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Chapter 8: Problem 22

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
The specification of an average penetration of at most 50 mils has not been met.

Step by step solution

01

Define the Hypotheses

We need to state both the null hypothesis (\( H_0 \)) and the alternative hypothesis (\( H_a \)) for this test. Here, the null hypothesis is that the true average penetration is 50 mils or less, \( H_0: \mu \leq 50 \). The alternative hypothesis we want to test is that the true average penetration is more than 50 mils, \( H_a: \mu > 50 \).
02

Select the Test Statistic

Since the sample size is relatively large (45 specimens), we can use the t-test for the sample mean as an approximation. The test statistic \( t \) is calculated using the formula: \[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \] where \( \bar{x} = 52.7 \), \( \mu_0 = 50 \), \( s = 4.8 \), and \( n = 45 \).
03

Calculate the Test Statistic Value

Plug the given values into the formula: \[ t = \frac{52.7 - 50}{4.8/\sqrt{45}} \]. Calculating this, \( t = \frac{2.7}{4.8/\sqrt{45}} = \frac{2.7}{0.7152} \approx 3.775 \). So, the calculated t-value is approximately 3.775.
04

Determine the Critical Value and Compare

For a one-tailed test at a significance level (usually \( \alpha = 0.05 \)), the critical value for \( t \) with \( n-1 = 44 \) degrees of freedom can be found in t-distribution tables or using statistical software. The critical t-value for \( \alpha = 0.05 \) and 44 degrees of freedom is approximately 1.680. Compare the calculated t-value (3.775) to the critical value (1.680).
05

Make a Decision

Since the calculated t-value (3.775) is greater than the critical t-value (1.680), we reject the null hypothesis \( H_0 \). This means there is sufficient evidence at the \( \alpha = 0.05 \) level to conclude that the true average penetration is 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
In statistics, the t-test is a method used to determine if there is a significant difference between the means of two groups. It's particularly useful when dealing with smaller sample sizes, although in this case it was applied to a reasonably large sample of steel conduit specimens. When you perform a t-test, you compute the t-statistic using the formula:
\[ t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} \]
where \( \bar{x} \) is the sample mean, \( \mu_0 \) is the known value of the population mean you're testing against, \( s \) is the sample standard deviation, and \( n \) is the number of samples in your study.
  • In our exercise, the t-statistic helps determine if the actual average penetration of steel conduits differs from the claimed specification of 50 mils.
  • A higher absolute value of the t-statistic suggests a more considerable difference between the sample mean and the known population mean.
null hypothesis
The null hypothesis, denoted as \( H_0 \), acts as the starting assumption for statistical testing. It's a statement that suggests there is no effect or no difference, and we try to test against this assumption. By seeking to reject \( H_0 \), researchers aim to provide evidence that a significant effect exists in their study or observation.
In this context:
  • For the steel penetration exercise, the null hypothesis asserts that the true average penetration is at most 50 mils, expressed as \( H_0: \mu \leq 50 \).
  • This hypothesis is put in place to challenge the conduit's expected performance limiting to 50 mils of corrosion penetration.
The goal in hypothesis testing is often to determine if there's sufficient evidence to reject \( H_0 \), suggesting that the real-world performance deviates from the expected, laid-out benchmarks.
alternative hypothesis
The alternative hypothesis, denoted as \( H_a \), complements the null hypothesis. It is a statement that proposes a different scenario, usually representing the effect or difference the researcher is attempting to verify. If evidence supports \( H_a \), it implies a significant finding that challenges the initial dogma expressed by \( H_0 \).
In this exercise:
  • The alternative hypothesis claims that the true average penetration exceeds 50 mils, expressed as \( H_a: \mu > 50 \).
  • This statement reflects the concern over the conduit's performance potentially not meeting the corrosion-resistance expectation, suggesting penetration might indeed be deeper than desired.
When the calculated t-value exceeds the critical value in hypothesis testing, it indicates enough evidence to support \( H_a \), leading to a decision to reject \( H_0 \).
significance level
The significance level in hypothesis testing is denoted by \( \alpha \) and represents the probability of rejecting the null hypothesis when it is actually true. It is a threshold set by the researcher to define how confident they need to be in their results before concluding a significant finding. Commonly, \( \alpha \) is set at 0.05, indicating a 5% risk of making a Type I error, where the null hypothesis is incorrectly rejected.
For the steel conduit exercise, the significance level (
  • With \( \alpha \) set at 0.05, it means researchers are prepared to accept a 5% chance of concluding that the average penetration is greater than 50 mils when it actually is not.
  • The decision is based on comparing the calculated statistical value to the critical value derived from statistical tables or software. If the t-value obtained exceeds the critical t-value, the result is deemed statistically significant at the chosen \( \alpha \) level. Thus, \( \alpha = 0.05 \) becomes a pivotal factor in deciding whether the null hypothesis should be rejected.
This outcome provides the statistical evidence needed to make informed engineering or manufacturing decisions regarding steel conduit suitability.

