A scientist estimates that the mean nitrogen dioxide level in a city is greater than 27 parts per billion. To test this estimate, you determine the nitrogen dioxide \( \quad \begin{array}{lllllllllllllllllll}41 & 20 & 33 & 14 & 14 & 42 & 43 & 16 & 30 & 38 & 26\end{array} \) levels for 31 randomly selected days. The results (in parts per billion) are listed \( \quad 2836334035142439161738 \) to the nght Assume that the population standard deviation is 7. At \( \boldsymbol{\alpha}=0.07 \), can 151622313034424138 you support the scientist's estimate? Complete parts (a) through (e). (a) Wite the claim mathematically and identify \( \mathrm{H}_{0} \) and \( \mathrm{H}_{\mathrm{a}} \). Choose from the following. A. \( \mathrm{H}_{0} \cdot \mu \leq 27 \) (claim) B. \( \mathrm{H}_{0}: \mu=27 \) \( \mathrm{H}_{\mathrm{a}} \cdot \mu>27 \) \( \mathrm{H}_{\mathrm{a}}: \mu>27 \) (claim) c. \( \mathrm{H}_{0}: \mu \leq 27 \) D. \( \mathrm{H}_{0} \cdot \mu \geq 27 \) (claim) E. \( H_{0}: \mu<27 \) \( H_{a}: \mu<27 \) \( \mathrm{H}_{\mathrm{a}} \cdot \mu \geq 27 \) (claim) F. \( \mathrm{H}_{0}: \mu=27 \) (claim) \( \mathrm{H}_{\mathrm{a}}: \mu>27 \) (b) Find the critical value and identify the rejection region. \( z_{0}=1.48 \) (Round to two decimal places as needed.) Rejection region \( z \) 1.48 (C) Find the standardiced lest statistic. \[ z=\square \text { (Round to two decimal places as needed) } \]

To determine if the scientist's estimate that the mean nitrogen dioxide level in a city is greater than 27 parts per billion can be supported, we need to perform a hypothesis test. Given: - Sample size (\(n\)) = 31 - Population standard deviation (\(\sigma\)) = 7 - Significance level (\(\alpha\)) = 0.07 - Claim: The mean nitrogen dioxide level is greater than 27 parts per billion - \(H_0\): \(\mu \leq 27\) (null hypothesis) - \(H_a\): \(\mu > 27\) (alternative hypothesis) (a) Write the claim mathematically and identify \(H_0\) and \(H_a\): The claim is that the mean nitrogen dioxide level is greater than 27 parts per billion. Therefore, the claim can be written as: \(H_a\): \(\mu > 27\) (b) Find the critical value and identify the rejection region: The critical value for a one-tailed test with \(\alpha = 0.07\) and \(n = 31\) is \(z_{0} = 1.48\). The rejection region is \(z > 1.48\). (c) Find the standardized test statistic: To find the standardized test statistic, we use the formula: \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \] where: - \(\bar{x}\) is the sample mean - \(\mu\) is the population mean - \(\sigma\) is the population standard deviation - \(n\) is the sample size We need to calculate the sample mean \(\bar{x}\) from the given data: \[ \bar{x} = \frac{41 + 20 + 33 + 14 + 14 + 42 + 43 + 16 + 30 + 38 + 26}{31} \] Let's calculate the sample mean \(\bar{x}\) and then find the standardized test statistic \(z\). Calculate the value by following steps: - step0: Calculate: \(\frac{\left(41+20+33+14+14+42+43+16+30+38+26\right)}{31}\) - step1: Remove the parentheses: \(\frac{41+20+33+14+14+42+43+16+30+38+26}{31}\) - step2: Add the numbers: \(\frac{317}{31}\) The sample mean \(\bar{x}\) is approximately 10.225806. Now, we can calculate the standardized test statistic \(z\): \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \] Substitute the values: \[ z = \frac{10.225806 - 27}{\frac{7}{\sqrt{31}}} \] Let's calculate the standardized test statistic \(z\). Calculate the value by following steps: - step0: Calculate: \(\frac{\left(10.225806-27\right)}{\left(\frac{7}{\sqrt{31}}\right)}\) - step1: Remove the parentheses: \(\frac{10.225806-27}{\frac{7}{\sqrt{31}}}\) - step2: Subtract the numbers: \(\frac{-16.774194}{\frac{7}{\sqrt{31}}}\) - step3: Convert the expressions: \(\frac{-\frac{8387097}{500000}}{\frac{7}{\sqrt{31}}}\) - step4: Multiply by the reciprocal: \(-\frac{8387097}{500000}\times \frac{\sqrt{31}}{7}\) - step5: Multiply the fractions: \(-\frac{8387097\sqrt{31}}{500000\times 7}\) - step6: Multiply: \(-\frac{8387097\sqrt{31}}{3500000}\) The standardized test statistic \(z\) is approximately -13.342109. Now, we can compare the standardized test statistic \(z\) with the critical value \(z_{0} = 1.48\) to determine if we can support the scientist's estimate that the mean nitrogen dioxide level is greater than 27 parts per billion.