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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
The data suggests the average penetration exceeds 50 mils.

Step by step solution

01

Define the Hypotheses

We need to test whether the true average penetration exceeds the specified maximum of 50 mils. Thus, our null hypothesis \( H_0 \) is that the true mean \( \mu = 50 \), and the alternative hypothesis \( H_a \) is that \( \mu > 50 \).
02

Determine the Test Statistic

The test statistic for a one-sample t-test is calculated as \( t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \), where \( \bar{x} = 52.7 \), \( \mu_0 = 50 \), \( s = 4.8 \), and \( n = 45 \). Substitute these values into the formula to find \( t \): \[ t = \frac{52.7 - 50}{4.8 / \sqrt{45}} \approx 3.96 \]
03

Find the Critical Value

Since we're conducting a one-tailed t-test at a common significance level \( \alpha = 0.05 \), we find the critical value \( t_{\alpha, n-1} \) from t-distribution tables, where \( n-1 = 44 \). The critical value for \( t_{0.05, 44} \approx 1.68 \).
04

Compare Test Statistic and Critical Value

The calculated t-statistic \( t = 3.96 \) is greater than the critical value \( 1.68 \). This means the test statistic falls in the rejection region.
05

Make the Decision

Since the test statistic exceeds the critical value, we reject the null hypothesis \( H_0 \). This indicates that there is sufficient evidence to conclude that the true average penetration is more than 50 mils.

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Key Concepts

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

One-sample t-test
Understanding the one-sample t-test is key to analyzing a sample鈥檚 mean when you have a small sample size, or when the population variance is unknown. This statistical method helps compare the sample mean to a known value or a population mean. In the case of our exercise, the one-sample t-test is utilized to verify if the average penetration exceeds the specified 50 mils.

Here's how it works: You compute a test statistic using your sample data, which tells you how much your sample mean deviates from the known value (e.g., the specified 50 mils mean). These deviations are scaled using the sample standard deviation and the sample size.

The formula used is:
  • Test Statistic, \( t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \)
This calculation also considers the number of specimens (sample size), making the result specific to your scenario. If your computed t-value is large, it suggests a significant difference between your sample mean and the known value.
Null Hypothesis
The foundation of hypothesis testing starts with the null hypothesis. This is a neutral statement indicating that no effect or no difference exists. In statistical tests, it's assumed true until evidence suggests otherwise. In our exercise, the null hypothesis \( H_0 \) asserts that the true mean penetration is equal to 50 mils.

The role of the null hypothesis is critical. It provides a baseline or reference point for statistical testing. By testing it, you can determine whether the data provides enough evidence to reject this assumption. Keeping this concept simple: the null hypothesis is basically saying, "Everything is as expected," until proven otherwise.
Alternative Hypothesis
Opposite to the null hypothesis is the alternative hypothesis. Here, you're testing whether there's an effect or a difference. In this exercise, the alternative hypothesis \( H_a \) posits that the true mean penetration exceeds 50 mils.

This hypothesis is what you're trying to support with your test data. If the analysis suggests rejecting the null hypothesis, the alternative hypothesis is typically accepted. It's important to specify the direction of the test in your hypotheses鈥攚hether you're testing for a mean greater than, less than, or different from the specified value. In our case, we are checking specifically if the penetration surpasses 50, establishing a "greater than" condition.
Critical Value
Critical values are essential in hypothesis testing as they define the threshold for decision-making. Once you calculate your test statistic, you need to compare it to the critical value to decide whether to reject the null hypothesis.

A critical value depends on the significance level (commonly \( \alpha = 0.05 \)) and the degrees of freedom, which is derived from your sample size (\( n - 1 \)). This value can be obtained from a t-distribution table.
  • If your calculated statistic (e.g., \( t = 3.96 \)) exceeds the critical value (e.g., around 1.68 for \( df = 44 \)), you reject the null hypothesis.
This boundary helps you determine how extreme a test statistic must be before you're confident that the null hypothesis doesn't hold. Essentially, the further your statistic is beyond the critical value, the stronger the evidence against the null hypothesis.

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Most popular questions from this chapter

Annual holdings turnover for a mutual fund is the percentage of a fund's assets that are sold during a particular year. Generally speaking, a fund with a low value of turnover is more stable and risk averse, whereas a high value of turnover indicates a substantial amount of buying and selling in an attempt to take advantage of short-term market fluctuations. Here are values of turnover for a sample of 20 large-cap blended funds (refer to Exercise \(1.53\) for a bit more information) extracted from Morningstar.com: \(\begin{array}{llllllllll}1.03 & 1.23 & 1.10 & 1.64 & 1.30 & 1.27 & 1.25 & 0.78 & 1.05 & 0.64 \\ 0.94 & 2.86 & 1.05 & 0.75 & 0.09 & 0.79 & 1.61 & 1.26 & 0.93 & 0.84\end{array}\) a. Would you use the one-sample \(t\) test to decide whether there is compelling evidence for concluding that the population mean turnover is less than \(100 \%\) ? Explain. b. A normal probability plot of the \(20 \ln\) (turnover) values shows a very pronounced linear pattern, suggesting it is reasonable to assume that the turnover distribution is lognormal. Recall that \(X\) has a lognormal distribution if \(\ln (X)\) is normally distributed with mean value \(\mu\) and variance \(\sigma^{2}\). Because \(\mu\) is also the median of the \(\ln (X)\) distribution, \(e^{\mu}\) is the median of the \(X\) distribution. Use this information to decide whether there is compelling evidence for concluding that the median of the turnover population distribution is less than \(100 \%\).

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

Let the test statistic \(T\) have a \(t\) distribution when \(H_{0}\) is true. Give the significance level for each of the following situations: a. \(H_{\mathrm{a}}: \mu>\mu_{0}, \mathrm{df}=15\), rejection region \(t \geq 3.733\) b. \(H_{\mathrm{a}}: \mu<\mu_{0}, n=24\), rejection region \(t \leq-2.500\) c. \(H_{\mathrm{a}}: \mu \neq \mu_{0}, n=31\), rejection region \(t \geq 1.697\) or \(t \leq-1.697\)

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 \(\\{z: z \leq-2.88\\}\) ?
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