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The amount of shaft wear after a fixed mileage was determined for each of seven randomly selected internal combustion engines, resulting in a mean of 0.0372 inch and a standard deviation of 0.0125 inch. a. Assuming that the distribution of shaft wear is normal, test at level .05 the hypotheses \(H_{0}: \mu=.035\) versus \(H_{\dot{a}}: \boldsymbol{\mu}>.035 .\) b. Using \(\sigma=0.0125, \alpha=.05,\) and Appendix Table 5, what is the approximate value of \(\beta,\) the probability of a Type II error, when \(\mu=.04\) ? c. What is the approximate power of the test when \(\mu=.04\) and \(\alpha=.05 ?\)

Short Answer

Expert verified
The ingoing interpretation of the z-score tells whether or not null hypothesis should be rejected. The Type II error (\(\beta\)) defines the probability of accepting the null hypothesis when it's false. Lastly, the power of the test gives the chance to reject null hypothesis when it's actually false.

Step by step solution

01

Formulating the hypotheses

The null hypothesis \(H_{0}\) is \(\mu=0.035\) and the alternative hypothesis \(H_{a}\) is \(\mu > 0.035\). Here, \(\mu\) refers to population mean.
02

Perform the z-test

Use the given statistics to calculate the z-score using the formula \(z=\frac{(\bar{x}-\mu_0)}{\frac{s}{\sqrt{n}}}\), where \(\bar{x}\) represents sample mean, \(\mu_0\) is the value from null hypothesis, \(s\) is the standard deviation, and \(n\) is the sample size. From the problem, we know that \(\bar{x}=0.0372\), \(\mu_0=0.035\), \(s=0.0125\), and \(n=7\). \n Plug in these values into the formula to get the z-score.
03

Deciding Upon Hypotheses

With the calculated z-score, refer to a Z-table (also known as standard normal distribution table). Since the z-test is a one-tailed test (as per \(H_a: \mu > 0.035\)), look for the resulting probability associated with the z-score and compare it with the significance level ('alpha' value) of 0.05. If the resulting probability is less than 0.05, reject the null hypothesis i.e, there is sufficient evidence that the mean shaft wear is more than 0.035 inches.
04

Calculating probability of Type II error (\(\beta\))

Type II error (\(\beta\)) is the error that occurs when we fail to reject null hypothesis when it is actually false. Given that \(\mu = 0.04\) (alternative truth), use this value in the formula \(\beta = P(Z < \frac{\mu_0 - \mu_a}{\sigma/\sqrt{n}})\), where \(\mu_a\) is the actual population mean. Looking up this value in the z-table, we get the probability of Type II error.
05

Calculating Power of the Test

The power of the test is 1-\(\beta\). Use the \(\beta\) value from the previous step to calculate power of the test.

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Key Concepts

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

Type II Error
A Type II error, represented by \(\beta\), occurs in hypothesis testing when a false null hypothesis is not rejected. It's the mistake of missing an actual effect or difference. For instance, in the context of the exercise, if the true mean shaft wear (\(\mu\)) was greater than 0.035 inch, but the test failed to reject the hypothesis that \(\mu = 0.035\), this would be a Type II error.

Calculating the probability of a Type II error involves understanding what happens when the alternative hypothesis is true. In the exercise, this was explored by considering a true mean of 0.04 inch and calculating the probability that the test statistic would still not be significant enough to reject the null hypothesis. This is critical because it reveals the chances of the test missing the detection of meaningful data.

Understanding and minimizing Type II errors are fundamental to ensuring that the conclusions drawn from a statistical test are reliable. To improve the assessment of Type II errors, it's essential to consider factors such as the effect size, sample size, and variability of the data. A larger sample or a more significant effect size can help reduce the risk of a Type II error.
z-test
The z-test is a statistical method used to determine whether there is a significant difference between sample and population means. It's appropriate when the sample size is large, or the population variance is known, and the data is approximately normally distributed. The z-test compares the sample mean to the population mean and assesses whether any observed differences are due to random chance.

In the given exercise, a z-test was performed to test the hypothesis about the mean shaft wear of engines. By calculating the z-score and comparing it to a critical value from the standard normal distribution (commonly found in z-tables), you can make a decision to either reject or fail to reject the null hypothesis.

