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The amount of shaft wear (.0001 in.) after a fixed mileage was determined for each of \(n=8\) internal combustion engines having copper lead as a bearing material, resulting in \(\bar{x}=3.72\) and \(s=1.25 .\) a. Assuming that the distribution of shaft wear is normal with mean \(\mu\), use the \(t\) test at level .05 to test \(H_{0}: \mu=\) \(3.50\) versus \(H_{\mathrm{a}}: \mu>3.50\). b. Using \(\sigma=1.25\), what is the type II error probability \(\beta\left(\mu^{\prime}\right)\) of the test for the alternative \(\mu^{\prime}=4.00\) ?

Short Answer

Expert verified
The null hypothesis is not rejected (t = 0.498). Type II error probability \( \beta \) for \( \mu' = 4.00 \) is about 0.15.

Step by step solution

01

Understanding the Problem

We are given sample statistics for shaft wear in engines with a mean of \( \bar{x} = 3.72 \) and standard deviation \( s = 1.25 \). We need to perform a hypothesis test on whether the true mean \( \mu \) of this wear is significantly greater than 3.50. Additionally, knowing the standard deviation, we evaluate the type II error probability for the alternative hypothesis where \( \mu' = 4.00 \).
02

Setting up the Hypothesis

The null hypothesis for this problem is \( H_0: \mu = 3.50 \). The alternative hypothesis, which we are testing against, is \( H_a: \mu > 3.50 \). We will use a t-test to evaluate this at a significance level of \( \alpha = 0.05 \).
03

Calculate the Test Statistic

The t-test statistic is calculated using the formula:\[t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}\]Substitute the given values:\[t = \frac{3.72 - 3.50}{1.25/\sqrt{8}} = \frac{0.22}{0.4419} \approx 0.498\]
04

Determine the Critical Value

Since we are using a one-tailed test with \( \alpha = 0.05 \) and degrees of freedom \( n-1 = 7 \), we find the critical value from a t-distribution table is approximately 1.895.
05

Compare Test Statistic with Critical Value

Compare the calculated t statistic (0.498) with the critical value (1.895). Since 0.498 is less than 1.895, we do not reject the null hypothesis \( H_0 \). This means there is not sufficient evidence at the 0.05 significance level to support that \( \mu > 3.50 \).
06

Calculating Type II Error Probability for Given Alternative

The type II error probability \( \beta \) is calculated for \( \mu' = 4.00 \) using:\[\beta = P\left( T_{\text{calculated}} < 1.895 \mid \mu = 4.00 \right)\]The noncentrality parameter \( \delta \) can be computed as:\[\delta = \frac{\mu' - 3.50}{s/\sqrt{n}} = \frac{4.00 - 3.50}{1.25/\sqrt{8}} \approx 1.13\]Using a noncentral t-distribution, the probability that the calculated t-score is less than the critical value can be computed using statistical software to find \( \beta \approx 0.15 \).
07

Conclusion

For the test with \( \mu' = 4.00 \), there is a type II error probability of approximately 0.15.

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Key Concepts

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

Type I and Type II Errors
Statistical hypothesis testing often deals with two main types of errors: Type I and Type II. Understanding these errors is crucial for making informed decisions about statistical conclusions. A Type I error occurs when we reject a true null hypothesis. In simple terms, it's when our test suggests there is an effect or a difference when there isn't one. The probability of making a Type I error is denoted by the significance level \(\alpha\), usually set at 0.05. This means we are willing to accept a 5% chance of rejecting a true null hypothesis.

On the other hand, a Type II error happens when we fail to reject a false null hypothesis. This is like missing a signal that is actually there. The probability of making a Type II error is denoted by \(\beta\). Unlike the significance level, which we choose beforehand, \(\beta\) depends on several factors, including the sample size, the true population parameter, and the significance level used. It's important to consider both types of errors when interpreting the results of any statistical test, as each has its own implications on the conclusions drawn.
T-test
A t-test is a statistical test used to compare the sample mean to a known value or another sample mean, especially when the sample size is small, and the population standard deviation is unknown. It is quite handy when dealing with small sample sizes, typically fewer than 30, and when the data is approximately normally distributed.

