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Based on information from the Rocky Mountain News, a random sample of \(n_{1}=12\) winter days in Denver gave a sample mean pollution index of \(\bar{x}_{1}=43\). Previous studies show that \(\sigma_{1}=21\). For Englewood (a suburb of Denver), a random sample of \(n_{2}=14\) winter days gave a sample mean pollution index of \(\bar{x}_{2}=36\). Previous studies show that \(\sigma_{2}=15\). Assume the pollution index is normally distributed in both Englewood and Denver. Do these data indicate that the mean population pollution index of Englewood is different (either way) from that of Denver in the winter? Use a \(1 \%\) level of significance.

Short Answer

Expert verified
The data do not show a significant difference in pollution indices between Denver and Englewood at the 1% level.

Step by step solution

01

Set Up the Hypotheses

We want to test if there is a difference between the mean pollution indices of Denver and Englewood. Let \( \mu_1 \) be the mean pollution index for Denver and \( \mu_2 \) be for Englewood. The null hypothesis \( H_0: \mu_1 = \mu_2 \) states there is no difference, while the alternative hypothesis \( H_a: \mu_1 eq \mu_2 \) states there is a difference.
02

Determine the Significance Level

The problem specifies a significance level \( \alpha = 0.01 \), which means we will reject the null hypothesis if our test statistic falls in the most extreme 1% of its possible values under the null hypothesis.
03

Calculate the Test Statistic

Since both population standard deviations are known, we use the formula for the test statistic under the assumption of normality: \[ z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \] Substitute the given values: \[ z = \frac{43 - 36}{\sqrt{\frac{21^2}{12} + \frac{15^2}{14}}} \] This calculates to: \[ z = \frac{7}{\sqrt{36.75 + 16.07}} = \frac{7}{7.38} \approx 0.949 \]
04

Find the Critical Values

For a two-tailed test at the \(1\%\) significance level, the critical z-values are \(\pm 2.576\). These values define the rejection regions for the null hypothesis.
05

Make a Decision

The calculated \( z \)-value of 0.949 does not exceed the critical values \(-2.576\) or \(2.576\). Therefore, we do not reject the null hypothesis because the \( z \)-value does not fall in the rejection region.
06

State the Conclusion

There is not enough statistical evidence at the \(1\%\) significance level to conclude that the mean pollution indices for Denver and Englewood are different.

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Key Concepts

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

Significance Level
In hypothesis testing, the significance level is a critical component that defines how extreme the data must be to reject the null hypothesis. You can think of the significance level, denoted as \( \alpha \), as a threshold percentage. Typically, common values are 0.05, 0.01, or even 0.10, but it's set at 0.01 in our exercise. This means we're willing to accept a 1% chance of incorrectly rejecting the null hypothesis when it's actually true (this is known as a Type I error).
A lower significance level indicates a more stringent criterion for making a decision. In our exercise, using a 1% significance level means that we need strong evidence against the null hypothesis to decide that the pollution index differs between Denver and Englewood. Simply put, the more remarkable the findings must be to challenge the null hypothesis at such a low \( \alpha \).
  • \(\alpha = 0.01\) implies extremely strong evidence is needed to claim a difference.
  • Significance levels help frame how confident we need to be in order to reject \( H_0 \).
Test Statistic
The test statistic is a standardized value that is calculated from sample data during a hypothesis test. It is used to decide whether to support or reject the null hypothesis. Our specific situation involves comparing means with known standard deviations. Given this, we use a z-score as our test statistic.
In our exercise, the formula to compute the z-score is:\[z = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \]The numerator \( (\bar{x}_1 - \bar{x}_2) \) represents the observed difference in sample means, while the denominator adjusts this difference for the sample variability.
Once calculated, this z-score tells us how many standard deviations our observed difference is from the null hypothesis's assumption (here, that the means are equal). In our example, a z-score of approximately 0.949 suggests the observed difference isn't exceptionally unusual, given the null hypothesis. This forms a critical part of deciding whether to reject or retain \( H_0 \).
  • Used to compare observed data to what's expected under \( H_0 \).
  • A higher absolute z-score means the data is more unusual under \( H_0 \).
Null Hypothesis
The null hypothesis \((H_0)\) is a starting assumption for statistical testing. It proposes that there is no effect or difference in a particular context. In terms of our exercise, \(H_0\) states that the mean pollution indices for Denver and Englewood are equal, or mathematically, \(\mu_1 = \mu_2\).
This hypothesis serves as a default statement that the data has no compelling effect or pattern to observe. It remains our reference point throughout the testing process, unless our calculations provide sufficient evidence to conclude otherwise. It's crucial here because all of the test statistic's interpretations hinge on our understanding of this \( H_0 \).
  • \(H_0: \mu_1 = \mu_2\) suggests no difference in real terms.
  • This is the hypothesis we aim to test and potentially reject.
Alternative Hypothesis
The alternative hypothesis \((H_a)\) suggests that a pattern or difference does exist and contradicts the null hypothesis. In our example, \(H_a:\mu_1 eq \mu_2\) posits that the pollution levels in Denver and Englewood differ, without specifying which is higher or lower.
This alternative is crucial as it is the hypothesis we look to support with our data. It provides an insight or conclusion we're trying to prove. If statistical analysis (test statistic) supports the evidence that contradicts \( H_0 \), we turn to \( H_a \) as the likely reality.
In practical applications, considering \( H_a \) helps provide the framework for creating test statistics and calculating probabilities that ultimately confirm (or deny) significant differences.
  • \(H_a: \mu_1 eq \mu_2\) allows for any substantial difference.
  • It's what we wish to prove in opposition to \( H_0 \).

