Questions: Test a claim that the mean amount of lead in the air in U.S. cities is less than 0.037 microgram per cubic meter. It was found that the mean amount of lead in the air for the random sample of 56 U.S. cities is 0.039 microgram per cubic meter and the standard deviation is 0.068 microgram per cubic meter. At α=0.01, can the claim be supported? Complete parts (a) through (e) below. Assume the population is normally distributed. (a) Identify the claim and state H0 and Ha. H0: μ ≥ 0.037 Ha: μ < 0.037 (Type integers or decimals. Do not round.) The claim is the alternative hypothesis. (b) Find the critical value(s) and identify the rejection region(s). The critical value(s) is/are t0= (Use a comma to separate answers as needed. Round to two decimal places as needed.)

Solution Steps

We are testing the claim that the mean amount of lead in the air in U.S. cities is less than $0.037$ microgram per cubic meter. The hypotheses are defined as follows:

• Null Hypothesis ($H_0$): $\mu \geq 0.037$
• Alternative Hypothesis ($H_a$): $\mu < 0.037$

The standard error ($SE$) is calculated using the formula:

$$SE = \frac{\sigma}{\sqrt{n}} = \frac{0.068}{\sqrt{56}} \approx 0.0091$$

The test statistic ($Z_{test}$) is calculated using the formula:

$$Z_{test} = \frac{\bar{x} - \mu_0}{SE} = \frac{0.039 - 0.037}{0.0091} \approx 0.2201$$

For a left-tailed test, the P-value is determined as follows:

$$P = T(z) \approx 0.5871$$

For a left-tailed test at a significance level of $\alpha = 0.01$, the critical value ($t$) is found using the t-distribution with $n - 1 = 55$ degrees of freedom:

$$\text{Critical Value} \approx -2.3961$$

The rejection region for this left-tailed test is defined as:

$$t < -2.3961$$

We compare the test statistic to the critical value:

• Test Statistic: $0.2201$
• Critical Value: $-2.3961$

Since $0.2201$ is not less than $-2.3961$, we fail to reject the null hypothesis.

Final Answer