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Chapter 8: Problem 26

To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil for a 2-year period. The maximum penetration (in mils) for each specimen is then measured, yielding a sample average penetration of \(\bar{x}=52.7\) and a sample standard deviation of \(s=4.8\). The conduits were manufactured with the specification that true average penetration be at most 50 mils. They will be used unless it can be demonstrated conclusively that the specification has not been met. What would you conclude?

Short Answer

Expert verified
Reject the null hypothesis; conclude the true average penetration is above 50 mils.

Step by step solution

01

Define the Hypotheses

We are conducting a hypothesis test to determine if the true average penetration is greater than 50 mils. Set up the null hypothesis as \( H_0: \mu = 50 \) and the alternative hypothesis as \( H_a: \mu > 50 \).
02

Determine the Significance Level

Typically, a 5% significance level \( \alpha = 0.05 \) is used for hypothesis tests. This will be our threshold to determine whether the null hypothesis should be rejected.
03

Calculate the Test Statistic

Use the formula for the t-test statistic for a sample mean: \[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \] Substitute \( \bar{x} = 52.7 \), \( \mu_0 = 50 \), \( s = 4.8 \), and \( n = 45 \):\[ t = \frac{52.7 - 50}{4.8/\sqrt{45}} \]Calculate the result.
04

Calculate the Degrees of Freedom

The degrees of freedom for this test is \( n - 1 = 45 - 1 = 44 \).
05

Determine the Critical Value and Compare

Using a t-distribution table, find the critical value for \( t \) with 44 degrees of freedom at \( \alpha = 0.05 \) significance level. If the calculated \( t \) statistic is greater than this critical value, reject the null hypothesis.
06

Conclusion

If the calculated \( t \)-statistic exceeds the critical value from the table, it indicates sufficient evidence to reject the null hypothesis, concluding that the true average penetration is indeed greater than 50 mils.

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Key Concepts

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

t-test
A t-test is a statistical test used to compare the means of different groups or to test a hypothesis about a mean when the sample size is small or the population standard deviation is unknown.
In our scenario, the aim is to determine whether the average penetration of the steel conduits exceeds the desired specification of 50 mils.
Types of t-test:
  • One-Sample t-test: Compares the mean of a single group against a known mean.
  • Independent Two-Sample t-test: Compares the means of two independent groups.
  • Paireld t-test: Used for comparing means from the same group at different times.
For our exercise, we're using a one-sample t-test as we're comparing the mean penetration against a fixed specification. The formula is:\[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \]where:
  • \( \bar{x} \) is the sample mean
  • \( \mu_0 \) is the population mean under the null hypothesis
  • \( s \) is the sample standard deviation
  • \( n \) is the sample size
significance level
The significance level, often denoted by \( \alpha \), is a threshold that determines when to reject a null hypothesis. Commonly set at 0.05 or 5%, it indicates the probability of rejecting a true null hypothesis.
The choice of level reflects how cautious you want to be when claiming an effect exists.
  • 0.05 Significance Level: This means there's a 5% risk of concluding that a difference exists when there is none.
  • Lower Significance Level: Reduces the chance of a Type I error, but may increase the chance of a Type II error.
  • Higher Significance Level: Increases the chance of finding a significant effect (even if it doesn't exist).
In our example, we use \( \alpha = 0.05 \), meaning if the calculated \( t \)-statistic is beyond the critical value corresponding to this level, we reject the null hypothesis with a 95% confidence.
critical value
The critical value is a point on the test distribution used to decide whether to reject the null hypothesis. It depends on the significance level \( \alpha \) and the degrees of freedom. In hypothesis testing, if the test statistic exceeds the critical value, the null hypothesis is rejected.
For our t-test, the critical value is derived from the t-distribution table for 44 degrees of freedom (since our sample size is 45) and a significance level of 0.05.
  • Calculating Critical Value: Use a t-distribution table or software to find the specific critical value for the degrees of freedom and \( \alpha \).
The concept is straightforward:
  • If \( |t| > \text{critical value} \), reject \( H_0 \).
  • If \( |t| \leq \text{critical value} \), do not reject \( H_0 \).
degrees of freedom
Degrees of freedom refer to the number of values in a calculation that can vary freely. In t-tests, they help in determining the shape of the t-distribution, which in turn affects the critical value.
For a sample size of \( n \), the degrees of freedom for the one-sample t-test is calculated as \( n-1 \). In the context of our exercise, where 45 specimens are analyzed:
  • Degrees of Freedom = \( 45 - 1 = 44 \)
Degrees of freedom are crucial for identifying the right distribution to use when looking up the critical value:
  • Smaller samples: Result in wider distributions, reflecting more variability among the sample means.
  • Larger samples: Result in narrower distributions, indicating more stability and less variability in the sample means.
Understanding degrees of freedom ensures that your test reflects the correct level of statistical confidence based on the sample size.

