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 \(.10\) by first determining the \(P\)-value. Would your conclusion be different if \(\alpha=.05 \mathrm{had}\) 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

State the Hypotheses

We need to determine if the true average percentage of organic matter is different from 3. Therefore, the null hypothesis \( H_0 \) is that the mean \( \mu = 3 \% \), and the alternative hypothesis \( H_a \) is that \( \mu eq 3 \% \). This is a two-tailed test.

02

Calculate the Test Statistic

The test statistic \( t \) can be calculated using the formula: \[ t = \frac{\bar{x} - \mu}{s / \sqrt{n}} \] where \( \bar{x} = 2.481 \), \( \mu = 3 \), \( s = 1.616 \), and \( n = 30 \). Plug in these values to get: \[ t = \frac{2.481 - 3}{1.616 / \sqrt{30}} \approx -1.76 \]

03

Determine the Degrees of Freedom

The degrees of freedom \( df \) for this test is \( n - 1 \), where \( n = 30 \). So, \( df = 30 - 1 = 29 \).

04

Find the P-Value

Using a \( t\)-distribution table or calculator, find the \( P \)-value for \( t = -1.76 \) with \( df = 29 \). Since this is a two-tailed test, we look at both tails. The \( P \)-value is approximately 0.088.

05

Decision at Significance Level 0.10

Since the \( P \)-value of 0.088 is less than \( \alpha = 0.10 \), we reject the null hypothesis \( H_0 \). This suggests that there is evidence to say the true average is different than 3\%.

06

Decision at Significance Level 0.05

Now consider \( \alpha = 0.05 \). Here, the \( P \)-value of 0.088 is greater than \( \alpha = 0.05 \), so we fail to reject \( H_0 \). Thus, there is not enough evidence to conclude that the true average is different than 3\%.

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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 crucial concept in statistics. It represents the average value in a set of data. For our soil specimens data, the sample mean is calculated as 2.481. This means that, on average, the organic matter content in the soil samples analyzed is 2.481%.

Here's how to understand it:

• First, add up all the values in the data set.
• Next, divide the total by the number of data points (in this case, 30 soil specimens).

The sample mean is important because it serves as an estimate of the population mean. It gives us a starting point for making inferences about the soil's organic matter content in general.

Standard Deviation

The standard deviation provides insight into the variability or spread of a data set. In simpler terms, it tells us how much the individual data points deviate from the sample mean. For our soil data, the standard deviation is 1.616.

Here's a breakdown:

• If the standard deviation is low, it means the data points are close to the mean.
• If it's high, the data points are spread out over a wider range.

Calculating this involves determining the square root of the variance, where variance measures the average squared differences from the mean. Understanding standard deviation helps us comprehend the consistency of organic matter in the soil samples.

T-Distribution

The t-distribution is a type of probability distribution that is used when we are dealing with small sample sizes or when the population standard deviation is unknown. For our exercise, a t-distribution is suitable given the sample size of 30.

Key characteristics include:

• It's symmetrical and bell-shaped, similar to the normal distribution, but with heavier tails.
• This means it has more probability mass in its tails, suitable for small sample sizes as it gives a more conservative estimate compared to the normal distribution.

In hypothesis testing, the t-distribution allows us to calculate the t-statistic and subsequently the p-value, which helps in decision making.

Significance Level

The significance level, often denoted as alpha (\( \alpha \)), is a threshold we set to decide if we should reject the null hypothesis during hypothesis testing. In our problem, we're dealing with two significance levels: 0.10 and 0.05.

Here's what it involves:

• It's the probability of incorrectly rejecting a true null hypothesis (a Type I error).
• A lower significance level means stricter criteria for rejecting the null hypothesis.

For the 0.10 level, our calculated p-value of 0.088 led us to reject the null hypothesis, suggesting our data provides evidence against the mean being 3%. However, at the more stringent 0.05 level, we fail to reject it, showing the importance of understanding and choosing suitable significance levels in research and decision-making.