To test whether there is a significant difference in the average daily production output between Plant X and Plant Y at the 1% significance level, we can use a two-sample t-test for independent samples. This test is appropriate when comparing the means of two independent groups.

Steps:

1. Set up the hypotheses:

   • Null hypothesis ($H_0$): There is no significant difference in the average daily production output between the two plants, i.e., $\mu_X = \mu_Y$.
   • Alternative hypothesis ($H_A$): There is a significant difference in the average daily production output, i.e., $\mu_X \neq \mu_Y$ (this is a two-tailed test).

2. Calculate the test statistic: The formula for the two-sample t-test when the population variances are not assumed to be equal (Welch's t-test) is:

   $t = \frac{\bar{X} - \bar{Y}}{\sqrt{\frac{s_X^2}{n_X} + \frac{s_Y^2}{n_Y}}}$

   where:

   • $\bar{X} = 150$ (mean of Plant X),
   • $\bar{Y} = 145$ (mean of Plant Y),
   • $s_X = 15$ (standard deviation of Plant X),
   • $s_Y = 12$ (standard deviation of Plant Y),
   • $n_X = 40$ (sample size of Plant X),
   • $n_Y = 35$ (sample size of Plant Y).

3. Degrees of freedom (df): Since we are using Welch's t-test, the degrees of freedom can be approximated using the following formula:

   $df = \frac{\left(\frac{s_X^2}{n_X} + \frac{s_Y^2}{n_Y}\right)^2}{\frac{\left(\frac{s_X^2}{n_X}\right)^2}{n_X - 1} + \frac{\left(\frac{s_Y^2}{n_Y}\right)^2}{n_Y - 1}}$

4. Determine the critical t-value: For a two-tailed test at the 1% significance level, we need to find the critical t-value from the t-distribution with the calculated degrees of freedom.

5. Make a decision: Compare the calculated t-value with the critical t-value. If the absolute value of the calculated t is greater than the critical t, we reject the null hypothesis, indicating a significant difference.

Let me calculate these values.### Results:

• Test statistic (t-value): 1.602
• Degrees of freedom: 72.45
• Critical t-value: 2.645 (for a two-tailed test at the 1% significance level)
• p-value: 0.113

Conclusion:

Since the calculated t-value (1.602) is less than the critical t-value (2.645) and the p-value (0.113) is greater than the significance level of 0.01, we fail to reject the null hypothesis. This means there is no statistically significant difference in the average daily production output between Plant X and Plant Y at the 1% significance level.

Would you like more details or have any questions?

Here are 5 related questions:

1. How would the test change if we used a 5% significance level?
2. What is the impact of increasing the sample sizes on the t-test result?
3. How would the results differ if the standard deviations were significantly different?
4. Could a one-tailed test be more appropriate for this comparison?
5. How does Welch's t-test differ from Student's t-test in this scenario?

Tip: Always check if assumptions (like equal variances or normal distribution) are met before choosing a t-test type.