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Lead Levels in Drinking Water Analyses of drinking water samples for 100 homes in each of two different sections of a city gave the following means and standard deviations of lead levels (in parts per million): a. Calculate the test statistic and its \(p\) -value (observed significance level) to test for a difference in the two population means. Use the \(p\) -value to evaluate the statistical 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 \(\mathrm{b},\) is the statistical significance in part a of practical significance to the city engineers? Explain.

Short Answer

Expert verified
Answer: The practical significance of the difference in mean lead levels between the two city sections is not enough to be of concern for the city engineers, as the possible difference is between -3.554 and -0.246 parts per million, which is less than their threshold of concern (5 parts per million).

Step by step solution

01

Define the hypotheses

We want to test for a difference in population means. The null hypothesis \((H_0)\) is that there is no difference in the population means, and the alternative hypothesis \((H_a)\) is that there is a difference: $$H_0: \mu_1 - \mu_2 = 0$$ $$H_a: \mu_1 - \mu_2 \neq 0$$
02

Calculate the test statistic

The test statistic for independent sample t-test is given by: $$t = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}}$$ Where \(s_p\) is the pooled standard deviation, which is calculated as: $$s_{p}=\sqrt{\frac{(n_{1}-1) s_{1}^{2}+(n_{2}-1) s_{2}^{2}}{n_{1}+n_{2}-2}}$$ For our data, we have $$n_1 = n_2 = 100$$ $$\bar{x}_1 = 34.1$$ $$\bar{x}_2 = 36.0$$ $$s_1 = 5.9$$ $$s_2 = 6.0$$ Now, calculate \(s_p\) and the test statistic: $$s_p=\sqrt{\frac{(100-1)5.9^2 +(100-1)6.0^2}{100+100-2}} = 5.95025$$ $$t = \frac{(34.1 - 36.0) - 0}{5.95025 \sqrt{\frac{1}{100} + \frac{1}{100}}} = -2.54841$$
03

Calculate the p-value

Now, we need to find the p-value. Since this is a two-tailed test, we look for the probability associated with the test statistic in the t-distribution with df = \(n_1+n_2-2 = 198\). The p-value is: $$p = P(T_{198} \le -2.54841) * 2 = 0.01169$$
04

Evaluate the statistical significance

The p-value (0.01169) is less than the significance level (0.05), so we reject the null hypothesis and conclude that there is a significant difference between the population means of the lead level in the two city sections. #b. Use a 95% confidence interval to estimate the difference in the mean lead levels#
05

Calculate the confidence interval

A 95% confidence interval is given by: $$\bar{x}_1 - \bar{x}_2 \pm t^* s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}$$ Where \(t^*\) is the critical value in the t-distribution with df = 198 and a significance level of 0.05. In this case, \(t^*= 1.972\). Now, we can calculate the confidence interval: $$34.1 - 36.0 \pm 1.972 * 5.95025 \sqrt{\frac{1}{100} + \frac{1}{100}}$$ $$= -1.9 \pm 1.972 * 0.83883$$ $$= (-3.554, -0.246)$$
06

Interpret the confidence interval

The 95% confidence interval for the difference in mean lead levels between the two city sections is (-3.554, -0.246). This indicates that we are 95% confident that the true difference between the two population means lies between -3.554 and -0.246 parts per million. #c. Evaluate the practical significance of the results#
07

Compare confidence interval to the threshold of practical significance

The city environmental engineers are concerned if the difference in lead levels is more than 5 parts per million. However, our confidence interval states that the possible difference is between -3.554 and -0.246 parts per million. This means that it is highly unlikely that the difference is more than 5 parts per million. Thus, the results are not of practical significance for the city engineers.

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Key Concepts

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

Null Hypothesis
In statistical testing, the null hypothesis (denoted as \(H_0\)) is a statement that suggests there is no effect or no difference between groups in a study. It serves as a starting point for statistical analysis. For example, in the case of analyzing lead levels in drinking water, the null hypothesis would state that the average lead levels in the two sections of the city are the same:
  • \(H_0: \mu_1 - \mu_2 = 0\)
Here, \(\mu_1\) and \(\mu_2\) represent the population means of lead levels in the two different city sections. The purpose of testing is to see if there is enough statistical evidence to reject this hypothesis.
Rejecting the null hypothesis implies that there is a statistically significant difference between the groups being compared. Conversely, failing to reject the null hypothesis suggests there is not enough evidence to demonstrate a difference.
In our case, if the calculated values show a significant difference, we would reject the null hypothesis.
Confidence Interval
A confidence interval gives a range of values that is likely to contain a population parameter with a certain level of confidence. Typically, a 95% confidence interval is used, meaning we are 95% confident that our interval contains the true population parameter.
For the lead levels example, the confidence interval aims to estimate the difference in means between the two city sections. Calculating it involves:
  • Identifying the sample means and pooled standard deviation.
  • Using the appropriate critical value for the t-distribution based on the desired confidence level (95%).
  • Calculating the interval by subtracting and adding the margin of error to the sample means difference.
The calculation showed a confidence interval of \(-3.554\) to \(-0.246\) parts per million. This means there's a 95% chance that the real difference in means falls within this range.
This interval helps understand not only whether a difference exists, but also its possible value.
p-value
The \(p\)-value is a measure that helps determine the strength of the evidence against the null hypothesis. It represents the probability of observing the test results, or something more extreme, assuming the null hypothesis is true.
In simple terms, a low \(p\)-value (typically less than 0.05) indicates strong evidence against the null hypothesis, leading to its rejection. In the lead levels analysis, the calculated \(p\)-value was \(0.01169\). This value is:
  • Lower than the common threshold of \(0.05\), providing sufficient evidence to reject the null hypothesis.
Thus, the low \(p\)-value suggests a statistically significant difference between the lead levels in different city sections.
The use of the \(p\)-value is crucial, as it quantifies the level of surprise or extremeness of the observed data under the assumption that the null hypothesis is true.
Statistical Significance
Statistical significance refers to the likelihood that a relationship or difference observed in a study sample is not due to chance. It is primarily assessed using a \(p\)-value:
  • If the \(p\)-value is below a predetermined threshold level (usually \(0.05\)), the result is considered statistically significant.
In the lead levels case, the findings were statistically significant because the \(p\)-value was \(0.01169\), smaller than \(0.05\). Rejecting the null hypothesis implies that there is enough evidence to say the lead levels differ between the city sections.
However, statistically significant results may not always be practically significant. Practical significance considers whether the difference is large enough to be of concern or meaningful in real-world terms. Despite the statistical significance in this case, the difference was not practically significant, as it did not meet the concern threshold for the engineers (over 5 parts per million). Understanding this distinction is vital for making informed decisions based on study results.

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