To test whether there is a significant difference in the average daily production output between the two plants, we can conduct a two-sample t-test for the difference in means. Given that the sample sizes are relatively large (40 and 35), and we assume the populations are normally distributed, a two-sample t-test is appropriate.

Step-by-Step Solution:

1. Hypotheses:

   • Null hypothesis ($H_0$): There is no significant difference in the average production output between the two plants. $H_0: \mu_X = \mu_Y \quad \text{or} \quad \mu_X - \mu_Y = 0$
   • Alternative hypothesis ($H_1$): There is a significant difference in the average production output between the two plants. $H_1: \mu_X \neq \mu_Y$ This is a two-tailed test because we are testing whether the means are different, regardless of direction.

2. Significance Level:

   • $\alpha = 0.01$ (1% significance level)

3. Test Statistic (t-score):

   The formula for the t-score for two independent samples is: $t = \frac{(\bar{X} - \bar{Y})}{\sqrt{\frac{s_X^2}{n_X} + \frac{s_Y^2}{n_Y}}}$ where:

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

4. Degrees of Freedom (df):

   The degrees of freedom for the two-sample t-test are calculated using the following formula (Welch-Satterthwaite equation for unequal variances): $df = \frac{\left( \frac{s_X^2}{n_X} + \frac{s_Y^2}{n_Y} \right)^2}{\frac{\left( \frac{s_X^2}{n_X} \right)^2}{n_X - 1} + \frac{\left( \frac{s_Y^2}{n_Y} \right)^2}{n_Y - 1}}$

   Now, let's compute the values.

I'll calculate the test statistic $t$ and degrees of freedom $df$.The calculated t-statistic is approximately $t = 1.60$, and the degrees of freedom (df) are approximately 72.45.

5. Critical Value and p-value:

Since this is a two-tailed test at the 1% significance level ($\alpha = 0.01$), we will look for the critical t-value corresponding to $\alpha/2 = 0.005$ in a t-distribution with approximately 72 degrees of freedom.

Now, I will find the critical t-value and compare it to the calculated t-statistic to draw a conclusion.The critical t-value at the 1% significance level (two-tailed) is approximately $t_{critical} = 2.65$. The p-value corresponding to the calculated t-statistic is approximately $p = 0.113$.

Conclusion:

• The calculated t-statistic ($t = 1.60$) is less than the critical t-value ($t_{critical} = 2.65$).
• The p-value ($p = 0.113$) is greater than the significance level ($\alpha = 0.01$).

Since the t-statistic does not exceed the critical value and the p-value is greater than 0.01, we fail to reject the null hypothesis. There is no significant difference in the average daily production output between Plant X and Plant Y at the 1% significance level.

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Would you like more details or have any questions? Here are some related questions to explore:

1. What assumptions are required for a two-sample t-test?
2. How would the results change if we used a 5% significance level?
3. What is the interpretation of the p-value in hypothesis testing?
4. How do we handle cases where the sample sizes or variances are significantly different?
5. What are the implications of failing to reject the null hypothesis in a business setting?

Tip: When conducting hypothesis tests, always ensure the assumptions of the test (e.g., normality, independence) are met for valid results.