To estimate the required sample size $n$ for a 95% confidence interval with a given standard error $SE = 0.34$ and a known population variance $\sigma^2 = 7.3$, we use the formula for the standard error of the mean:

$SE = \frac{\sigma}{\sqrt{n}}$

where:

• $SE$ is the standard error,
• $\sigma$ is the standard deviation of the population,
• $n$ is the sample size.

Given that $\sigma^2 = 7.3$, we can calculate $\sigma$ as:

$\sigma = \sqrt{7.3}$

Next, we rearrange the formula to solve for $n$:

$n = \left( \frac{\sigma}{SE} \right)^2$

Substitute the given values:

$n = \left( \frac{\sqrt{7.3}}{0.34} \right)^2$

Let's compute the value of $n$.The required sample size $n$ to estimate the population mean with a standard error of 0.34 using a 95% confidence interval is approximately 63 observations (rounding up to the nearest whole number).

Would you like more details on this calculation or have any other questions?

Here are some related questions that might interest you:

1. How does the confidence level affect the required sample size?
2. What is the relationship between standard error and standard deviation?
3. How do you interpret a confidence interval in statistics?
4. What happens if the population variance is unknown?
5. How would the required sample size change if the standard error were different?

Tip: Always round up when calculating sample size, as partial samples are not practical in statistics.