Question

a scientist estimates that the mean nitrogen dioxide level in a city is greater than 26 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 = 0.01 \\), 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 = 26 \\)
\\( h _ { a } : \mu > 26 \\) (claim)
\\( \bigcirc \\) b. \\( h _ { 0 } : \mu \leq 26 \\) (claim)
\\( h _ { a } : \mu > 26 \\)
\\( \bigcirc \\) c. \\( h _ { 0 } : \mu = 26 \\) (claim)
\\( h _ { a } : \mu > 26 \\)
\\( \bigcirc \\) d. \\( h _ { 0 } : \mu \leq 26 \\)
\\( h _ { a } : \mu > 26 \\) (claim)
\\( \bigcirc \\) e. \\( h _ { 0 } : \mu \geq 26 \\) (claim)
\\( h _ { a } : \mu < 26 \\)
\\( \bigcirc \\) f. \\( h _ { 0 } : \mu < 26 \\)
\\( h _ { a } : \mu \geq 26 \\) (claim)
(b) find the critical value and identify the rejection region.
\\( z _ { 0 } = 2.33 \\) (round to two decimal places as needed.)
rejection region: \\( z > 2.33 \\)
(c) find the standardized test statistic.
\\( z = 1.14 \\) (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 \(z\) falls in the rejection region, we reject \(H_0\). The rejection region is \(z>2.33\) (from part (b)).

Step2: Compare the test - statistic with the rejection region

We have \(z = 1.14\) (from part (c)). Since \(1.14<2.33\), the test statistic does not fall in the rejection region.

Answer:

Fail to reject \(H_0\)