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Most popular questions from this chapter

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. \(H_{0}: p_{1}-p_{2}=-.1, H_{\mathrm{a}}: p_{1}-p_{2}<-.1\)

Water samples are taken from water used for cooling as it is being discharged from a power plant into a river. It has been determined that as long as the mean temperature of the discharged water is at most \(150^{\circ} \mathrm{F}\), there will be no negative effects on the river's ecosystem. To investigate whether the plant is in compliance with regulations that prohibit a mean discharge-water temperature above \(150^{\circ}, 50\) water samples will be taken at randomly selected times, and the temperature of each sample recorded. The resulting data will be used to test the hypotheses \(H_{0}: \mu=150^{\circ}\) versus \(H_{\mathrm{a}}: \mu>150^{\circ}\). In the context of this situation, describe type I and type II errors. Which type of error would you consider more serious? Explain.

One method for straightening wire before coiling it to make a spring is called "roller straightening." The article "The Effect of Roller and Spinner Wire Straightening on Coiling Performance and Wire Properties" (Springs, 1987: 27-28) reports on the tensile properties of wire. Suppose a sample of 16 wires is selected and each is tested to determine tensile strength \(\left(\mathrm{N} / \mathrm{mm}^{2}\right)\). The resulting sample mean and standard deviation are 2160 and 30 , respectively. a. The mean tensile strength for springs made using spinner straightening is \(2150 \mathrm{~N} / \mathrm{mm}^{2}\). What hypotheses should be tested to determine whether the mean tensile strength for the roller method exceeds 2150 ? b. Assuming that the tensile strength distribution is approximately normal, what test statistic would you use to test the hypotheses in part (a)? c. What is the value of the test statistic for this data? d. What is the \(P\)-value for the value of the test statistic computed in part (c)? e. For a level \(.05\) test, what conclusion would you reach?

Two different companies have applied to provide cable television service in a certain region. Let \(p\) denote the proportion of all potential subscribers who favor the first company over the second. Consider testing \(H_{0}: p=.5\) versus \(H_{\mathrm{a}}: p \neq .5\) based on a random sample of 25 individuals. Let \(X\) denote the number in the sample who favor the first company and \(x\) represent the observed value of \(X\). a. Which of the following rejection regions is most appropriate and why? $$ \begin{aligned} &R_{1}=\\{x: x \leq 7 \text { or } x \geq 18\\}, R_{2}=\\{x: x \leq 8\\}, \\ &R_{3}=\\{x: x \geq 17\\} \end{aligned} $$ b. In the context of this problem situation, describe what type I and type II errors are. c. What is the probability distribution of the test statistic \(X\) when \(H_{0}\) is true? Use it to compute the probability of a type I error. d. Compute the probability of a type II error for the selected region when \(p=.3\), again when \(p=.4\), and also for both \(p=.6\) and \(p=.7\). e. Using the selected region, what would you conclude if 6 of the 25 queried favored company 1 ?

The times of first sprinkler activation for a series of tests with fire prevention sprinkler systems using an aqueous film-forming foam were (in sec) \(\begin{array}{lllllllllllll}27 & 41 & 22 & 27 & 23 & 35 & 30 & 33 & 24 & 27 & 28 & 22 & 24\end{array}\) (see "Use of AFFF in Sprinkler Systems," Fire Technology, 1976: 5). The system has been designed so that true average activation time is at most \(25 \mathrm{sec}\) under such conditions. Does the data strongly contradict the validity of this design specification? Test the relevant hypotheses at significance level \(.05\) using the \(P\)-value approach.
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