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State DMV records indicate that of all vehicles undergoing emissions testing during the previous year, \(70 \%\) passed on the first try. A random sample of 200 cars tested in a particular county during the current year yields 124 that passed on the initial test. Does this suggest that the true proportion for this county during the current year differs from the previous statewide proportion? Test the relevant hypotheses using \(\alpha=.05\).

Short Answer

Expert verified
Reject the null hypothesis; the proportion differs from the previous year.

Step by step solution

01

Define Hypotheses

We start by defining the null and alternative hypotheses. The null hypothesis, denoted as \(H_0\), assumes the proportion of cars passing the test is equal to the previous statewide proportion, \(p = 0.70\). The alternative hypothesis, \(H_a\), suggests a difference: \(p eq 0.70\).
02

Collect Sample Data

From the exercise, we have \(n = 200\) cars, and \(x = 124\) that passed. The sample proportion is calculated as \( \hat{p} = \frac{124}{200} = 0.62 \).
03

Calculate Test Statistic

Use the formula for the test statistic for proportions:\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1-p_0)}{n}}} \]Substitute \( \hat{p} = 0.62\), \( p_0 = 0.70 \), and \( n = 200 \):\[ z = \frac{0.62 - 0.70}{\sqrt{\frac{0.70 \times 0.30}{200}}} \approx -2.83 \]
04

Determine Critical Value

Since this is a two-tailed test with \( \alpha = 0.05 \), use a standard normal distribution to find the critical z-values. These critical values are approximately \(-1.96\) and \(1.96\).
05

Make a Decision

Compare the test statistic \(z = -2.83\) to the critical values. Since \(-2.83 \leq -1.96\), we reject the null hypothesis \(H_0\).
06

Conclusion

The conclusion is that there is sufficient evidence at the \( \alpha = 0.05 \) significance level to suggest that the true proportion of cars passing the emissions test in the current year differs from the previous statewide proportion.

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Key Concepts

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

Proportion Test
A proportion test is a statistical method used to analyze whether a sample proportion differs from a known or hypothesized population proportion. This kind of test is very helpful in understanding whether a characteristic or feature in a sample is characteristic of the entire population or if it is due to random chance. In this exercise, for example, the goal is to determine if the 62% proportion of cars passing emissions in a county is different from the known 70% statewide proportion.
The test involves comparing the sample proportion \(\hat{p}\) to the population proportion \(p_0\). If the difference is statistically significant, we can conclude that the sample provides evidence that the true population proportion differs from the hypothesized proportion. This test frequently uses a normal distribution to approximate the distribution of the sample proportions, especially when the sample size is large. This makes the proportion test a powerful tool in hypothesis testing, providing clear insights into whether observed differences are significant or not.
Null and Alternative Hypotheses
In hypothesis testing, establishing the null and alternative hypotheses is the first crucial step. The null hypothesis, denoted as \(H_0\), represents the status quo or a statement of no effect or no difference. It typically reflects the assumption that any kind of deviation from the expected norm is just due to random variation.
In the context of this exercise, the null hypothesis is that the proportion of cars passing emissions is the same as the previous statewide proportion, which is 70%. Thus, we state given \(H_0 : p = 0.70\).
The alternative hypothesis, denoted as \(H_a\), embodies the assertion that opposes the null hypothesis. It suggests that there is a significant difference that needs to be addressed. Here, \(H_a: p eq 0.70\) suggests that the proportion of cars passing emissions actually differs from the 70% proportion seen previously statewide. This two-tailed approach tests for deviations in both directions, indicating possible increases or decreases in the proportion.
Significance Level
The significance level, denoted by \(\alpha\), is a threshold set by the researcher to decide if the observed effect is statistically significant. It represents the probability of rejecting the null hypothesis when it is actually true - a type I error.
For this exercise, a significance level of \(\alpha = 0.05\) is used. This implies that there is a 5% risk of concluding there is a difference in proportions when actually, none exists. Choosing a significance level involves a trade-off between risk and confidence. A lower significance level means more confidence in results, but it might also increase the chance of missing a true effect (a type II error).
Commonly used \(\alpha\) values include 0.01, 0.05, and 0.10. The critical values derived from these \(\alpha\) levels help determine the rejection region for the null hypothesis. In a two-tailed test with \(\alpha = 0.05\), the critical z-values from the standard normal distribution are approximately -1.96 and 1.96. Observing a test statistic beyond these critical values leads to the rejection of the null hypothesis.
Test Statistic Calculation
Calculating the test statistic involves applying a specific formula to determine how far the sample proportion is from the hypothesized population proportion in terms of standard deviations. For a proportion test, the formula for the test statistic \(z\) is given by:\[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1-p_0)}{n}}} \]where
  • \(\hat{p}\) is the sample proportion
  • \(p_0\) is the hypothesized population proportion
  • \(n\) is the sample size
In this exercise:
  • The sample proportion \(\hat{p}\) is 0.62
  • The hypothesized proportion \(p_0\) is 0.70
  • The sample size \(n\) is 200
By substituting these values into the formula, the test statistic is calculated as approximately \(-2.83\). This value indicates how many standard deviations the observed proportion is from the hypothesized proportion. Since this test statistic falls into the rejection region (beyond the critical values of -1.96 and 1.96), we conclude that the sample provides sufficient evidence against the null hypothesis, suggesting a real difference in proportions.

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Most popular questions from this chapter

Chapter 7 presented a CI for the variance \(\sigma^{2}\) of a normal population distribution. The key result there was that the rv \(\chi^{2}=(n-1) S^{2} / \sigma^{2}\) has a chi-squared distribution with \(n-1\) df. Consider the null hypothesis \(H_{0}: \sigma^{2}=\sigma_{0}^{2}\) (equivalently, \(\sigma=\sigma_{0}\). Then when \(H_{0}\) is true, the test statistic \(\chi^{2}=(n-1) \quad S^{2} / \sigma_{0}^{2}\) has a chi-squared distribution with \(n-1\) df. If the relevant alternative is \(H_{\mathrm{a}}: \sigma^{2}>\sigma_{0}^{2}\), rejecting \(H_{0}\) if \((n-1) s^{2} / \sigma_{0}^{2} \geq \chi_{\alpha, n-1}^{2}\) gives a test with significance level \(\alpha\). To ensure reasonably uniform characteristics for a particular application, it is desired that the true standard deviation of the softening point of a certain type of petroleum pitch be at most \(.50^{\circ} \mathrm{C}\). The softening points of ten different specimens were determined, yielding a sample standard deviation of \(.58^{\circ} \mathrm{C}\). Does this strongly contradict the uniformity specification? Test the appropriate hypotheses using \(\alpha=.01\).

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. \(H_{\mathrm{a}}: \mu>\mu_{0}\), 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 "Caffeine Knowledge, Attitudes, and Consumption in Adult Women" ( \(J\). of Nutrition Educ., 1992: 179-184) reports the following summary data on daily caffeine consumption for a sample of adult women: \(n=47, \bar{x}=215 \mathrm{mg}\), \(s=235 \mathrm{mg}\), and range \(=5-1176 .\) a. Does it appear plausible that the population distribution of daily caffeine consumption is normal? Is it necessary to assume a normal population distribution to test hypotheses about the value of the population mean consumption? Explain your reasoning. b. Suppose it had previously been believed that mean consumption was at most \(200 \mathrm{mg}\). Does the given data contradict this prior belief? Test the appropriate hypotheses at significance level .10 and include a \(P\)-value in your analysis.

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. \(.148\)

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