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Chapter 8: Problem 76

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
The P-value is \(0.0005\); we reject \(H_0\) and conclude \(p < 0.75\).

Step by step solution

01

Define the Hypotheses

We begin by defining the null and alternative hypotheses for this statistical test. The null hypothesis \(H_0\) states that the true proportion \(p\) of mechanics who can identify the problem is \(.75\). The alternative hypothesis \(H_a\) is that \(p < 0.75\). Thus, \(H_0: p = 0.75\) and \(H_a: p < 0.75\).
02

Calculate Sample Proportion

The sample proportion \(\hat{p}\) is found by dividing the number of successful identifications by the total number of mechanics surveyed. Thus, \(\hat{p} = \frac{42}{72} = 0.5833\).
03

Compute Standard Error

Calculate the standard error of the sample proportion using the formula \(SE = \sqrt{\frac{p(1-p)}{n}}\), where \(p = 0.75\) and \(n = 72\). So, \(SE = \sqrt{\frac{0.75 \times 0.25}{72}} = 0.0506\).
04

Calculate Z-Score

Determine the Z-score using the formula \(Z = \frac{\hat{p} - p}{SE}\). Substituting the values, we get \(Z = \frac{0.5833 - 0.75}{0.0506} = -3.2937\).
05

Find P-value

The P-value is the probability that a standard normal variable is less than the calculated Z-score. Using standard normal distribution tables or a calculator, the P-value for \(Z = -3.2937\) is approximately \(0.0005\).
06

Conclusion

Since the P-value \(0.0005\) is less than the typical significance level of \(0.05\), we reject the null hypothesis \(H_0\). This indicates there is strong evidence 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 Statistical Significance
Statistical significance is a crucial concept in hypothesis testing. It helps us determine whether what we observe in a sample reflects a true effect in the population or is simply due to random chance.
In hypothesis testing, we usually have a threshold, often set at 0.05 or 5%, which we call the significance level. This means we are willing to accept a 5% chance of being wrong when we claim results are significant. When a test result shows statistical significance, it indicates that the observed data is unlikely to have occurred purely by chance and provides evidence against the null hypothesis.

In the exercise, we tested the ability of mechanics to diagnose a problem. Our conclusion was significant because the calculated P-value was lower than the significance level, suggesting our sample had substantially less ability to diagnose than assumed.
P-Value: The Indicator of Evidence
The P-value is a valuable indicator in hypothesis testing. It quantifies the probability that the observed data (or something more extreme) would occur if the null hypothesis were true.
If the P-value is low, this indicates stronger evidence against the null hypothesis. A common threshold used is 0.05.
  • If the P-value is less than 0.05, we reject the null hypothesis, suggesting the findings are statistically significant.
  • If it's more than 0.05, the data does not give enough evidence to reject the null hypothesis.
In the mechanics exercise, the P-value was 0.0005. This very low number indicated strong evidence against the null hypothesis that 75% of mechanics could identify the problem.
The Role of the Z-Score
A Z-score is a measure of how many standard deviations an element is from the mean. In hypothesis testing, it helps us understand the position of our sample statistic in relation to the null hypothesis.
When you calculate a Z-score, you're looking to see how far away your sample result is from what was expected, assuming the null hypothesis is correct.
  • A Z-score close to 0 means the sample result is close to the null hypothesis prediction.
  • A very high or low Z-score (beyond certain critical values) suggests that your sample result is statistically unusual under the null hypothesis.
For this exercise, we calculated a Z-score of -3.2937. This indicated the proportion of mechanics who identified the problem was much lower than the presumed proportion because the Z-score showed a significant difference.
Standard Error and Its Importance
Standard error (SE) is a statistical metric that quantifies the variability or dispersion of a sample statistic. It tells us how much the sample proportion is expected to deviate from the true population proportion.
In hypothesis testing, SE is crucial as it provides a scale for the sample's variability and is used to calculate the Z-score. The smaller the standard error, the more reliable the sample mean represents the population mean.
For our exercise, the standard error was computed as 0.0506, giving us information about the variability in the proportion of mechanics who correctly identified the problem. This value was used in the Z-score formula to assess whether the observed sample proportion significantly deviated from the assumed population proportion of 0.75, leading to the conclusion that it did.

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Most popular questions from this chapter

An aspirin manufacturer fills bottles by weight rather than by count. Since each bottle should contain 100 tablets, the average weight per tablet should be 5 grains. Each of 100 tablets taken from a very large lot is weighed, resulting in a sample average weight per tablet of \(4.87\) grains and a sample standard deviation of \(.35\) grain. Does this information provide strong evidence for concluding that the company is not filling its bottles as advertised? Test the appropriate hypotheses using \(\alpha=.01\) by first computing the \(P\)-value and then comparing it to the specified significance level.

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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 \(H_{\mathrm{a}}: p \neq .5\) based on a random sample of 25 individuals. Let \(X\) denote the number in the sample who favor the first company and \(x\) represent the observed value of \(X\). a. Which of the following rejection regions is most appropriate and why? $$ \begin{aligned} &R_{1}=\\{x: x \leq 7 \text { or } x \geq 18\\}, R_{2}=\\{x: x \leq 8\\}, \\ &R_{3}=\\{x: x \geq 17\\} \end{aligned} $$ b. In the context of this problem situation, describe what type I and type II errors are. c. What is the probability distribution of the test statistic \(X\) when \(H_{0}\) is true? Use it to compute the probability of a type I error. d. Compute the probability of a type II error for the selected region when \(p=.3\), again when \(p=.4\), and also for both \(p=.6\) and \(p=.7\). e. Using the selected region, what would you conclude if 6 of the 25 queried favored company 1 ?

The article "Orchard Floor Management Utilizing SoilApplied Coal Dust for Frost Protection" (Agri. and Forest Meteorology, 1988: 71-82) reports the following values for soil heat flux of eight plots covered with coal dust. \(\begin{array}{llllllll}34.7 & 35.4 & 34.7 & 37.7 & 32.5 & 28.0 & 18.4 & 24.9\end{array}\) The mean soil heat flux for plots covered only with grass is 29.0. Assuming that the heat-flux distribution is approximately normal, does the data suggest that the coal dust is effective in increasing the mean heat flux over that for grass? Test the appropriate hypotheses using \(\alpha=.05\).

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