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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

Give as much information as you can about the \(P\)-value of a \(t\) test in each of the following situations: a. Upper-tailed test, df \(=8, t=2.0\) b. Lower-tailed test, \(\mathrm{df}=11, t=-2.4\) c. Two-tailed test, df \(=15, t=-1.6\) d. Upper-tailed test, df \(=19, t=-.4\) e. Upper-tailed test, \(\mathrm{df}=5, t=5.0\) f. Two-tailed test, \(\mathrm{df}=40, t=-4.8\)

Scientists think that robots will play a crucial role in factories in the next several decades. Suppose that in an experiment to determine whether the use of robots to weave computer cables is feasible, a robot was used to assemble 500 cables. The cables were examined and there were 15 defectives. If human assemblers have a defect rate of \(.035\) \((3.5 \%)\), does this data support the hypothesis that the proportion of defectives is lower for robots than for humans? Use a .01 significance level.

The calibration of a scale is to be checked by weighing a \(10-\mathrm{kg}\) test specimen 25 times. Suppose that the results of different weighings are independent of one another and that the weight on each trial is normally distributed with \(\sigma=.200 \mathrm{~kg}\). Let \(\mu\) denote the true average weight reading on the scale. a. What hypotheses should be tested? b. Suppose the scale is to be recalibrated if either \(\bar{x} \geq 10.1032\) or \(\bar{x} \leq 9.8968\). What is the probability that recalibration is carried out when it is actually unnecessary? c. What is the probability that recalibration is judged unnecessary when in fact \(\mu=10.1\) ? When \(\mu=9.8\) ? d. Let \(z=(\bar{x}-10) /(\sigma / \sqrt{n})\). For what value \(c\) is the rejection region of part (b) equivalent to the "two-tailed" region of either \(z \geq c\) or \(z \leq-c\) ? e. If the sample size were only 10 rather than 25 , how should the procedure of part (d) be altered so that \(\alpha=.05\) ? f. Using the test of part (e), what would you conclude from the following sample data? $$ \begin{array}{rrrrr} 9.981 & 10.006 & 9.857 & 10.107 & 9.888 \\ 9.728 & 10.439 & 10.214 & 10.190 & 9.793 \end{array} $$ g. Reexpress the test procedure of part (b) in terms of the standardized test statistic \(Z=(\bar{X}-10) /(\sigma / \sqrt{n})\).

For the following pairs of assertions, indicate which do not comply with our rules for setting up hypotheses and why (the subscripts 1 and 2 differentiate between quantities for two different populations or samples): a. \(H_{0}: \mu=100, H_{\mathrm{a}}: \mu>100\) b. \(H_{0}: \sigma=20, H_{\mathrm{a}}: \sigma \leq 20\) c. \(H_{0}: p \neq .25, H_{\mathrm{a}}: p=.25\) d. \(H_{0}: \mu_{1}-\mu_{2}=25, H_{\mathrm{a}}: \mu_{1}-\mu_{2}>100\) e. \(H_{0}: S_{1}^{2}=S_{2}^{2}, H_{\mathrm{a}}: S_{1}^{2} \neq S_{2}^{2}\) f. \(H_{0}: \mu=120, H_{\mathrm{a}}: \mu=150\) g. \(H_{0}: \sigma_{1} / \sigma_{2}=1, H_{\mathrm{a}}: \sigma_{1} / \sigma_{2} \neq 1\) h. \(H_{0}: p_{1}-p_{2}=-.1, H_{\mathrm{a}}: p_{1}-p_{2}<-.1\)

A mixture of pulverized fuel ash and Portland cement to be used for grouting should have a compressive strength of more than \(1300 \mathrm{KN} / \mathrm{m}^{2}\). The mixture will not be used unless experimental evidence indicates conclusively that the strength specification has been met. Suppose compressive strength for specimens of this mixture is normally distributed with \(\sigma=60\). Let \(\mu\) denote the true average compressive strength. a. What are the appropriate null and alternative hypotheses? b. Let \(\bar{X}\) denote the sample average compressive strength for \(n=10\) randomly selected specimens. Consider the test procedure with test statistic \(\bar{X}\) and rejection region \(\bar{x} \geq 1331.26\). What is the probability distribution of the test statistic when \(H_{0}\) is true? What is the probability of a type I error for the test procedure? c. What is the probability distribution of the test statistic when \(\mu=1350\) ? Using the test procedure of part (b), what is the probability that the mixture will be judged unsatisfactory when in fact \(\mu=1350\) (a type II error)? d. How would you change the test procedure of part (b) to obtain a test with significance level .05? What impact would this change have on the error probability of part (c)? e. Consider the standardized test statistic \(Z=(\bar{X}-1300) /(\sigma / \sqrt{n})=(\bar{X}-1300) / 13.42\). What are the values of \(Z\) corresponding to the rejection region of part (b)?
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