1. The water pollution readings at State Park Beach seem to be lower than those of the prior year. A sample of 8 readings (in coliform/100mL) was randomly selected from the records of this year's daily readings: 3.5 3.9 2.8 3.1 3.3 3.4 4.8 3.2 Does this sample provide sufficient evidence to conclude that the mean of this year's pollution readings is significantly lower than last year's mean of 3.8 at a = 0.05? Please show all steps of classical approach clearly and interpret your conclusion.

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### Comparing Water Pollution Readings

#### Problem Statement:

The water pollution readings at State Park Beach seem to be lower than those of the prior year. A sample of 8 readings (in coliform/100mL) was randomly selected from the records of this year’s daily readings:

```
3.5, 3.9, 2.8, 3.1, 3.3, 3.4, 4.8, 3.2
```

#### Objective:

The goal is to determine if this sample provides sufficient evidence to conclude that the mean of this year’s pollution readings is significantly lower than last year’s mean of 3.8 at a significance level of \(\alpha = 0.05\). The analysis should be conducted using the classical approach of hypothesis testing.

---

### Step-by-Step Classical Approach:

1. **Formulate the Hypotheses:**
   - Null Hypothesis (\(H_0\)): \(\mu = 3.8\) (The mean pollution reading this year is equal to last year's mean)
   - Alternative Hypothesis (\(H_1\)): \(\mu < 3.8\) (The mean pollution reading this year is less than last year's mean)

2. **Significance Level:**
   - \(\alpha = 0.05\)

3. **Calculate the Sample Mean (\(\bar{x}\)) and Sample Standard Deviation (s):**
   - Sample Mean (\(\bar{x}\)) = \(\frac{3.5 + 3.9 + 2.8 + 3.1 + 3.3 + 3.4 + 4.8 + 3.2}{8}\) 
   - Calculate each step in the computation to find \(\bar{x}\).
   - Sample Standard Deviation (s) can be computed using the formula:
     \[
     s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}}
     \]
     where \( x_i \) are the individual sample points, \( \bar{x} \) is the sample mean, and \( n \) is the sample size.

4. **Compute the Test Statistic:**
   - Test Statistic (t) can be calculated using:
     \[
     t = \frac{\bar{x} - \mu_0}{s / \sqrt{n
Transcribed Image Text:### Comparing Water Pollution Readings #### Problem Statement: The water pollution readings at State Park Beach seem to be lower than those of the prior year. A sample of 8 readings (in coliform/100mL) was randomly selected from the records of this year’s daily readings: ``` 3.5, 3.9, 2.8, 3.1, 3.3, 3.4, 4.8, 3.2 ``` #### Objective: The goal is to determine if this sample provides sufficient evidence to conclude that the mean of this year’s pollution readings is significantly lower than last year’s mean of 3.8 at a significance level of \(\alpha = 0.05\). The analysis should be conducted using the classical approach of hypothesis testing. --- ### Step-by-Step Classical Approach: 1. **Formulate the Hypotheses:** - Null Hypothesis (\(H_0\)): \(\mu = 3.8\) (The mean pollution reading this year is equal to last year's mean) - Alternative Hypothesis (\(H_1\)): \(\mu < 3.8\) (The mean pollution reading this year is less than last year's mean) 2. **Significance Level:** - \(\alpha = 0.05\) 3. **Calculate the Sample Mean (\(\bar{x}\)) and Sample Standard Deviation (s):** - Sample Mean (\(\bar{x}\)) = \(\frac{3.5 + 3.9 + 2.8 + 3.1 + 3.3 + 3.4 + 4.8 + 3.2}{8}\) - Calculate each step in the computation to find \(\bar{x}\). - Sample Standard Deviation (s) can be computed using the formula: \[ s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n-1}} \] where \( x_i \) are the individual sample points, \( \bar{x} \) is the sample mean, and \( n \) is the sample size. 4. **Compute the Test Statistic:** - Test Statistic (t) can be calculated using: \[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n
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