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 Sci., 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 by first determining the \(P\)-value. 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

We will perform a two-tailed test. The null hypothesis \(H_0\) is that the population mean \(\mu\) is equal to 3\%, and the alternative hypothesis \(H_a\) is that the population mean is not equal to 3\%. Mathematically, this can be expressed as: \ H_0: \mu = 3, \ H_a: \mu eq 3.

02

Calculate the Test Statistic

The test statistic for the mean is given by the formula: \ \text{Test Statistic} = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}, \ where \(\bar{x}\) is the sample mean, \(\mu_0 = 3\) is the hypothesized population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size. First, calculate \(n=30\), as there are 30 data points. Plug the values into the formula to get the test statistic: \ \text{Test Statistic} = \frac{2.481 - 3}{0.295} \approx -1.761.

03

Determine the P-value

Since the problem specifies a two-tailed test, we use the calculated test statistic \(-1.761\) to find the P-value. Using a t-distribution table with \(n-1=29\) degrees of freedom, we find the P-value by looking at the area in both tails. The P-value is approximately 0.0897.

04

Compare P-value with Alpha Level

Compare the P-value (0.0897) with the significance level \(\alpha = 0.10\). Since the P-value is less than 0.10, we reject the null hypothesis at this level. However, with \(\alpha = 0.05\), the P-value is greater, so we would not reject the null hypothesis at this stricter significance level.

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Key Concepts

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

P-value

In hypothesis testing, the P-value plays a crucial role in helping us make decisions about our hypotheses. It represents the probability of obtaining results as extreme as the observed data, assuming the null hypothesis is true. In simpler terms, it gives us an idea of how likely it is to observe the data we have if the null hypothesis is correct.

• A low P-value (typically less than the chosen significance level, such as 0.05 or 0.10) suggests that the observed data is unlikely under the null hypothesis. This can lead us to reject the null hypothesis.
• A high P-value suggests that the observed data is relatively likely under the null hypothesis, often leading us to not reject it.

In our example, the P-value was 0.0897. This value is compared against a pre-determined significance level, called \( \alpha \), which is the threshold for making our decision. If the P-value is smaller than \( \alpha \), we reject the null hypothesis. Here, at \( \alpha = 0.10 \), the P-value is less, so we reject the null. But at \( \alpha = 0.05 \), we don't reject it because 0.0897 is greater.

Two-tailed test

When considering hypothesis tests, a two-tailed test is used when we're interested in deviations in both directions—either lower or higher than a specific value. In our soil example, we want to test if the true mean is not equal to 3%, so a two-tailed test is appropriate.

• This type of test checks for the possibility of the sample mean being significantly lower or higher than the population mean.
• It uses both tails of the probability distribution to assess the deviation of the sample statistic from the hypothesized parameter.

In the example, the null hypothesis asserts that the population mean is exactly 3%, while the alternative hypothesis considers it might be any value other than 3%. Thus, we look at both extremes of the data distribution to decide whether to reject or not reject the null hypothesis.

Sample mean

The sample mean is a fundamental concept in statistics and serves as a key summary measure of a data set. It is calculated by summing all data points and dividing by the number of observations. The sample mean provides an estimate of the central tendency of the data.

• In hypothesis testing, the sample mean \( \bar{x} \) is used in constructing the test statistic.
• It provides a point estimate of the population mean, which we often use to make inferences about a population in statistical testing.

In our dataset of soil specimens, the sample mean is 2.481. This value becomes essential in calculating the test statistic, helping us determine how far this observed mean is from the hypothesized population mean of 3%. The difference between the sample mean and the hypothesized mean impacts our decision in hypothesis testing.

t-distribution

The t-distribution is a probability distribution used in statistics, particularly when dealing with small sample sizes or when the population standard deviation is unknown. It is similar to the normal distribution but has heavier tails, providing a more accurate estimation in such cases.

• The t-distribution accounts for the extra variability expected when estimating population parameters with smaller samples.
• Its shape depends on the degrees of freedom, which is related to the sample size (degrees of freedom = sample size - 1).

In our scenario, with a sample of 30 soil specimens, degrees of freedom are 29. We used the t-distribution to find the P-value corresponding to the calculated test statistic of -1.761. The t-distribution allowed us to make a probability-based assessment about the likelihood of our sample mean deviating this much from the hypothesized mean, under the assumption that the null hypothesis is true.