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Most popular questions from this chapter

Consider independent random samples from two populations that are normal or approximately normal, or the case in which both sample sizes are at least \(30 .\) Then, if \(\sigma_{1}\) and \(\sigma_{2}\) are unknown but we have reason to believe that \(\sigma_{1}=\sigma_{2}\), we can pool the standard deviations. Using sample sizes \(n_{1}\) and \(n_{2}\), the sample test statistic \(\bar{x}_{1}-\bar{x}_{2}\) has a Student's \(t\) distribution, where \(t=\frac{\bar{x}_{1}-\bar{x}_{2}}{s \sqrt{\frac{1}{n_{1}}+\frac{1}{n_{2}}}}\) with degrees of freedom d.f. \(=n_{1}+n_{2}-2\) and where the pooled standard deviation \(s\) is $$s=\sqrt{\frac{\left(n_{1}-1\right) s_{1}^{2}+\left(n_{2}-1\right) s_{2}^{2}}{n_{1}+n_{2}-2}}$$ Note: With statistical software, select the pooled variance or equal variance options. (a) There are many situations in which we want to compare means from populations having standard deviations that are equal. This method applies even if the standard deviations are known to be only approximately equal (see Section \(11.4\) for methods to test that \(\sigma_{1}=\sigma_{2}\) ). Consider Problem 17 regarding average incidence of fox rabies in two regions. For region I, \(n_{1}=16\), \(\bar{x}_{1}=4.75\), and \(s_{1} \approx 2.82\) and for region II, \(n_{2}=15, \bar{x}_{2} \approx 3.93\), and \(s_{2} \approx\) 2.43. The two sample standard deviations are sufficiently close that we can assume \(\sigma_{1}=\sigma_{2}\). Use the method of pooled standard deviation to redo Problem 17 , where we tested if there was a difference in population mean average incidence of rabies at the \(5 \%\) level of significance. (b) Compare the \(t\) value calculated in part (a) using the pooled standard deviation with the \(t\) value calculated in Problem 17 using the unpooled standard deviation. Compare the degrees of freedom for the sample test statistic. Compare the conclusions.

Is there a relationship between confidence intervals and two-tailed hypothesis tests? Let \(c\) be the level of confidence used to construct a confidence interval from sample data. Let \(\alpha\) be the level of significance for a two-tailed hypothesis test. The following statement applies to hypothesis tests of the mean. For a two-tailed hypothesis test with level of significance \(\alpha\) and null hypothesis \(H_{0}: \mu=k\), we reject \(H_{0}\) whenever \(k\) falls outside the \(c=1-\alpha\) confidence interval for \(\mu\) based on the sample data. When \(k\) falls within the \(c=1-\alpha\) confidence interval, we do not reject \(H_{0}\). (A corresponding relationship between confidence intervals and two-tailed hypothesis tests also is valid for other parameters, such as \(p, \mu_{1}-\mu_{2}\), or \(p_{1}-p_{2}\) which we will study in Sections \(9.3\) and \(9.5 .\) ) Whenever the value of \(k\) given in the null hypothesis falls outside the \(c=1-\alpha\) confidence interval for the parameter, we reject \(H_{0} .\) For example, consider a two-tailed hypothesis test with \(\alpha=0.01\) and \(H_{0}: \mu=20 \quad H_{1}: \mu \neq 20\) A random sample of size 36 has a sample mean \(\bar{x}=22\) from a population with standard deviation \(\sigma=4\). (a) What is the value of \(c=1-\alpha\) ? Using the methods of Chapter 8 , construct a \(1-\alpha\) confidence interval for \(\mu\) from the sample data. What is the value of \(\mu\) given in the null hypothesis (i.e., what is \(k)\) ? Is this value in the confidence interval? Do we reject or fail to reject \(H_{0}\) based on this information? (b) Using methods of Chapter 9 , find the \(P\) -value for the hypothesis test. Do we reject or fail to reject \(H_{0}\) ? Compare your result to that of part (a).

When using a Student's \(t\) distribution for a paired differences test with \(n\) data pairs, what value do you use for the degrees of freedom?

Weatherwise magazine is published in association with the American Meteorological Society. Volume 46 , Number 6 has a rating system to classify Nor'easter storms that frequently hit New England states and can cause much damage near the ocean coast. A severe storm has an average peak wave height of \(16.4\) feet for waves hitting the shore. Suppose that a Nor'easter is in progress at the severe storm class rating. (a) Let us say that we want to set up a statistical test to see if the wave action (i.e., height) is dying down or getting worse. What would be the null hypothesis regarding average wave height? (b) If you wanted to test the hypothesis that the storm is getting worse, what would you use for the alternate hypothesis? (c) If you wanted to test the hypothesis that the waves are dying down, what would you use for the alternate hypothesis? (d) Suppose you do not know if the storm is getting worse or dying out. You just want to test the hypothesis that the average wave height is different (either higher or lower) from the severe storm class rating. What would you use for the alternate hypothesis? (e) For each of the tests in parts (b), (c), and (d), would the area corresponding to the \(P\) -value be on the left, on the right, or on both sides of the mean? Explain your answer in each case.

A random sample of 46 adult coyotes in a region of northern Minnesota showed the average age to be \(\bar{x}=2.05\) years, with sample standard deviation \(s=0.82\) years (based on information from the book Coyotes: Biology, Bebavior and Management by M. Bekoff, Academic Press). However, it is thought that the overall population mean age of coyotes is \(\mu=1.75 .\) Do the sample data indicate that coyotes in this region of northern Minnesota tend to live longer than the average of \(1.75\) years? Use \(\alpha=0.01\).

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