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Resuelveel siguiente ejercicios de estadística

This problem is asking you to perform a statistical test to determine whether the distribution of the adsorption levels (in ppm of heavy metals) is uniform. This can be done using a **Chi-Square Goodness of Fit Test** to compare the observed frequencies of samples in each adsorption range with the expected frequencies under the assumption of uniform distribution.

### Steps to Solve:
1. **Null Hypothesis (H₀)**: The distribution of adsorption levels is uniform.
   
2. **Alternative Hypothesis (H₁)**: The distribution of adsorption levels is not uniform.

3. **Observed Frequencies (O)**: These are given in the problem:
   - 200 to 300 ppm: 234 samples
   - 300 to 400 ppm: 325 samples
   - 400 to 500 ppm: 267 samples
   - 500 to 600 ppm: 345 samples

4. **Expected Frequencies (E)**: If the distribution is uniform, each adsorption range should have approximately the same number of samples. 
   - The total number of samples is:
     \[
     234 + 325 + 267 + 345 = 1171
     \]
   - Since there are 4 ranges, the expected frequency for each range is:
     \[
     \frac{1171}{4} = 292.75
     \]

5. **Chi-Square Statistic (χ²)**: The formula is:
   \[
   χ² = \sum \frac{(O - E)^2}{E}
   \]
   where \(O\) is the observed frequency and \(E\) is the expected frequency.

6. **Degrees of Freedom**: The degrees of freedom for this test is the number of categories minus 1. There are 4 categories (ranges), so the degrees of freedom is:
   \[
   df = 4 - 1 = 3
   \]

7. **Chi-Square Test**: We will calculate the value of the chi-square statistic and compare it to the critical value from the chi-square distribution table at a given significance level (usually \( \alpha = 0.05 \)).

Let me calculate the chi-square value next.The calculated chi-square statistic is approximately \(26.93\), with 3 degrees of freedom.

To determine whether this result is statistically significant, we can compare this chi-square statistic to the critical value from the chi-square distribution table for 3 degrees of freedom at a 0.05 significance level.

Let me calculate the critical value next.The critical value for the chi-square distribution with 3 degrees of freedom at the 0.05 significance level is approximately \(7.81\).

### Conclusion:
Since the calculated chi-square statistic (\(26.93\)) is much greater than the critical value (\(7.81\)), we reject the null hypothesis. This means there is strong evidence to suggest that the distribution of adsorption levels is not uniform.

Would you like more details or have any further questions?

Here are 5 related questions to expand your understanding:
1. How would the test result change if the adsorption values were more evenly distributed?
2. What is the significance of the degrees of freedom in the chi-square test?
3. How would the chi-square test differ if there were more than 4 adsorption levels?
4. Can this test be applied to non-numerical categories?
5. What are some limitations of the chi-square test?

**Tip:** Always check the assumptions of the chi-square test, including sufficient sample size and independence of observations, to ensure valid results.

En el desarrollo de métodos para purificar agua contaminada con residuos metálicos pesados, la empresa ColdWather prueba bacterias para un proceso de bio-lixiviación. La prueba reporta un nivel de adsorción (ppm) de metales pesados según lo siguiente: 200 a 300 ppm: 234 muestras, 300 a 400 ppm: 325 muestras, 400 a 500 ppm: 267 muestras, 500 a 600 ppm: 345 muestras. Los especialistas estiman que la distribución se comporta de manera uniforme. Realice una prueba adecuada para probar esta afirmación.

Explore how the Chi-Square Goodness of Fit Test is used to analyze the uniformity of metal adsorption levels in ppm across different ranges. This detailed guide includes the steps, formulas, and calculation of the chi-square statistic to understand the distribution.

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