Question

5.12 Auto exhaust and lead exposure. 

Researchers interested in lead exposure due to car exhaust sampled the blood of 52 police officers subjected to constant inhalation of automobile exhaust fumes while working traffic enforcement in a primarily urban environment. The blood samples of these officers had an average lead concentration of 124.32μg/l and a SD of 37.74 μg/l; a previous study of individuals from a nearby suburb, with no history of exposure, found an average blood level concentration of 35 μg/l.

(a) Write down the hypotheses that would be appropriate for testing if the police officers appear to have been exposed to a higher concentration of lead.

(b) Explicitly state and check all conditions necessary for inference on these data.

(c) Test the hypothesis that the downtown police officers have a higher lead exposure than the group in the previous study. Interpret your results in context.

(d) Based on your preceding result, without performing a calculation, would a 99% confidence interval for the average blood concentration level of police officers contain 35 μg/l?

Answer 1

Concepts and reason

Parametric t and z tests are used to test the means of populations. The calculation method differs according to the nature of the samples. A distinction is made between independent samples or paired samples. The t and z tests are known as parametric because the assumption is made that the samples are normally distributed. For the t-test, the variance of the population is presumed to be unknown. The problem deals with the concepts of test for one sample t.

Fundamentals

When the population mean is not known, the population is normally distributed or $n > 30$ and the sample is simple random sample, one sample t test is used. The formula for the test statistic is,

$t = \frac{{\bar x - \mu }}{{s/\sqrt n }}$

Here, $\mu$ is the hypothesized mean, $\bar x$ is the sample mean s is the sample standard deviation, and n is the sample size.

Here, the test statistic value is compared with the t-critical value. If the test statistic value falls in the rejection region, the test results are significant at the desired level of significance.

(a)

The null and alternative hypotheses are,

Null hypothesis, ${H_0}:$ Police officers have not been exposed to a higher concentration of lead.

Symbolically, ${H_0}:\mu \le 35$

Alternative hypothesis, ${H_1}:$ Police officers appear to have been exposed to a higher concentration of lead.

Symbolically, ${H_1}:\mu > 35$

Here, $\mu$ is the hypothesized mean of lead concentration.

(b)

The necessary conditions or requirements are:

1.The sample is a simple random sample.

2.The value of the population standard deviation $\sigma$ is not known.

3.The population is normally distributed or $n > 30$ .

In this problem the sample size is 52, and the population standard deviation is not known. So, the requirements are satisfied. So, there is no problem to conduct one sample t-test.

(c)

The null and alternative hypotheses are,

Null hypothesis, ${H_0}:$ Downtown Police officers do not have higher exposure to lead

Symbolically, ${H_0}:\mu \le 35$

Alternative hypothesis, ${H_1}:$ Downtown Police officers have higher exposure to lead

Symbolically, ${H_1}:\mu > 35$

Here, $\mu$ is the hypothesized mean of lead concentration.

Let the sample mean be $\bar x = 124.32$ .

Let the sample standard deviation be $s = 37.74$ .

Let the sample size be $n = 52$ .

The test statistic value is,

$\begin{array}{c}\\t = \frac{{\bar x - \mu }}{{s/\sqrt n }}\\\\ = \frac{{124.32 - 35}}{{37.74/\sqrt {52} }}\\\\ = 17.067\\\end{array}$

So, the test statistic value is $t = 17.067$ .

The degrees of freedom is,

$\begin{array}{c}\\df = n - 1\\\\ = 52 - 1\\\\ = 51\\\end{array}$

Let the level of significance be $\alpha = 0.05$ .

Critical value: Using the t-distribution tables for 51 degrees of freedom for $\alpha = 0.05$ the right tail critical value is ${t_{0.05,51}} = 1.675$ .

Rejection region:

Reject the null hypothesis, if $t > 1.675$ otherwise not.

Conclusion: The test statistic value (17.067) is greater than the critical value (1.675). Reject the null hypothesis. Therefore, it can be concluded that the downtown police officers have a higher lead exposure than the group in the previous study.

(d)

No, 99% confidence interval for the average blood concentration level of police officers does not contain 35 \mu g /L. This is because we have already conducted the hypothesis test which confirms police officers have higher lead exposure than the group in the previous study which is greater than 35 \mu g /L with large test statistic (17.067) value. It gives strong evidence is support of the hypothesis for any significance level.

Ans: Part a

The null and alternative hypotheses are,

$\begin{array}{l}\\{H_0}:\mu \le 35\\\\{H_1}:\mu > 35\\\end{array}$

Here, $\mu$ is the hypothesized mean of lead concentration.

Part b

In this problem the sample size is 52, and the population standard deviation is not known. So, the requirements are satisfied. So, there is no problem to conduct one sample t-test.

Part c

The test statistic value is greater than the critical value. Reject the null hypothesis. Therefore, it can be concluded that the downtown police officers have a higher lead exposure than the group in the previous study.

Part d

No, 99% confidence interval for the average blood concentration level of police officers do not contain 35 \mu g /L.