A scientist estimates that the mean nitrogen dioxide level in a city is greater than 32 parts per billion. To test this estimate, you determine the nitrogen dioxide \( \quad \) 39 43 levels for 31 randomly selected days. The results (in parts per billion) are listed to the right. Assume that the population standard deviation is \( 7 . A t \boldsymbol{\alpha}=0.05 \), can 173716343522303334 you support the scientist's estimate? Complete parts (a) through (e). (a) Write the claim mathematically and identify \( \mathrm{H}_{0} \) and \( \mathrm{H}_{a} \). Choose from the following. A. \( H_{0} \cdot \mu=32 \) B \[ \mathrm{H}_{\mathrm{a}} . \mu>32 \text { (claim) } \] \[ \begin{array}{l} H_{0} \mu \leq 32 \\ H_{a}: \mu>32 \text { (claim) } \end{array} \] D. \( \mathrm{H}_{0}-\mu<32 \) E \( \mathrm{H}_{0}-\mu \leq 32 \) (claim) \( \mathrm{H}_{\mathrm{a}}-\mu \geq 32 \) (claim) \( \mathrm{H}_{\mathrm{a}} \cdot \mu=32 \) c. \( H_{0}: \mu=32 \) (claim) \( \mathrm{H}_{\mathrm{a}}: \mu>32 \) F. \( \mathrm{H}_{0}: \mu \geq 32 \) (claim) \( \mathrm{H}_{\mathrm{a}} \cdot \mu<32 \) (b) Find the critical value and identify the rejection region. \( z_{0}=1.65 \) (Round to two decimal places as needed.) Rejection region: z \( \square \) 1.65 (c) Find the standardized test statistic. \( \mathrm{z}= \) \( \square \) (Round to two decimal places as needed.)

To solve this problem, we will follow the steps outlined in the prompt. ### (a) Write the claim mathematically and identify \( H_0 \) and \( H_a \). The scientist's claim is that the mean nitrogen dioxide level in the city is greater than 32 parts per billion. This can be expressed mathematically as: - Null Hypothesis (\( H_0 \)): The mean nitrogen dioxide level is less than or equal to 32 parts per billion. - Alternative Hypothesis (\( H_a \)): The mean nitrogen dioxide level is greater than 32 parts per billion. Thus, we can write: \[ H_0: \mu \leq 32 \] \[ H_a: \mu > 32 \text{ (claim)} \] The correct choice from the options provided is: - **B.** \[ H_0: \mu \leq 32 \] \[ H_a: \mu > 32 \text{ (claim)} \] ### (b) Find the critical value and identify the rejection region. Given that this is a one-tailed test with \( \alpha = 0.05 \), we will find the critical value \( z_0 \) for a right-tailed test. The critical value for \( z \) at \( \alpha = 0.05 \) is approximately \( 1.645 \) (rounded to two decimal places, it is \( 1.65 \)). Thus, the rejection region is: \[ z > 1.65 \] ### (c) Find the standardized test statistic. To find the standardized test statistic \( z \), we will use the formula: \[ z = \frac{\bar{x} - \mu_0}{\sigma / \sqrt{n}} \] Where: - \( \bar{x} \) = sample mean - \( \mu_0 \) = hypothesized mean (32) - \( \sigma \) = population standard deviation (7) - \( n \) = sample size (31) First, we need to calculate the sample mean \( \bar{x} \) from the provided nitrogen dioxide levels. Let's assume the nitrogen dioxide levels for the 31 days are as follows (you will need to provide the actual data for accurate calculations): Assuming the nitrogen dioxide levels are: \[ 39, 43, \ldots \] (31 values) Let's calculate the sample mean \( \bar{x} \) and then use it to find the standardized test statistic \( z \). 1. Calculate the sample mean \( \bar{x} \): \[ \bar{x} = \frac{\text{sum of all nitrogen dioxide levels}}{n} \] 2. Substitute \( \bar{x} \), \( \mu_0 \), \( \sigma \), and \( n \) into the formula for \( z \). Please provide the complete list of nitrogen dioxide levels so I can calculate \( \bar{x} \) and subsequently the standardized test statistic \( z \).