The z-score measures how many standard deviations a data point (or sample mean) is from the population mean. In the context of hypothesis testing, the z-score is used when the population standard deviation is known, or when the sample size is large enough (typically $n > 30$) to approximate the t-distribution with a normal distribution.

Given the problem statement, let's compute the z-score using the formula:

$z = \frac{\bar{x} - \mu_0}{\frac{\sigma}{\sqrt{n}}}$

However, in this problem, we are using the sample standard deviation $s$ instead of the population standard deviation $\sigma$. Since the sample size is large, we can use the z-score as an approximation. So:

$z = \frac{\bar{x} - \mu_0}{\frac{s}{\sqrt{n}}}$

Where:

• $\bar{x} = 495495 \, \text{mg}$ (sample mean)
• $\mu_0 = 500500 \, \text{mg}$ (population mean)
• $s = 2020 \, \text{mg}$ (sample standard deviation)
• $n = 5555$ (sample size)

Plug in the values:

$z = \frac{495495 - 500500}{\frac{2020}{\sqrt{5555}}}$

Now let's compute it:

$z = \frac{-5005}{\frac{2020}{\sqrt{5555}}}$ $z \approx \frac{-5005}{26.95} \approx -185.66$

So the z-score is approximately $-185.66$.

This is an extremely large (negative) z-score, indicating that the sample mean is far below the hypothesized population mean.

Would you like more details or have any further questions?

Further Questions:

1. What does such an extreme z-score imply about the data?
2. How does the z-score relate to the p-value in hypothesis testing?
3. What is the critical z-score for a 0.05 significance level in a two-tailed test?
4. How would the z-score change if the sample mean were closer to 500500 mg?
5. What are the implications of using the z-score instead of the t-score in this case?

Tip: The z-score can be used as an approximation for the t-score when the sample size is large, making the results more straightforward to interpret using the standard normal distribution.