Question

a scientist estimates that the mean nitrogen dioxide level in a city is greater than 30 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 7. at \\( \alpha = 0.11 \\), can you support the scientists estimate? complete parts (a) through (e).
following
a. \\( h _ { 0 } : \mu \leq 30 \\)
\\( h _ { a } : \mu > 30 \\) (claim)
b. \\( h _ { 0 } : \mu \leq 30 \\) (claim)
\\( h _ { a } : \mu > 30 \\)
c. \\( h _ { 0 } : \mu < 30 \\)
\\( h _ { a } : \mu \geq 30 \\) (claim)
d. \\( h _ { 0 } : \mu \geq 30 \\) (claim)
\\( h _ { a } : \mu < 30 \\)
e. \\( h _ { 0 } : \mu = 30 \\)
\\( h _ { a } : \mu > 30 \\) (claim)
f. \\( h _ { 0 } : \mu = 30 \\) (claim)
\\( h _ { a } : \mu > 30 \\)
(b) find the critical value and identify the rejection region.
\\( z _ { 0 } = 1.23 \\) (round to two decimal places as needed.)
rejection region: \\( z > 1.23 \\)
(c) find the standardized test statistic.
\\( z = - 1.69 \\) (round to two decimal places as needed.)
(d) decide whether to reject or fail to reject the null hypothesis.
reject \\( h _ { 0 } \\)
fail to reject \\( h _ { 0 } \\)

Explanation:

Step1: Recall the decision rule for hypothesis testing

If the standardized test statistic \(z\) is in the rejection region, we reject \(H_0\). If \(z\) is not in the rejection region, we fail to reject \(H_0\).

Step2: Compare the standardized test statistic and the critical value

The rejection region is \(z> 1.23\). The standardized test statistic \(z=-1.69\). Since \(-1.69<1.23\) (i.e., \(z\) is not in the rejection region).

Answer:

Fail to reject \(H_0\)