Question

test a claim that the mean amount of carbon monoxide in the air in u.s. cities is less than 2.32 parts per million. it was found that the mean amount of carbon monoxide in the air for the random sample of 65 cities is 2.38 parts per million and the standard deviation is 2.11 parts per million. at $\alpha=0.01$, can the claim be supported? complete parts (a) through (e) below. assume the population is normally distributed.
$h_0: \mu < 2.32$
(type integers or decimals. do not round.)
the claim is the alternative hypothesis.
(b) use technology to find the critical value(s) and identify the rejection region(s).
the critical value(s) is/are $t_0 = -2.38$
(use a comma to separate answers as needed. round to two decimal places as needed.)
choose the graph which shows the rejection region.
oa. $t>t_0$
ob. $toc. $-t_0 < t < t_0$
od. $t < -t_0, t > t_0$

Explanation:

Step1: Determine the test statistic formula

For a one - sample \(t\) - test, the formula 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

Given \(\bar{x} = 2.38\), \(\mu=2.42\), \(s = 2.11\), \(n = 65\).

Step3: Calculate the test statistic

$$
LATEXBLOCK0
$$

Step4: Compare the test statistic with the critical value

The critical value \(t_{0}=-2.38\) (left - tailed test). Since \(-0.15>-2.38\) (the test statistic does not fall in the rejection region \(t < t_{0}\)).

Answer:

Since the test statistic \(t\approx - 0.15\) is not less than the critical value \(t_{0}=-2.38\), we fail to reject the null hypothesis. So, there is not enough evidence at the \(\alpha = 0.01\) level of significance to support the claim that the mean amount of carbon monoxide in the air in U.S. cities is less than \(2.42\) parts per million.