Conclusion

At \(\alpha = 0.10\), we reject the null hypothesis suggesting the mean is different from 3%. At \(\alpha = 0.05\), we do not reject the null hypothesis.

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Key Concepts

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

Sample Mean

The sample mean is a fundamental measure in statistics and represents the average of a set of values. After collecting the data from soil specimens, we calculate the sample mean to get an estimation of the central tendency. The sample mean, denoted by \( \bar{x} \), serves as the first point of comparison with any hypothesized population mean.
For this soil study, the sample mean is \(2.481\), calculated by summing the measured organic matter percentages and dividing by the total number of samples (30). This value is crucial as it represents the average percentage of organic matter in the sampled soil and is used later in hypothesis testing. Understanding the sample mean provides insight into the overall composition of the soil, and it acts as a representation of what might be expected on average from the broader population of similar soil specimens.

T-Test

The t-test is a statistical method used to determine if there is a significant difference between the sample mean and a known value of the population mean. This test is particularly useful when the sample size is small or the population standard deviation is unknown. It operates under assumptions such as normally distributed data, which in this exercise is justified by an acceptable normal probability plot and a reasonably large sample size.
To perform the t-test here, the test statistic is calculated using the formula: \[t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \]where \( \bar{x} \) is the sample mean, \( \mu_0 \) is the hypothesized population mean (3% in this case), \( s \) is the sample standard deviation, and \( n \) is the sample size. The computed t-statistic is approximately \(-1.759\), which guides us in determining the significance of the difference between our sample mean and the hypothesized mean.

Critical Value

In hypothesis testing, the critical value acts as a threshold to determine whether the test statistic indicates a significant result. It is derived from the t-distribution table corresponding to the chosen significance level and degrees of freedom.
For this example, at a significance level of \(\alpha = 0.10\) and \(df = 29\), the critical values are approximately \( \pm 1.699\) for a two-tailed test. These values define the rejection regions. If the calculated t-statistic falls beyond these critical values (either too high or too low), it suggests that the observed effect is significant, leading to the rejection of the null hypothesis. When conducting tests, understanding how to find and interpret critical values is crucial for making informed decisions about the null hypothesis.

Significance Level

The significance level, denoted by \(\alpha\), reflects the probability of rejecting the null hypothesis when it is true, known as a Type I error. Common levels used are 0.05, 0.10, etc., representing a 5% or 10% risk of such errors.
In this soil data exercise, two significance levels are considered: \(\alpha = 0.10\) and \(\alpha = 0.05\). At \(\alpha = 0.10\), the decision is to reject the null hypothesis, suggesting the sample mean is significantly different from 3%. However, at a stricter \(\alpha = 0.05\), the evidence is insufficient to reject the null hypothesis.
Thus, the choice of significance level affects the conclusion of hypothesis tests, highlighting its importance in the context of interpreting statistical results.