91Ó°ÊÓ

Please provide the following information for Problems. (a) What is the level of significance? State the null and alternate hypotheses. (b) Check Requirements What sampling distribution will you use? What assumptions are you making? Compute the sample test statistic and corresponding \(z\) or \(t\) value as appropriate. (c) Find (or estimate) the \(P\) -value. Sketch the sampling distribution and show the area corresponding to the \(P\) -value. (d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \(\alpha ?\) (e) Interpret your conclusion in the context of the application. Note: For degrees of freedom \(d . f .\) not in the Student's \(t\) table, use the closest \(d . f .\) that is smaller. In some situations, this choice of \(d . f .\) may increase the \(P\) -value a small amount and therefore produce a slightly more "conservative" answer. Environment: Pollution Index Based on information from The Denver Post, 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.

Step by step solution

01

Define the Level of Significance and Hypotheses

The level of significance is set at \( \alpha = 0.01 \). The null hypothesis \( H_0 \) states that the mean pollution index of Englewood is equal to that of Denver: \( H_0: \mu_1 = \mu_2 \). The alternate hypothesis \( H_a \) states that the mean pollution indices are different: \( H_a: \mu_1 eq \mu_2 \).

02

Check Requirements and Determine Appropriate Distribution

Since both samples are large enough, and previous studies provide the population standard deviations, the normal distribution (or Z-distribution) is appropriate for this hypothesis test. Key assumptions include a normal distribution of the pollution index in both areas.

03

Compute the Test Statistic

The test statistic for comparing means from two independent samples is calculated using the formula: \[ z = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \]Substituting the given values: \[ z = \frac{(43 - 36)}{\sqrt{\frac{21^2}{12} + \frac{15^2}{14}}} \approx 0.87 \]

04

Find the P-value and Sketch the Distribution

The P-value can be calculated using a standard normal distribution table. With \( z = 0.87 \), the P-value for a two-tailed test is about 0.384. The sketch would depict a standard normal curve with areas in both tails corresponding to this two-tailed P-value.

05

Decision Rule and Conclusion

Since the P-value (0.384) is greater than \( \alpha = 0.01 \), we fail to reject the null hypothesis. The data are not statistically significant at the 1% level, indicating insufficient evidence to conclude that the means are different.

06

Interpretation of Results

In the context of the application, there is not enough evidence to suggest that the mean pollution index during winter in Englewood differs from that in Denver. Therefore, we cannot claim a significant difference between the two locations based on this data.

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Key Concepts

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

Level of Significance

The level of significance, often represented by \( \alpha \), is a threshold set by the researcher to determine the probability of rejecting the null hypothesis when it is actually true. In simpler terms, it's the risk you are willing to take of making a Type I error. A common level of significance used is 5%, but for stricter testing, such as our example, a 1% level of significance is employed. This means there is only a 1% risk of incorrectly declaring a significant difference when there is none. Setting a low \( \alpha \) level is particularly important in fields where incorrect conclusions can have serious implications, ensuring that findings are robust and truly represent a significant difference.

Null and Alternate Hypotheses

Hypothesis testing involves setting up two conflicting statements—the null hypothesis \( (H_0) \) and the alternate hypothesis \( (H_a) \). The null hypothesis is a statement of no effect or no difference, and it is what we assume to be true until evidence suggests otherwise. In our exercise, the null hypothesis asserts that there is no difference in the mean pollution indices between Englewood and Denver: \( H_0: \mu_1 = \mu_2 \).

The alternate hypothesis, on the other hand, is what you would believe if the null hypothesis is false. For this study, it suggests that the pollution indices for the two regions are not equal: \( H_a: \mu_1 eq \mu_2 \). By conducting the test, we seek evidence to reject the null hypothesis in favor of the alternate.

Sampling Distribution

The sampling distribution refers to the probability distribution of a given statistic based on a random sample. For hypothesis testing, this distribution helps us determine how the sample statistic relates to the population parameter. In our discussed problem, we compare means from two independent samples.

Given the available sample sizes and prior knowledge of population standard deviations, the normal distribution (or Z-distribution) is our choice for this test. This is feasible because:

• The normal distribution assumption holds thanks to prior studies indicating normality in pollution indices in both regions.
• Population standard deviations are known, allowing for accurate calculations.

The chosen distribution aids in determining how extreme a sample result is under the null hypothesis.

P-value

A P-value is a measure that helps us determine the strength of the evidence against the null hypothesis. It tells us the probability of observing a test statistic as extreme as or more extreme than the observed value, assuming that the null hypothesis is true. In our situation, with \( z = 0.87 \), the calculated P-value is approximately 0.384.

Since we are performing a two-tailed test at a 1% significance level, we compare this P-value against our significance level. Because 0.384 is much larger than the 0.01 level, we do not reject the null hypothesis. This high P-value suggests there is considerable overlap between the sampling distributions of both sample means, indicating insufficient evidence to declare a difference between Englewood and Denver's pollution indices.