In the context of our exercise, the one-sample t-test helps determine if the average shaft wear in engines with copper lead as a bearing material is greater than 3.50. To conduct this test, we calculate the t-test statistic using the formula \[t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}\]. Here, \(\bar{x}\) represents the sample mean, \(\mu_0\) is the hypothesized population mean, \(s\) is the standard deviation, and \(n\) is the sample size.

The calculated t-value is compared against a critical value from the t-distribution table, which depends on the chosen significance level and degrees of freedom. Decisions about rejecting or not rejecting the null hypothesis hinge on this comparison.
Normal Distribution
The normal distribution, also known as the bell curve, is a probability distribution that is symmetric around the mean. It is a foundational concept in statistics because of its appealing properties. Many statistical tests, including the t-test, assume data is normally distributed because of how common this distribution appears in nature.

Key characteristics of a normal distribution include:
  • Mean, median, and mode are all equal, situated at the center of the distribution.
  • The curve is symmetric around these values, with tails that extend outward infinitely.
  • Approximately 68% of data falls within one standard deviation of the mean, around 95% within two standard deviations, and about 99.7% within three.
In the provided exercise, we assume that the distribution of shaft wear follows a normal distribution, allowing us to apply the t-test. This assumption simplifies the process, as it permits the use of test statistics and critical values derived under normality.
Significance Level
The significance level, often denoted as \(\alpha\), is a critical concept in hypothesis testing. It's essentially the threshold we set for deciding when the results of an experiment are statistically significant. In simpler terms, it shows how much risk of making a Type I error we are willing to accept.

Commonly, a significance level of 0.05 is chosen, implying a 5% risk of rejecting the null hypothesis when it is actually true. This balance allows researchers to minimize errors and make confident conclusions.

For our hypothesis test about shaft wear, a 0.05 level means that we will consider the effect or difference significant if our p-value is less than 0.05. If the p-value is below this threshold, we conclude that the results are statistically significant, suggesting that the average wear is indeed greater than 3.50. However, if the t-test calculated p-value exceeds 0.05, as in our exercise, we do not have strong enough evidence to support the alternative hypothesis.

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Most popular questions from this chapter

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 melting point is normal with \(\sigma=1.20\). a. Test \(H_{0}: \mu=95\) versus \(H_{\mathrm{a}}: \mu \neq 95\) using a two- tailed 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 \(\alpha=.01\) ?

For which of the given \(P\)-values would the null hypothesis be rejected when performing a level \(.05\) test? a. \(.001\) b. \(.021\) c. \(.078\) d. \(.047\) e. \(.148\)

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

A new method for measuring phosphorus levels in soil is described in the article "A Rapid Method to Determine Total Phosphorus in Soils" (Soil Sci. Amer. J., 1988: 1301-1304). Suppose a sample of 11 soil specimens, each with a true phosphorus content of \(548 \mathrm{mg} / \mathrm{kg}\), is analyzed using the new method. The resulting sample mean and standard deviation for phosphorus level are 587 and 10 , respectively. a. Is there evidence that the mean phosphorus level reported by the new method differs significantly from the true value of \(548 \mathrm{mg} / \mathrm{kg}\) ? Use \(\alpha=.05\). b. What assumptions must you make for the test in part (a) to be appropriate?

Exercise 36 in Chapter 1 gave \(n=26\) observations on escape time (sec) for oil workers in a simulated exercise, from which the sample mean and sample standard deviation are \(370.69\) and \(24.36\), respectively. Suppose the investigators had believed a priori that true average escape time would be at most \(6 \mathrm{~min}\). Does the data contradict this prior belief? Assuming normality, test the appropriate hypotheses using a significance level of .05.

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