https://r2.erweima.ai/mathbot/image/1724589374352rym6nzma.jpg

https://r2.erweima.ai/mathbot/image/1724589374352kd8vfm0w.jpg

Con una confianza del 95%, ¿existe homocedasticidad entre las varianzas de las medias muestrales? Demuestre su respuesta. 

To determine if there is homoscedasticity (equal variances) between the sample means of the two variables at a 95% confidence level, we can perform an F-test. This test compares the variances of the two samples to determine if they are significantly different.

### Steps:
1. **Calculate the sample means \(\bar{X}_1\) and \(\bar{X}_2\) for each variable.**
2. **Calculate the sample variances \(S^2_{X_1}\) and \(S^2_{X_2}\).**
3. **Perform the F-test using the formula:**

   \[
   F = \frac{S^2_{X_1}}{S^2_{X_2}}
   \]

   where \(S^2_{X_1}\) is the larger variance.

4. **Determine the critical value from the F-distribution table at the 95% confidence level.**
5. **Compare the calculated F value to the critical F value:**
   - If \( F \leq \text{F-critical}\), the variances are not significantly different, implying homoscedasticity.
   - If \( F > \text{F-critical}\), the variances are significantly different, implying heteroscedasticity.

Let's calculate the necessary statistics using the provided data.

### Data:
- Variable 1: 250.87, 290.58, 240.88, 190.37, 260.19, 160.08, 360.28, 230.19, 240.54
- Variable 2: 330.86, 270.01, 230.97, 290.05, 390.28, 350.03, 300.24, 290.33, 310.60, 290.55

I'll calculate the means, variances, and the F-test value now.### Calculations:
1. **Sample Means:**
   - Mean of Variable 1 (\(\bar{X}_1\)): \(247.11\)
   - Mean of Variable 2 (\(\bar{X}_2\)): \(305.39\)

2. **Sample Variances:**
   - Variance of Variable 1 (\(S^2_{X_1}\)): \(3255.53\)
   - Variance of Variable 2 (\(S^2_{X_2}\)): \(1930.43\)

3. **F-test Statistic:**
   \[
   F = \frac{S^2_{X_1}}{S^2_{X_2}} = \frac{3255.53}{1930.43} = 1.69
   \]

4. **F-critical value at 95% confidence level**:
   \[
   F_{\text{critical}} = 3.23
   \]

### Conclusion:
Since the calculated F value (\(1.69\)) is less than the F-critical value (\(3.23\)), we **do not reject** the null hypothesis of equal variances. This implies that there is homoscedasticity between the variances of the sample means at the 95% confidence level.

Would you like more details or have any questions?

### Related Questions:
1. How is the F-distribution used in hypothesis testing?
2. What are the assumptions underlying the F-test?
3. How can we interpret the results of an F-test?
4. What would happen if the F value was greater than the critical value?
5. How do you calculate the degrees of freedom in an F-test?

### Tip:
Always verify the assumptions of normality and independence before using the F-test for variances to ensure the validity of the results.

Learn how to determine homoscedasticity using an F-test at a 95% confidence level. Calculate sample means, variances, and interpret results for statistical analysis.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation