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Chapter 8: Problem 58

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 \(\alpha=.05\) had been used? [Note: A normal probability plot of the data shows an acceptable pattern in light of the reasonably large sample size.]

Short Answer

Expert verified
At \(\alpha = 0.10\), reject \(H_0\); at \(\alpha = 0.05\), do not reject \(H_0\).

Step by step solution

01

State Hypotheses

We need to formulate our null and alternative hypotheses. The null hypothesis (\(H_0\)) is that the true average percentage of organic matter is 3%, while the alternative hypothesis (\(H_a\)) is that it is not 3%.\[ H_0: \mu = 3 \]\[ H_a: \mu eq 3 \]
02

Determine Test Statistic

The test statistic for the hypothesis test can be calculated using the following formula: \[ z = \frac{\bar{x} - \mu_0}{(s / \sqrt{n})} \]where \(\bar{x} = 2.481\) is the sample mean, \(\mu_0 = 3\) is the hypothesized population mean, \(s = 1.616\) is the sample standard deviation, and \(n = 30\) (the sample size). The standard error given is \(0.295\).
03

Compute the Test Statistic

Substitute the values into the formula:\[ z = \frac{2.481 - 3}{0.295} \approx -1.759 \]
04

Determine the P-value

The P-value is the probability of obtaining a test statistic as or more extreme than the observed one, assuming the null hypothesis is true. Since it is a two-tailed test (as \( H_a \) involves \( eq \)), the P-value is calculated as:Two-tailed P-value = \(2 \times P(z < -1.759)\).Lookup the standard normal table or use a calculator to find \(P(z < -1.759)\). P-value \( \approx 2 \times 0.0393 = 0.0786 \).
05

Compare P-value to Significance Levels

Compare the P-value to the significance levels \(\alpha = 0.10\) and \(\alpha = 0.05\):- For \( \alpha = 0.10 \): The P-value (0.0786) is less than 0.10, so we reject \(H_0\).- For \( \alpha = 0.05 \): The P-value (0.0786) is greater than 0.05, so we do not reject \(H_0\).
06

State Conclusion

At the \(\alpha = 0.10\) significance level, we reject the null hypothesis and conclude that there is evidence to suggest the true average is not 3%. At the \(\alpha = 0.05\) significance level, we do not reject the null hypothesis and conclude that there is not enough evidence to suggest the true average differs from 3%.

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Key Concepts

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

Sample Mean
When working with data, we often use the concept of a sample mean, which helps us organize and make decisions based on the data. The sample mean is simply the average of the values in a sample. To find it, add up all the values and then divide by the number of values. In the context of this exercise, the sample mean reflects the average percentage of organic matter in the collected soil specimens. It is denoted as \( \bar{x} \).
The formula to calculate the sample mean is:
  • \( \bar{x} = \frac{\sum{x_i}}{n} \)
where \( \sum{x_i} \) is the sum of all the data points, and \( n \) is the number of data points.
In this exercise:
  • The sample mean \( \bar{x} \) is given as 2.481.
Understanding the sample mean is crucial, as it is often used as the best estimate of the true population mean in hypothesis testing.
Standard Deviation
Understanding variability in data is vital, and standard deviation helps us with this. Standard deviation measures how spread out the numbers in a data set are. A small standard deviation suggests that the data points are close to the mean, while a large one indicates more spread.
The formula for calculating standard deviation in a sample is:
  • \( s = \sqrt{\frac{\sum{(x_i - \bar{x})^2}}{n-1}} \)
where
  • \( s \) is the sample standard deviation
  • \( x_i \) are the individual data points
  • \( \bar{x} \) is the sample mean
  • \( n \) is the number of data points
In this exercise:
  • The sample standard deviation \( s \) is given as 1.616.
This value helps us understand the variability of organic matter percentage in the soil specimens and plays a significant role in the hypothesis test, as it is used to compute the test statistic.
Significance Level
The significance level, denoted as \( \alpha \), is a threshold we set to decide whether to reject the null hypothesis. It represents the probability of making a Type I error, which means rejecting a true null hypothesis. Typical significance levels used are 0.01, 0.05, and 0.10.
In this problem:
  • A significance level of 0.10 was initially used.
  • Alternately, it asks to consider \( \alpha = 0.05 \).
The selection of \( \alpha \) affects the conclusion of the hypothesis test. At \( \alpha = 0.10 \), we reject the null hypothesis if the p-value is less than 0.10, suggesting that there's convincing evidence to believe the true average differs from 3\%. Conversely, at \( \alpha = 0.05 \), the evidence must be stronger (p-value less than 0.05) to reject the null hypothesis.
P-value
In hypothesis testing, the p-value plays a crucial role in deciding whether to reject the null hypothesis. It measures the probability of observing a test statistic as extreme as the one obtained, assuming the null hypothesis is true.
The interpretation process is straightforward:
  • If the p-value is less than or equal to the chosen significance level \( \alpha \), reject the null hypothesis, indicating sufficient evidence against it.
  • If the p-value is greater, do not reject the null hypothesis.
In this exercise:
  • The computed p-value is approximately 0.0786.
Comparisons with significance levels:
  • For \( \alpha = 0.10 \): The p-value (0.0786) is less, so we reject \( H_0 \).
  • For \( \alpha = 0.05 \): The p-value is greater, so we do not reject \( H_0 \).
Understanding p-value is crucial, as it provides insight into the strength of the evidence against the null hypothesis.