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Most popular questions from this chapter

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 \(H_{a}=p \neq .5\) based on a random sample of 25 individuals. Let the test statistic \(X\) be the number in the sample who favor the first company and \(x\) represent the observed value of \(X\). a. Describe type I and II errors in the context of this problem situation. b. Suppose that \(x=6\). Which values of \(X\) are at least as contradictory to \(H_{0}\) as this one? c. What is the probability distribution of the test statistic \(X\) when \(H_{0}\) is true? Use it to compute the \(P\)-value when \(x=6\). d. If \(H_{0}\) is to be rejected when \(P\)-value \(\leq .044\), compute the probability of a type II error when \(p=.4\), again when \(p=.3\), and also when \(p=.6\) and \(p=.7\). [Hint: \(P\)-value \(>.044\) is equivalent to what inequalities involving \(x\) (see Example 8.4)?] e. Using the test procedure of (d), what would you conclude if 6 of the 25 queried favored company 1 ?

To determine whether the pipe welds in a nuclear power plant meet specifications, a random sample of welds is selected, and tests are conducted on each weld in the sample. Weld strength is measured as the force required to break the weld. Suppose the specifications state that mean strength of welds should exceed \(100 \mathrm{Ib} / \mathrm{in}^{2}\), the inspection team decides to test \(H_{0}: \mu=100\) versus \(H_{a}: \mu>100\). Explain why it might be preferable to use this \(H_{\mathrm{a}}\) rather than \(\mu<100\).

Pairs of \(P\)-values and significance levels, \(\alpha\), are given. For each pair, state whether the observed \(P\)-value would lead to rejection of \(H_{0}\) at the given significance level. a. \(P\)-value \(=.084, \alpha=.05\) b. \(P\)-value \(=.003, \alpha=.001\) c. \(P\)-value \(=.498, \alpha=.05\) d. \(P\)-value \(=.084, \alpha=.10\) e. \(P\)-value \(=.039, \alpha=.01\) f. \(P\)-value \(=.218, \alpha=.10\)

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. With the sample mean itself as the test statistic, what is the \(P\)-value when \(\bar{x}=9.85\), and what would you conclude at significance level .01? c. For a test with \(\alpha=.01\), what is the probability that recalibration is judged unnecessary when in fact \(\mu=\) 10.1? When \(\mu=9.8\) ?

The melting point of each of 16 samples of a certain brand of hydrogenated vegetable oil was determined, resulting in \(\bar{x}=94.32\). Assume that the distribution of the melting point is normal with \(\sigma=1.20\). a. Test \(H_{0}: \mu=95\) versus \(H_{a}: \boldsymbol{\mu} \neq 95\) using a twotailed level .01 test. b. If a level .01 test is used, what is \(\beta(94)\), the probability of a type II error when \(\mu=94\) ? c. What value of \(n\) is necessary to ensure that \(\beta(94)=.1\) when \(\alpha=.01\) ?
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