An environmental researcher claims that the mean amount of sulfur dioxide in the air in U.S. cities is 1.15 parts per billion. In a random sample of 134 U.S. cities, the mean amount of sulfur dioxide in the air is 0.93 parts per billion. Assume the population standard deviation is 2.62 parts per billion. At a = 0.01, is there enough evidence to reject the claim? %3D

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### Statistics Question: Sulfur Dioxide in U.S. Cities

#### Problem Statement:

An environmental researcher claims that the mean amount of sulfur dioxide in the air in U.S. cities is 1.15 parts per billion. In a random sample of 134 U.S. cities, the mean amount of sulfur dioxide in the air is 0.93 parts per billion. Assume the population standard deviation is 2.62 parts per billion. At \( \alpha = 0.01 \), is there enough evidence to reject the claim?

#### Step-by-Step Solution:

1. **Null and Alternative Hypothesis**:
   - Null Hypothesis (\(H_0\)): The mean amount of sulfur dioxide in the air in U.S. cities is 1.15 parts per billion.
     \[
     H_0: \mu = 1.15 \, \text{parts per billion}
     \]
   - Alternative Hypothesis (\(H_A\)): The mean amount of sulfur dioxide in the air in U.S. cities is not 1.15 parts per billion.
     \[
     H_A: \mu \ne 1.15 \, \text{parts per billion}
     \]

2. **Significance Level**:
   \[
   \alpha = 0.01
   \]

3. **Sample Statistics**:
   - Sample size (\(n\)): 134
   - Sample mean (\(\bar{x}\)): 0.93 parts per billion
   - Population standard deviation (\(\sigma\)): 2.62 parts per billion

4. **Test Statistic**:
   Since the sample size is large (\(n > 30\)), we use the Z-test for the population mean.
   \[
   Z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}}
   \]
   Plug in the values:
   \[
   Z = \frac{0.93 - 1.15}{\frac{2.62}{\sqrt{134}}} = \frac{-0.22}{\frac{2.62}{11.57}} = \frac{-0.22}{0.2264} \approx -0.97
   \]

5. **Critical Value**:
   For a two-tailed test with \( \alpha = 0.01 \), the critical z-values
Transcribed Image Text:### Statistics Question: Sulfur Dioxide in U.S. Cities #### Problem Statement: An environmental researcher claims that the mean amount of sulfur dioxide in the air in U.S. cities is 1.15 parts per billion. In a random sample of 134 U.S. cities, the mean amount of sulfur dioxide in the air is 0.93 parts per billion. Assume the population standard deviation is 2.62 parts per billion. At \( \alpha = 0.01 \), is there enough evidence to reject the claim? #### Step-by-Step Solution: 1. **Null and Alternative Hypothesis**: - Null Hypothesis (\(H_0\)): The mean amount of sulfur dioxide in the air in U.S. cities is 1.15 parts per billion. \[ H_0: \mu = 1.15 \, \text{parts per billion} \] - Alternative Hypothesis (\(H_A\)): The mean amount of sulfur dioxide in the air in U.S. cities is not 1.15 parts per billion. \[ H_A: \mu \ne 1.15 \, \text{parts per billion} \] 2. **Significance Level**: \[ \alpha = 0.01 \] 3. **Sample Statistics**: - Sample size (\(n\)): 134 - Sample mean (\(\bar{x}\)): 0.93 parts per billion - Population standard deviation (\(\sigma\)): 2.62 parts per billion 4. **Test Statistic**: Since the sample size is large (\(n > 30\)), we use the Z-test for the population mean. \[ Z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}} \] Plug in the values: \[ Z = \frac{0.93 - 1.15}{\frac{2.62}{\sqrt{134}}} = \frac{-0.22}{\frac{2.62}{11.57}} = \frac{-0.22}{0.2264} \approx -0.97 \] 5. **Critical Value**: For a two-tailed test with \( \alpha = 0.01 \), the critical z-values
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