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This problem deals with comparing lead concentrations before and behind an industrial plant and testing for differences in means and variances between samples. Here's how we can approach each part:

### 1.1 Testing for Differences in Variability and Mean (α = 0.05)

#### Variability Test:
We are asked to check for differences in the variability (standard deviation) between the two samples. To do this, we conduct an **F-test** to compare the variances of the two samples.

The hypotheses for the F-test are:
- \( H_0 \): \( \sigma_1^2 = \sigma_2^2 \) (no difference in variances)
- \( H_A \): \( \sigma_1^2 \neq \sigma_2^2 \) (there is a difference in variances)

The F-statistic is calculated as:
\[
F = \frac{s_1^2}{s_2^2}
\]
Where:
- \( s_1 = 48 \) ppm and \( s_2 = 51 \) ppm are the standard deviations,
- \( n_1 = 301 \) and \( n_2 = 251 \) are the sample sizes.

After calculating \( F \), we compare it with the critical value of \( F_{\alpha/2, n_1-1, n_2-1} \) to decide whether to reject \( H_0 \).

#### Mean Test:
To compare the mean concentrations, we perform a **two-sample t-test** for the means. The hypotheses are:
- \( H_0 \): \( \mu_1 = \mu_2 \) (the means are equal)
- \( H_A \): \( \mu_1 \neq \mu_2 \) (the means are different)

The t-statistic is calculated as:
\[
t = \frac{\bar{y}_1 - \bar{y}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}
\]
Where \( \bar{y}_1 = 280 \) ppm and \( \bar{y}_2 = 330 \) ppm are the sample means.

This will allow us to check if there is a significant difference between the means.

### 1.2 Difference in Lead Concentration at Time 2

We are given the difference in means after six months, \( d = 50 \) ppm, with a standard deviation of \( s_d = 7 \) ppm. We need to test if this difference is significant.

This is a **one-sample t-test** where the null hypothesis \( H_0 \) is that the true mean difference is 0. We calculate the t-statistic:
\[
t = \frac{d}{s_d/\sqrt{n}}
\]
Where \( d = 50 \), \( s_d = 7 \), and \( n \) is the number of measurements.

### 1.3 Germination Rate Test for Two Treatments

Here, we compare germination rates for two treatments: 55% and 58%. We can use a **two-proportion z-test** to compare the germination rates.

The hypotheses are:
- \( H_0 \): The proportions are equal (\( p_1 = p_2 \)).
- \( H_A \): The proportions are different (\( p_1 \neq p_2 \)).

The test statistic for the z-test is calculated as:
\[
z = \frac{\hat{p}_1 - \hat{p}_2}{\sqrt{\hat{p}(1 - \hat{p}) \left( \frac{1}{n_1} + \frac{1}{n_2} \right)}}
\]
Where:
- \( \hat{p}_1 = 0.55 \) and \( \hat{p}_2 = 0.58 \),
- \( n_1 = n_2 = 400 \).

---

Would you like detailed calculations for each of these tests? Also, here are five related questions that can further expand your understanding:

1. How do you interpret the results of the F-test in terms of variance?
2. What is the effect size, and how can it be used to interpret the t-test for the means?
3. How can we visualize the confidence intervals of the two sample means?
4. What assumptions are needed for the t-test to be valid?
5. How does the z-test for proportions differ from a chi-square test for independence?

**Tip**: Always check the assumptions of normality and equal variances when performing t-tests.

The task involves comparing lead concentrations measured in front of and behind an industrial facility, analyzing differences in variability and mean concentrations, comparing lead concentrations over time, and testing germination rates for two treatments.

Learn how to analyze differences in lead concentrations around an industrial plant using F-tests, t-tests, and z-tests. Understand how to test for differences in variability, mean concentrations, and germination rates. Suitable for undergraduate students studying statistics.

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