To calculate the test statistic for this situation, we will use the formula for the Z-test when the population standard deviation is unknown, but a sample standard deviation is given:

$Z = \frac{\bar{X} - \mu}{\frac{s}{\sqrt{n}}}$

Where:

• $\bar{X}$ is the sample mean (541 grams),
• $\mu$ is the population mean (500 grams),
• $s$ is the sample standard deviation (3.3 grams),
• $n$ is the sample size (147 bags).

Let's calculate the test statistic.The value of the calculated test statistic is 150.64.

Would you like more details or have any questions about this? Here are some related questions to expand your understanding:

1. How do you interpret the result of the Z-test in this context?
2. What is the significance of using a Z-test over a t-test here?
3. How would you determine if the machinery is significantly overfilling the bags?
4. What is the critical Z-value for a 95% confidence level in a one-tailed test?
5. How would increasing the sample size affect the test statistic?

Tip: A large test statistic like this indicates a significant difference between the sample mean and the population mean.