A manufacturing company wants to compare the average production output of two of its plants. Plant X has a sample of 40 days with an average daily production of 150 units and a standard deviation of 15 units. Plant Y has a sample of 35 days with an average daily production of 145 units and a standard deviation of 12 units. Test at the 1% significance level whether there is a significant difference in the average daily production output between the two plants.  

To determine whether there is a significant difference between the average daily production outputs of Plant X and Plant Y, we can perform a two-sample \( t \)-test for the difference of means. Since the sample sizes are relatively large (both are greater than 30), we can approximate the population variances with the sample variances, and use the pooled variance approach if necessary.

### Step-by-Step Process

1. **Hypotheses**:
   - Null Hypothesis (\( H_0 \)): \( \mu_X = \mu_Y \) (There is no difference in the average production outputs of the two plants.)
   - Alternative Hypothesis (\( H_A \)): \( \mu_X \neq \mu_Y \) (There is a difference in the average production outputs of the two plants.)

2. **Significance Level**:
   - Given the significance level \( \alpha = 0.01 \) (1%).

3. **Sample Information**:
   - For Plant X:
     - Sample size (\( n_X \)) = 40
     - Sample mean (\( \bar{X} \)) = 150
     - Sample standard deviation (\( s_X \)) = 15
   - For Plant Y:
     - Sample size (\( n_Y \)) = 35
     - Sample mean (\( \bar{Y} \)) = 145
     - Sample standard deviation (\( s_Y \)) = 12

4. **Test Statistic** (Two-sample \( t \)-test):
   The test statistic is calculated as follows:
   \[
   t = \frac{\bar{X} - \bar{Y}}{\sqrt{\frac{s_X^2}{n_X} + \frac{s_Y^2}{n_Y}}}
   \]
   where:
   - \( \bar{X} \) and \( \bar{Y} \) are the sample means,
   - \( s_X \) and \( s_Y \) are the sample standard deviations,
   - \( n_X \) and \( n_Y \) are the sample sizes.

5. **Degrees of Freedom**:
   The degrees of freedom (using Welch’s approximation) can be calculated as:
   \[
   df = \frac{\left(\frac{s_X^2}{n_X} + \frac{s_Y^2}{n_Y}\right)^2}{\frac{\left(\frac{s_X^2}{n_X}\right)^2}{n_X - 1} + \frac{\left(\frac{s_Y^2}{n_Y}\right)^2}{n_Y - 1}}
   \]
   
6. **Critical Value**:
   Using a two-tailed test at the \( \alpha = 0.01 \) significance level, we compare the calculated test statistic to the critical value from the \( t \)-distribution with the appropriate degrees of freedom.

7. **Decision Rule**:
   - If \( |t| > t_{\alpha/2, df} \) (the critical value from the \( t \)-distribution), we reject the null hypothesis.
   - Otherwise, we fail to reject the null hypothesis.

### Let me calculate the test statistic and determine the conclusion.### Results:

- **Test statistic (t-value)**: \( t = 1.60 \)
- **Degrees of freedom**: \( df \approx 72.45 \)
- **Critical t-value**: \( t_{\alpha/2, df} \approx 2.65 \) (for a two-tailed test at \( \alpha = 0.01 \))

### Conclusion:
Since the calculated \( |t| = 1.60 \) is less than the critical value \( t_{\alpha/2} = 2.65 \), we fail to reject the null hypothesis at the 1% significance level. This means that there is **no significant difference** in the average daily production outputs of Plant X and Plant Y.

Would you like more details, or do you have any questions about this process?

### Further related questions:
1. How would the test change if the significance level were 5%?
2. What is the effect of increasing the sample size on the test's conclusion?
3. Can we conclude that the two plants have identical outputs if we fail to reject the null hypothesis?
4. What assumptions are required for the two-sample t-test?
5. How could we calculate the power of this test?

### Tip:
Always check the assumptions of normality and equal variance before relying on the results of a t-test, especially when sample sizes are smaller.

A manufacturing company wants to compare the average production output of two of its plants. Plant X has a sample of 40 days with an average daily production of 150 units and a standard deviation of 15 units. Plant Y has a sample of 35 days with an average daily production of 145 units and a standard deviation of 12 units. Test at the 1% significance level whether there is a significant difference in the average daily production output between the two plants.

Perform a two-sample t-test to compare the average production outputs of Plant X and Plant Y. This problem includes statistical hypothesis testing with sample data of 40 days for Plant X and 35 days for Plant Y. Learn how to calculate the t-statistic, degrees of freedom, and critical value for a test at the 1% significance level.

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