A population has a mean (q) and a standard deviation equal to 2.3
What is the smallest sample size that would give a 95% confidence interval for (q) to have a width less than 0.1?
Please help
Thankyou
To calculate the smallest sample size that would give a 95% confidence interval for the mean (q) with a width less than 0.1, you can use the formula:
n = (Z * σ / E)^2
Where:
n = sample size
Z = Z-score corresponding to the desired level of confidence (in this case, the 95% confidence level)
σ = standard deviation of the population
E = margin of error (half of the desired interval width)
Thus, to solve the problem, you need to determine the Z-score for a 95% confidence level and calculate the margin of error (E) based on the desired interval width of 0.1.
1. Find the Z-score for a 95% confidence level:
The Z-score is determined by the desired confidence level and can be found using a Z-table or a statistical calculator. For a 95% confidence level, the corresponding Z-score (often denoted as Zα/2) is approximately 1.96.
2. Calculate the margin of error (E):
The margin of error (E) is half of the desired interval width. In this case, the desired interval width is 0.1, so E = 0.1 / 2 = 0.05.
3. Plug the values into the formula:
n = (Z * σ / E)^2
Substituting the values:
n = (1.96 * 2.3 / 0.05)^2
Calculating the result:
n = (4.508 / 0.05)^2
n = 24.096^2
n ≈ 580.397
Therefore, rounding up to the nearest whole number, the smallest sample size required to achieve a 95% confidence interval for the mean (q) with a width less than 0.1 would be 581.