To improve understanding and application of the z-test, it's helpful to focus on the prerequisites—for instance, confirming that the sample size is adequate and the data follows a normal distribution—as well as on correct calculation and interpretation of the z-score.
Normal Distribution
Normal distribution, sometimes known as 'Gaussian distribution', is a bell-shaped curve that is symmetric about the mean. In the field of statistics, it is a fundamental concept because it describes how the values of a variable are distributed. The mean, median, and mode of a perfectly normal distribution are all the same. Many statistical tests, including the z-test, assume that the underlying data follows a normal distribution.

In the textbook exercise, the assumption is made that the shaft wear follows a normal distribution, which is a key condition for proceeding with the z-test. This assumption allows the use of z-scores and standard deviation to estimate the probability of the observed sample mean if the null hypothesis were true.

To enhance comprehension, it is beneficial to visualize the normal distribution curve and identify the mean and standard deviation. It's also crucial to utilize plotting tools or goodness-of-fit tests to verify that the sample data does not deviate significantly from normality.
Statistical Power
Statistical power of a test is the probability that the test will correctly reject a false null hypothesis. It is essentially a measure of a test's ability to detect an effect, if there is one. Power is affected by several factors, including the significance level (\(\alpha\)), sample size (\(n\)), population standard deviation (\(\sigma\)), and the effect size (difference between the null and true population means).

In the scenario from the exercise, calculating the power involved finding the probability of not making a Type II error when the true mean shaft wear is 0.04 inch. The closer the power is to 1, the stronger the test is in detecting true effects.

To improve the power of a test, you can increase the sample size, reduce variability (if possible), or increase the effect size to be detected. Moreover, understanding the trade-off between Type I error (false positives) and Type II error (false negatives) is essential in designing an experiment or study with high statistical power.

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Most popular questions from this chapter

The power of a test is influenced by the sample size and the choice of significance level. a. Explain how increasing the sample size affects the power (when significance level is held fixed). b. Explain how increasing the significance level affects the power (when sample size is held fixed).

A television manufacturer claims that (at least) \(90 \%\) of its TV sets will need no service during the first 3 years of operation. A consumer agency wishes to check this claim, so it obtains a random sample of \(n=100\) purchasers and asks each whether the set purchased needed repair during the first 3 years after purchase. Let \(\hat{p}\) be the sample proportion of responses indicating no repair (so that no repair is identified with a success). Let \(p\) denote the actual proportion of successes for all sets made by this manufacturer. The agency does not want to claim false advertising unless sample evidence strongly suggests that \(p<.9 .\) The appropriate hypotheses are then \(H_{0}: p=.9\) versus \(H_{a}: p<.9\). a. In the context of this problem, describe Type I and Type II errors, and discuss the possible consequences of each. b. Would you recommend a test procedure that uses \(\alpha=.10\) or one that uses \(\alpha=.01 ?\) Explain.

The article "Theaters Losing Out to Living Rooms" (San Luis Obispo Tribune, June 17,2005\()\) states that movie attendance declined in \(2005 .\) The Associated Press found that 730 of 1000 randomly selected adult Americans preferred to watch movies at home rather than at a movie theater. Is there convincing evidence that the majority of adult Americans prefer to watch movies at home? Test the relevant hypotheses using a .05 significance level.

Do state laws that allow private citizens to carry concealed weapons result in a reduced crime rate? The author of a study carried out by the Brookings Institution is reported as saying, "The strongest thing I could say is that I don't see any strong evidence that they are reducing crime" (San Luis Obispo Tribune, January 23 . 2003 ). a. Is this conclusion consistent with testing \(H_{0}:\) concealed weapons laws reduce crime versus \(H_{a}:\) concealed weapons laws do not reduce crime or with testing \(H_{0}:\) concealed weapons laws do not reduce crime versus \(H_{a}\) : concealed weapons laws reduce crime Explain. b. Does the stated conclusion indicate that the null hypothesis was rejected or not rejected? Explain.

A certain university has decided to introduce the use of plus and minus with letter grades, as long as there is evidence that more than \(60 \%\) of the faculty favor the change. A random sample of faculty will be selected, and the resulting data will be used to test the relevant hypotheses. If \(p\) represents the proportion of all faculty that favor a change to plus-minus grading, which of the following pair of hypotheses should the administration test: $$ H_{0}: p=.6 \text { versus } H_{a}: p<.6 $$ or $$ H_{0}: p=.6 \text { versus } H_{a}: p>.6 $$ Explain your choice.

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