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 57 u.s. cities is 0.039 microgram per cubic meter and the standard deviation is 0.069 microgram per cubic meter. at \\( \alpha = 0.10 \\), can the claim be supported? complete parts (a) through (e) below. assume the population is normally distributed. (b) find the critical value(s) and identify the rejection region(s). the critical value(s) is/are \\( t _ { 0 } = - 1.29 \\). (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 - test statistic

The formula for the t - test statistic is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\), where \(\bar{x}\) is the sample mean, \(\mu\) is the hypothesized population mean, \(s\) is the sample standard deviation, and \(n\) is the sample size.
Given \(\bar{x} = 0.039\), \(\mu=0.037\), \(s = 0.069\), and \(n = 57\).

Step2: Substitute the values into the formula

First, calculate \(s/\sqrt{n}\):
\(\sqrt{n}=\sqrt{57}\approx7.55\), \(s/\sqrt{n}=\frac{0.069}{7.55}\approx0.0091\)
Then, calculate \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}=\frac{0.039 - 0.037}{0.0091}=\frac{0.002}{0.0091}\approx0.22\)

Answer:

\(t = 0.22\)