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To test the ability of auto mechanics to identify simple engine problems, an automobile with a single such problem was taken in turn to 72 different car repair facilities. Only 42 of the 72 mechanics who worked on the car correctly identified the problem. Does this strongly indicate that the true proportion of mechanics who could identify this problem is less than \(.75\) ? Compute the \(P\)-value and reach a conclusion accordingly.

Short Answer

Expert verified
Reject the null hypothesis; the true proportion is likely less than 0.75.

Step by step solution

01

Define the Hypotheses

For a proportion test, we set up the hypotheses as follows:- Null Hypothesis (\(H_0\)): The proportion of mechanics who can identify the problem, \(p\), is equal to \(0.75\). That is, \(p = 0.75\).- Alternative Hypothesis (\(H_1\)): The proportion, \(p\), is less than \(0.75\). That is, \(p < 0.75\). This is a left-tailed test.
02

Collect the Sample Data

Here, we have the sample size \(n = 72\) and the number of successes \(x = 42\). The sample proportion \(\hat{p}\) can be calculated as \(\hat{p} = \frac{x}{n} = \frac{42}{72} = 0.5833\).
03

Compute the Test Statistic

To find the test statistic, we use the formula for the Z-score in a proportion test:\[Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\]Where \(p_0 = 0.75\) is the hypothesized proportion. Substituting the values:\[Z = \frac{0.5833 - 0.75}{\sqrt{\frac{0.75 \times 0.25}{72}}} = \frac{-0.1667}{0.0515} \approx -3.24\]
04

Determine the P-Value

The \(P\)-value can be found using the standard normal distribution table. Look up \(Z = -3.24\) in the table to find the \(P\)-value.Since this is a left-tailed test, the \(P\)-value is about \(0.0006\).
05

Conclusion

Since the \(P\)-value \(0.0006\) is less than a typical significance level (e.g., \(0.05\)), we reject the null hypothesis. There is strong evidence to suggest that the true proportion of mechanics who can identify the problem is less than \(0.75\).

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Key Concepts

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

Understanding a Proportion Test
A proportion test is a statistical method used to determine if there is a significant difference between an observed proportion and a theoretical one. It is particularly useful in situations where you want to compare a sample proportion to a known population proportion.
In our exercise, we are testing if the proportion of mechanics who can correctly identify an engine problem is less than 0.75. This involves determining whether the observed sample proportion (42 out of 72 mechanics) significantly deviates from the hypothesized population proportion, which is 0.75 in this case.
The proportion test helps ensure that observed differences aren't due to random chance but rather reflect a real deviation from the expectation. It is one of several types of hypothesis tests used in statistics.
Calculating the Z-score
The Z-score is a crucial component in hypothesis testing, measuring how far an observed data point is from the expected value under the null hypothesis, in terms of standard deviations.
To calculate the Z-score for a proportion test, you use the formula:
  • \[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \]
Where:
  • \( \hat{p} \) is the observed sample proportion,
  • \( p_0 \) is the hypothesized proportion, and
  • \( n \) is the sample size.
In this exercise, the Z-score helps us quantify how unusual the observed sample proportion (0.5833) is compared to the hypothesized value (0.75). The resulting Z-score of -3.24 suggests that the observed proportion is unusually low if the true proportion were indeed 0.75.
Interpreting the P-value
The P-value in hypothesis testing helps us determine the significance of our results. It is the probability of obtaining a result as extreme as, or more extreme than, the observed data, assuming the null hypothesis is true.
A low P-value suggests that the observed data is unlikely under the null hypothesis, thus providing evidence against it.
  • In our scenario, the computed P-value is approximately 0.0006.
Given that this P-value is considerably lower than a typical significance level, like 0.05, it tells us that the observed result (that only 42 out of 72 mechanics could identify the problem) is highly unlikely if the true proportion were 0.75. Hence, we have strong evidence to reject the null hypothesis.
Role of the Alternative Hypothesis
The alternative hypothesis represents what we seek to prove or disprove with our test. It is the statement employed when the null hypothesis is rejected.
In our exercise, the alternative hypothesis is that the true proportion of mechanics who can identify the problem is less than 0.75. This is articulated as \( H_1: p < 0.75 \).
The direction of the alternative hypothesis ("less than") indicates a left-tailed test. This corresponds to checking if the observed sample proportion is significantly less than the hypothesized proportion.
  • The decision to accept or reject depends on the P-value relative to a chosen level of significance (usually 0.05).
Ultimately, the goal of setting an alternative hypothesis is to challenge the status quo (null hypothesis) and test if there is enough statistical evidence to support a new claim about the population.

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Most popular questions from this chapter

On the label, Pepperidge Farm bagels are said to weigh four ounces each ( \(113 \mathrm{~g}\) ). A random sample of six bagels resulted in the following weights (in grams): \(\begin{array}{llllll}117.6 & 109.5 & 111.6 & 109.2 & 119.1 & 110.8\end{array}\) a. Based on this sample, is there any reason to doubt that the population mean is at least \(113 \mathrm{~g}\) ? b. Assume that the population mean is actually \(110 \mathrm{~g}\) and that the distribution is normal with standard deviation \(4 \mathrm{~g}\). In a \(z\) test of \(H_{0}\) : \(\mu=113\) against \(H_{\mathrm{a}}: \mu<113\) with \(\alpha=.05\), find the probability of rejecting \(H_{0}\) with \(\operatorname{six}\) observations. c. Under the conditions of part (b) with \(\alpha=.05\), how many more observations would be needed in order for the power to be at least \(.95 ?\)

For a fixed alternative value \(\mu^{\prime}\), show that \(\beta\left(\mu^{\prime}\right) \rightarrow 0\) as \(n \rightarrow \infty\) for either a one-tailed or a two-tailed \(z\) test in the case of a normal population distribution with known \(\sigma\).

Because of variability in the manufacturing process, the actual yielding point of a sample of mild steel subjected to increasing stress will usually differ from the theoretical yielding point. Let \(p\) denote the true proportion of samples that yield before their theoretical yielding point. If on the basis of a sample it can be concluded that more than \(20 \%\) of all specimens yield before the theoretical point, the production process will have to be modified. a. If 15 of 60 specimens yield before the theoretical point, what is the \(P\)-value when the appropriate test is used, and what would you advise the company to do? b. If the true percentage of "early yields" is actually \(50 \%\) (so that the theoretical point is the median of the yield distribution) and a level \(.01\) test is used, what is the probability that the company concludes a modification of the process is necessary?

A test of whether a coin is fair will be based on \(n=50\) tosses. Let \(X\) be the resulting number of heads. Consider two rejection regions: \(R_{1}=\\{x\) : either \(x \leq 17\) or \(x \geq 33\\}\) and \(R_{2}=\\{x\) : either \(x \leq 18\) or \(x \geq 37\\}\) a. Determine the significance level (type I error probability) for each rejection region. b. Determine the power of each test when \(p=.49\). Is the test with rejection region \(R_{1}\) a uniformly most powerful level .033 test? Explain. c. Is the test with rejection region \(R_{2}\) unbiased? Explain. d. Sketch the power function for the test with rejection region \(R_{1}\), and then do so for the test with the rejection region \(R_{2}\). What does your intuition suggest about the desirability of using the rejection region \(R_{2}\) ?

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.

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