How large a sample should be selected to provide a 95% confidence interval with a margin of error of 10 (to the nearest whole number)? Assume that the population standard deviation is 40.

Try this formula:

n = [(z-value * sd)/E]^2

With your data:

n = [(1.96 * 40)/10]^2

I'll let you finish the calculation. Remember to round the answer to the nearest whole number.

61.4656

To determine the sample size needed to provide a 95% confidence interval with a margin of error of 10, we can use the formula for sample size calculation:

n = (Z * σ / E)^2

Where:
n = sample size
Z = Z-score corresponding to the desired level of confidence (for a 95% confidence level, Z = 1.96)
σ = population standard deviation
E = margin of error

Plugging in the given values, we have:

n = (1.96 * 40 / 10)^2

n = (78.4 / 10)^2

n = (7.84)^2

n ≈ 62

Therefore, a sample size of 62 (to the nearest whole number) should be selected to provide a 95% confidence interval with a margin of error of 10.

To determine the sample size needed for a 95% confidence interval with a margin of error of 10, we can use the following formula:

n = (Z * σ / E)²

Where:
- n represents the sample size.
- Z is the z-value corresponding to the desired confidence level. For a 95% confidence interval, the z-value is approximately 1.96.
- σ is the population standard deviation.
- E is the margin of error.

In this case, the population standard deviation (σ) is given as 40, and the margin of error (E) is 10. Therefore, we can calculate the sample size as follows:

n = (1.96 * 40 / 10)²
n = (78.4 / 10)²
n ≈ 7.84²
n ≈ 61.45

Since a sample size must be a whole number, we round up to the nearest whole number.

Therefore, a sample size of approximately 62 should be selected to provide a 95% confidence interval with a margin of error of 10.