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.
(a) identify the claim and state \\( h _ { 0 } \\) and \\( h _ { a } \\).
\\( h _ { 0 } : \mu \geq 0.037 \\)
\\( 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.
\\( \bigcirc \mathrm { a } \\).
\\( \bigcirc \mathrm { b } \\).
\\( \bigcirc \mathrm { c } \\).
\\( \bigcirc \mathrm { d } \\).

Explanation:

Step1: Calculate the test statistic

The formula for the \(t -\)test statistic is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\).
Here, \(\bar{x} = 0.038\), \(\mu=0.037\), \(s = 0.068\), \(n = 56\).

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Step2: Compare the test statistic with the critical value

The critical value \(t_{0}=- 2.40\) (left - tailed test).
Since \(t = 0.11>-2.40\) (the test statistic does not fall in the rejection region \(t < t_{0}\)).

Answer:

Since the test statistic \(t\approx0.11\) is not less than the critical value \(t_{0}=-2.40\), we fail to reject the null hypothesis \(H_{0}\). So, at the \(\alpha = 0.01\) significance level, there is not enough evidence to support the claim that the mean amount of lead in the air in U.S. cities is less than \(0.037\) microgram per cubic meter.