Question

a scientist estimates that the mean nitrogen dioxide level in a city is greater than 31 parts per billion. to test this estimate, you determine the nitrogen dioxide levels for 31 randomly selected days. the results (in parts per billion) are listed to the right. assume that the population standard deviation is 11. at \\( \alpha = 0.12 \\), can you support the scientists estimate? complete parts (a) through (e).
(a) write the claim mathematically and identify \\( h_0 \\) and \\( h_a \\). choose from the following.
\\( \bigcirc \\) a. \\( h_0: \mu = 31 \\)
\\( h_a: \mu>31 \\)
(claim)
\\( \bigcirc \\) b. \\( h_0: \mu\geq31 \\)
(claim)
\\( h_a: \mu<31 \\)
\\( \bigcirc \\) c. \\( h_0: \mu<31 \\)
\\( h_a: \mu\geq31 \\)
(claim)
\\( \bigcirc \\) d. \\( h_0: \mu = 31 \\)
(claim)
\\( h_a: \mu>31 \\)
\\( \bigcirc \\) e. \\( h_0: \mu\leq31 \\)
(claim)
\\( h_a: \mu>31 \\)
\\( \bigcirc \\) f. \\( h_0: \mu\leq31 \\)
\\( h_a: \mu>31 \\)
(claim)
(b) find the critical value and identify the rejection region.
\\( z_0= \\) (round to two decimal places as needed.)

Explanation:

Step1: Determine the type of test

Since the claim is that the mean nitrogen dioxide level is greater than 31, this is a right - tailed test.

Step2: Find the critical value

For a right - tailed test with significance level \(\alpha = 0.12\), we look up the \(z\) - value in the standard normal distribution table. The critical value \(z_0\) is the value such that \(P(Z>z_0)=\alpha\). Using the standard normal table or a calculator, \(z_0 = 1.17\) (because \(P(Z < 1.17)=0.88\) and \(P(Z>1.17)=1 - 0.88=0.12\)).
The rejection region is \(z>z_0\), i.e., \(z > 1.17\).

Answer:

\(z_0=1.17\)