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Let's work through this problem step by step. 

### Problem Overview:
You are asked to perform a Chi-squared test to verify whether a sample of 200 lots conforms to a binomial distribution with \( p = 0.1 \) and \( n = 3 \). The observed frequencies of defective units are given, and you need to calculate the expected frequencies, perform the Chi-squared test, and decide whether to reject the null hypothesis.

### Step 1: Calculate the Expected Frequencies
Given a binomial distribution \( X \sim Binomial(n=3, p=0.1) \), the probability mass function (PMF) is:
\[
P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}
\]
where \( k \) is the number of defective units (0, 1, 2, 3), \( n = 3 \), and \( p = 0.1 \).

Let's calculate the probabilities:

1. \( P(X = 0) = \binom{3}{0} \cdot (0.1)^0 \cdot (0.9)^3 \)
2. \( P(X = 1) = \binom{3}{1} \cdot (0.1)^1 \cdot (0.9)^2 \)
3. \( P(X = 2) = \binom{3}{2} \cdot (0.1)^2 \cdot (0.9)^1 \)
4. \( P(X = 3) = \binom{3}{3} \cdot (0.1)^3 \cdot (0.9)^0 \)

### Step 2: Compute the Expected Frequencies
Since there are 200 samples, multiply each probability by 200 to get the expected frequencies.

### Step 3: Perform the Chi-Squared Test
Calculate the Chi-squared statistic using the formula:
\[
\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}
\]
where \( O_i \) are the observed frequencies and \( E_i \) are the expected frequencies.

### Step 4: Compare with the Critical Value
Compare the calculated Chi-squared statistic with the critical value from the Chi-squared distribution with \( k - 1 \) degrees of freedom, where \( k \) is the number of categories (4 in this case).

### Step 5: Conclusion
Decide whether to reject the null hypothesis based on the comparison.

---

I'll perform the calculations now.### Results:

1. **Expected Frequencies:**
   - For 0 defective units: \( 145.80 \)
   - For 1 defective unit: \( 48.60 \)
   - For 2 defective units: \( 5.40 \)
   - For 3 defective units: \( 0.20 \)

2. **Chi-squared Statistic:**
   - The calculated Chi-squared statistic is \( 660.73 \).

### Step 4: Compare with Critical Value
Next, we need to compare the calculated Chi-squared statistic with the critical value from the Chi-squared distribution with 3 degrees of freedom at \( \alpha = 0.05 \).

The critical value \( \chi^2_{0.05, 3} \) is approximately \( 7.815 \).

### Conclusion:
- Since \( 660.73 \) is much greater than \( 7.815 \), we **reject the null hypothesis**. This suggests that the sample does not follow a binomial distribution with the given parameters.

Would you like more details or have any questions?

### 5 Related Questions:
1. How do you interpret a Chi-squared statistic?
2. What are the assumptions of the Chi-squared test?
3. How would the results change with different \( p \) values in the binomial distribution?
4. How can the degrees of freedom affect the critical value in a Chi-squared test?
5. What are some limitations of the Chi-squared test?

**Tip:** Always ensure that the expected frequencies are not too low when using the Chi-squared test; ideally, each expected frequency should be 5 or more.

Se sospecha que el número de unidades defectuosas, X, en un lote de tamaño 3, sigue una distribución Binomial. Realice una prueba Chi-cuadrada con α=0.05 para rechazar o no rechazar la hipótesis nula de que la siguiente muestra de 200 lotes proviene de una distribución Binomial con parámetro p. La frecuencia observada encontrada en la muestra se refiere al número de lotes de tamaño 3 en los que se encontraron 0, 1, 2 o 3 unidades defectuosas. Para los cálculos se deben redondear los valores de la frecuencia esperada a números enteros y completar la siguiente tabla (para el estadístico de prueba use dos cifras decimales y utilice ''.'' como separador de decimales).

This page walks through the steps to perform a Chi-squared test to determine if a sample of 200 lots follows a binomial distribution. It includes the calculation of expected frequencies, the Chi-squared statistic, and the decision to reject or not reject the null hypothesis, suitable for college-level students.

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