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 α = 0.01, can the claim be supported? complete parts (a) through (e) below. assume the population is normally distributed.
(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.
(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 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, \(s\) is the sample standard deviation, and \(n\) is the sample size.

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

Given that \(\bar{x} = 2.38\), \(\mu=2.32\), \(s = 2.11\), and \(n = 65\).

Step3: Substitute the values into the formula

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First, calculate \(\sqrt{65}\approx8.0623\), then \(2.11/\sqrt{65}\approx2.11\div8.0623\approx0.2617\).
Then \(t=\frac{0.06}{0.2617}\approx0.23\)

Answer:

\(0.23\)