Question

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 0.038 microgram per cubic meter and the standard deviation is 0.068 microgram per cubic meter. at \\( \alpha = 0.01 \\), can the claim be supported? complete parts (a) through (e) below. assume the population is normally distributed.
\\( h _ { a } : \mu < 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 \\( t _ { 0 } = - 2.40 \\).
(use a comma to separate answers as needed. round to two decimal places as needed.)
choose the graph which shows the rejection region.
(c) find the standardized test statistic, t.
the standardized test statistic is \\( t = \square \\).
(round to two decimal places as needed)

Explanation:

Step1: Recall the formula for the t - statistic

The formula for the t - statistic in a one - sample t - test is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(\mu\) is the population mean under the null hypothesis, \(s\) is the sample standard deviation, and \(n\) is the sample size.

Step2: Identify the values of \(\bar{x}\), \(\mu\), \(s\), and \(n\)

Given that \(\bar{x} = 0.038\), \(\mu=0.037\), \(s = 0.068\), and \(n = 56\).

Step3: Substitute the values into the formula

$$
LATEXBLOCK0
$$

First, calculate \(\sqrt{56}\approx7.483\), then \(0.068/\sqrt{56}\approx0.068\div7.483\approx0.0091\)

$$ t=\frac{0.001}{0.0091}\approx0.11 $$

Answer:

\(t = 0.11\)