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Most popular questions from this chapter

Let \(X_{1}, \ldots, X_{n}\) denote a random sample from a normal population distribution with a known value of \(\sigma\). a. For testing the hypotheses \(H_{0}: \mu=\mu_{0}\) versus \(H_{\mathrm{a}}: \mu>\mu_{0}\) (where \(\mu_{0}\) is a fixed number), show that the test with test statistic \(\bar{X}\) and rejection region \(\bar{x} \geq \mu_{0}+2.33 \sigma / \sqrt{n}\) has significance level \(.01\). b. Suppose the procedure of part (a) is used to test \(H_{0}: \mu \leq \mu_{0}\) versus \(H_{\mathrm{a}}: \mu>\mu_{0}\). If \(\mu_{0}=100, n=25\), and \(\sigma=5\), what is the probability of committing a type I error when \(\mu=99\) ? When \(\mu=98\) ? In general, what can be said about the probability of a type I error when the actual value of \(\mu\) is less than \(\mu_{0}\) ? Verify your assertion.

The article referenced in Example \(8.11\) also reported that in a sample of 106 wine consumers, \(22(20.8 \%)\) thought that screw tops were an acceptable substitute for natural corks. Suppose a particular winery decided to use screw tops for one of its wines unless there was strong evidence to suggest that fewer than \(25 \%\) of wine consumers found this acceptable. a. Using a significance level of \(.10\), what would you recommend to the winery? b. For the hypotheses tested in (a), describe in context what the type I and II errors would be, and say which type of error might have been committed.

A new design for the braking system on a certain type of car has been proposed. For the current system, the true average braking distance at \(40 \mathrm{mph}\) under specified conditions is known to be \(120 \mathrm{ft}\). It is proposed that the new design be implemented only if sample data strongly indicates a reduction in true average braking distance for the new design. a. Define the parameter of interest and state the relevant hypotheses. b. Suppose braking distance for the new system is normally distributed with \(\sigma=10\). Let \(\bar{X}\) denote the sample average braking distance for a random sample of 36 observations. Which of the following three rejection regions is appropriate: \(R_{1}=\\{\bar{x}: \bar{x} \geq 124.80\\}, \quad R_{2}=\\{\bar{x}: \bar{x} \leq 115.20\\}\), \(R_{3}=\\{\bar{x}\) : either \(\bar{x} \geq 125.13\) or \(\bar{x} \leq 114.87\\}\) ? c. What is the significance level for the appropriate region of part (b)? How would you change the region to obtain a test with \(\alpha=.001\) ? d. What is the probability that the new design is not implemented when its true average braking distance is actually \(115 \mathrm{ft}\) and the appropriate region from part (b) is used? e. Let \(Z=(\bar{X}-120) /(\sigma / \sqrt{n})\). What is the significance level for the rejection region \(\\{z: z \leq-2.33\\}\) ? For the region \(\\{z: z \leq-2.88\\}\) ?

The accompanying observations on residual flame time (sec) for strips of treated children's nightwear were given in the article "An Introduction to Some Precision and Accuracy of Measurement Problems" ( \(J\). of Testing and Eval., 1982: 132-140). Suppose a true average flame time of at most \(9.75\) had been mandated. Does the data suggest that this condition has not been met? Carry out an appropriate test after first investigating the plausibility of assumptions that underlie your method of inference. $$ \begin{array}{lllllll} 9.85 & 9.93 & 9.75 & 9.77 & 9.67 & 9.87 & 9.67 \\ 9.94 & 9.85 & 9.75 & 9.83 & 9.92 & 9.74 & 9.99 \\ 9.88 & 9.95 & 9.95 & 9.93 & 9.92 & 9.89 & \end{array} $$

Have you ever been frustrated because you could not get a container of some sort to release the last bit of its contents? The article "Shake, Rattle, and Squeeze: How Much Is Left in That Container?" (Consumer Reports, May 2009: 8) reported on an investigation of this issue for various consumer products. Suppose five \(6.0 \mathrm{oz}\) tubes of toothpaste of a particular brand are randomly selected and squeezed until no more toothpaste will come out. Then each tube is cut open and the amount remaining is weighed, resulting in the following data (consistent with what the cited article reported): \(.53, .65, .46, .50, .37\). Does it appear that the true average amount left is less than \(10 \%\) of the advertised net contents? a. Check the validity of any assumptions necessary for testing the appropriate hypotheses. b. Carry out a test of the appropriate hypotheses using a significance level of \(.05\). Would your conclusion change if a significance level of .01 had been used? c. Describe in context type I and II errors, and say which error might have been made in reaching a conclusion.
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