91Ó°ÊÓ

Analyses of drinking water samples for 100 homes in each of two different sections of a city gave the following information on lead levels (in parts per million): $$ \begin{array}{lcc} \hline & \text { Section 1 } & \text { Section 2 } \\ \hline \text { Sample Size } & 100 & 100 \\ \text { Mean } & 34.1 & 36.0 \\ \text { Standard Deviation } & 5.9 & 6.0 \end{array} $$ a. Calculate the test statistic and its \(p\) -value to test for a difference in the two population means. Use the \(p\) -value to evaluate the significance of the results at the \(5 \%\) level. b. Use a \(95 \%\) confidence interval to estimate the difference in the mean lead levels for the two sections of the city. c. Suppose that the city environmental engineers will be concerned only if they detect a difference of more than 5 parts per million in the two sections of the city. Based on your confidence interval in part b, is the statistical significance in part a of practical significance to the city engineers? Explain.

Step by step solution

01

a. Calculate the test statistic and p-value

To compare the means of the two populations, we can use the two-sample t-test. The test statistic is given by: $$ t = \frac{(\bar{X}_1 - \bar{X}_2) - (\mu_1 - \mu_2)}{\sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}}}, $$ where \(\bar{X}_1\) and \(\bar{X}_2\) are the sample means, \(\mu_1\) and \(\mu_2\) are the population means, \(S_1^2\) and \(S_2^2\) are the sample variances, and \(n_1\) and \(n_2\) are the sample sizes. Since we're testing for a difference, we can assume that \((\mu_1 - \mu_2)=0\). From the given data, we have \(\bar{X}_1 = 34.1\), \(\bar{X}_2 = 36.0\), \(S_1 = 5.9\), \(S_2 = 6.0\), \(n_1 = 100\), and \(n_2 = 100\). Plugging these values into the formula, we get: $$ t = \frac{(34.1 - 36.0)}{\sqrt{\frac{5.9^2}{100} + \frac{6.0^2}{100}}} = -1.51. $$ Next, we need to calculate the p-value. Using a t-distribution calculator with 198 degrees of freedom (the sum of the sample sizes minus 2), we find that the p-value is approximately 0.133.

02

a. Evaluate the significance at the 5% level

The p-value of 0.133 is greater than the significance level of 0.05. Therefore, we fail to reject the null hypothesis that there is no difference in the mean lead levels between the two sections. The results are not statistically significant at the 5% level.

03

b. Calculate a 95% confidence interval for the difference in the mean lead levels

To calculate a 95% confidence interval for the difference in the mean lead levels, we can use the following formula: $$ (\bar{X}_1 - \bar{X}_2) \pm t_{\alpha/2} \sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}}, $$ where \(t_{\alpha/2}\) is the t-value corresponding to the desired level of confidence with 198 degrees of freedom. For a 95% confidence interval, \(t_{\alpha/2} \approx 1.97\). Plugging in our values, we get: $$ (34.1 - 36.0) \pm 1.97\sqrt{\frac{5.9^2}{100} + \frac{6.0^2}{100}} = (-1.9 \pm 2.33) \text{ parts per million}. $$

04

c. Determine practical significance for the city engineers

The 95% confidence interval for the difference in mean lead levels is (-1.9 ± 2.33) parts per million. This interval does not exceed the minimum difference of 5 parts per million that would concern the city engineers. Therefore, although we found no statistical significance in part a, the result is also not practically significant for the city engineers.

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Key Concepts

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

Statistical Significance

The concept of statistical significance is pivotal in hypothesis testing. It allows researchers to determine if their findings can be attributed to a specific factor or if they occurred by chance. For example, in the drinking water analysis, the researchers used a two-sample t-test to investigate if there is a significant difference in lead levels between two sections of a city.

Statistical significance is generally determined by comparing the p-value to a predetermined significance level, commonly set at 0.05 (5%). If the p-value is less than the significance level, the results are said to be statistically significant, suggesting that the observed difference is unlikely to have occurred by chance. In our case, with a p-value of 0.133, we conclude that the difference in lead levels is not statistically significant, meaning that any observed difference could quite likely be due to random variation.

P-Value

The p-value quantifies the probability of observing the given data, or something more extreme, when the null hypothesis is true. It is a tool that helps us quantify the strength of our evidence against the null hypothesis.

In the exercise with water samples, the p-value calculated for the two-sample t-test was approximately 0.133. This means there is a 13.3% chance of finding a difference in lead levels as large or larger than the observed one if there was actually no difference between the two population means. Since this p-value is greater than the typical alpha level of 0.05, we retain the null hypothesis, thus indicating that there is no statistically significant difference in the mean lead levels between the two sections of the city.

Confidence Intervals

A confidence interval provides a range of values that likely contains the population parameter of interest. For instance, in the provided exercise, the 95% confidence interval for the difference in mean lead levels between the two sections was calculated to be between -1.9 and +0.43 parts per million. This interval can be interpreted to mean that we are 95% confident the true mean difference in lead levels falls within this range.

It is important to note that since the 95% confidence interval includes zero, it supports the notion that there is no significant difference in lead levels between the sections. Moreover, this interval does not include the 5 parts per million difference which the city engineers had established as a threshold for concern, suggesting the differences observed might not be of practical significance.

Hypothesis Testing

In performing a hypothesis testing procedure, researchers set up two opposing statements — the null hypothesis and the alternative hypothesis. The null hypothesis (\( H_0 \( H_0 \( H_0 \( H_0 \( H_0 \)) represents a theory that there is no effect or no difference, while the alternative hypothesis (\( H_A \( H_A \( H_A \( H_A \( H_A \)) suggests the opposite.

Using the data from the analysis of lead levels, the null hypothesis might assert that there is no difference in average lead levels between the two sections of the city (\( \bar{X}_1 = \bar{X}_2 \)). The t-test helps to determine whether the null hypothesis can be rejected in favor of the alternative hypothesis (a difference exists). Since the calculated p-value did not reach the level of significance required to reject the null hypothesis, the conclusion of the test was to not reject \( H_0 \( H_0 \( H_0 \( H_0 \( H_0 \). This decision-making process is an example of hypothesis testing in action.