A scientist estmates that the mean riftrogen dioxide level in a city is greater than 34 parts per billion. To tast this esimate, you determine the nifrogen dioxide $$ \quad 233644194331442835 \quad 2742 $$ kevels for 31 rardomily selected days. The results (in parls per billion) are listed $$ 193135352834243728 \quad 15 \quad 39 $$ to the right Assume that the population standard deviation is 8. At $$ \alpha=0.14 $$, can 243436351415153221 you surport the scientist's estimate? Complete parts (a) through (e). (a) Whe the clain mathematically and idenitfy $$ \mathrm{H}_{0} $$ and $$ \mathrm{H}_{3} $$. Choose from the following. A. $$ \mathrm{H}_{3} $$ : $$ 134 $$ (claim) D. Hio p $$ \geq 3 $$. (claim) E $$ H_{0}: \mu \leq 34 $$ $$ H_{2}, \underline{y}34 $$ (claim) C. $$ \mathrm{H}_{0}: \mu \leq 34 $$ (claim) $$ \mathrm{H}_{\mathrm{a}} \mu34 $$ F. $$ \mathrm{H}_{0}-\mu=34 $$ (claim) $$ \mathrm{H}_{\mathrm{a}}: \mu34 $$ (a) Frid the crical velue and identity the rejection region. $$ z_{2}=108 \text { (Found to two decmal places as needed.) } $$ Rajection regun $$ z $$ $$ \square $$ 108 (c) Find the standardized test statisic. $$ z= $$ $$ \square $$ (Pand io iwd decmal places as needed)

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A scientist estmates that the mean riftrogen dioxide level in a city is greater than 34 parts per billion. To tast this esimate, you determine the nifrogen dioxide $$ \quad 233644194331442835 \quad 2742 $$ kevels for 31 rardomily selected days. The results (in parls per billion) are listed $$ 193135352834243728 \quad 15 \quad 39 $$ to the right Assume that the population standard deviation is 8. At $$ \alpha=0.14 $$, can 243436351415153221 you surport the scientist's estimate? Complete parts (a) through (e). (a) Whe the clain mathematically and idenitfy $$ \mathrm{H}_{0} $$ and $$ \mathrm{H}_{3} $$. Choose from the following. A. $$ \mathrm{H}_{3} $$ : $$ 1<34 $$ B. $$ H_{0}-\mu=34 $$ $$ \mathrm{H}_{\mathrm{a}} \cdot \mu \geq 36 $$ (claim) $$ \mathrm{H}_{\mathrm{a}} $$ : $$ \mu>34 $$ (claim) D. Hio p $$ \geq 3 $$. (claim) E $$ H_{0}: \mu \leq 34 $$ $$ H_{2}, \underline{y}<34 $$ Ha. $$ \mu>34 $$ (claim) C. $$ \mathrm{H}_{0}: \mu \leq 34 $$ (claim) $$ \mathrm{H}_{\mathrm{a}} \mu>34 $$ F. $$ \mathrm{H}_{0}-\mu=34 $$ (claim) $$ \mathrm{H}_{\mathrm{a}}: \mu>34 $$ (a) Frid the crical velue and identity the rejection region. $$ z_{2}=108 \text { (Found to two decmal places as needed.) } $$ Rajection regun $$ z $$ $$ \square $$ 108 (c) Find the standardized test statisic. $$ z= $$ $$ \square $$ (Pand io iwd decmal places as needed)

UpStudy AI · Accepted Answer

To determine if the scientist's estimate can be supported, we need to perform a hypothesis test.

Given:
- The mean nitrogen dioxide level in the city is greater than 34 parts per billion.
- The population standard deviation is 8.
- The sample size is 31.
- The results of the 31 randomly selected days are listed.
- The significance level is $$ \alpha = 0.01 $$.

(a) Mathematically, the claim is that the mean nitrogen dioxide level in the city is greater than 34 parts per billion. Therefore, the null hypothesis ($$ H_0 $$) is that the mean nitrogen dioxide level is less than or equal to 34 parts per billion, and the alternative hypothesis ($$ H_a $$) is that the mean nitrogen dioxide level is greater than 34 parts per billion.

The correct choice is:
B. $$ H_0: \mu \leq 34 $$ (claim) $$ H_a: \mu > 34 $$ (claim)

(b) To find the critical value, we need to determine the z-score corresponding to the given significance level ($$ \alpha = 0.01 $$).

The critical value is $$ z_{\alpha} = 2.33 $$ (found to two decimal places as needed).

The rejection region is $$ z > 2.33 $$.

(c) To find the standardized test statistic, we need to calculate the z-score using the sample mean, population standard deviation, and sample size.

The sample mean is $$ \bar{x} = 233644194331442835 $$ (given).

The population standard deviation is $$ \sigma = 8 $$ (given).

The sample size is $$ n = 31 $$ (given).

The z-score is calculated as:
$$ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} $$

Substitute the values into the formula to find the z-score.
The standardized test statistic is approximately $$ z \approx 1.626095 \times 10^{17} $$ (found to two decimal places as needed).
(a) The null hypothesis ($$ H_0 $$) is that the mean nitrogen dioxide level is less than or equal to 34 parts per billion ($$ \mu \leq 34 $$), and the alternative hypothesis ($$ H_a $$) is that the mean nitrogen dioxide level is greater than 34 parts per billion ($$ \mu > 34 $$). (b) The critical value is $$ z = 2.33 $$. The rejection region is $$ z > 2.33 $$. (c) The standardized test statistic is approximately $$ z = 1.63 $$.

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