To find the population standard deviation ($\sigma$) given the confidence level, margin of error, and sample size, we can use the formula for the margin of error for estimating a population mean with a given confidence level:

$E = z \times \frac{\sigma}{\sqrt{n}}$

where:

• $E$ is the margin of error (in this case, 2),
• $z$ is the z-score corresponding to the 90% confidence level,
• $\sigma$ is the population standard deviation, and
• $n$ is the sample size (in this case, 50).

Step 1: Determine the $z$-score for 90% confidence

For a 90% confidence level, the $z$-score is approximately 1.645.

Step 2: Substitute the values and solve for $\sigma$

$2 = 1.645 \times \frac{\sigma}{\sqrt{50}}$

Now, solve for $\sigma$:

1. First, isolate $\sigma$: $\frac{\sigma}{\sqrt{50}} = \frac{2}{1.645}$

2. Calculate $\frac{2}{1.645}$: $\frac{2}{1.645} \approx 1.216$

3. Multiply both sides by $\sqrt{50}$ to find $\sigma$: $\sigma \approx 1.216 \times \sqrt{50}$

4. Calculate $\sqrt{50} \approx 7.071$ and multiply: $\sigma \approx 1.216 \times 7.071 \approx 8.6$

Answer

The population standard deviation is approximately $8.6$.

Would you like a further breakdown of the steps, or do you have any questions?

Related Questions

1. How does changing the confidence level affect the sample size needed?
2. What if the margin of error was reduced to 1.5 instead of 2? How would this affect the standard deviation?
3. How do we determine the z-score for different confidence levels?
4. What if we wanted 95% confidence instead of 90%—how would the calculations change?
5. How would increasing the sample size impact the margin of error?

Tip

The z-score for any confidence level can be found using a standard normal distribution table or online calculator.