Normal distribution

Question

Given a normal distribution with 𝜎 = 1.890. Find the required sample size for a 90% confidence level (estimating the mean) with a 1.08 margin-of-error.

Answer 1

There is a standard formula for this scenario:

(Minimum Sample Size) = ((Population StdDev)*(critical z-value) / (MarginOfError) ^ 2

You'll need to look up the critical value for a 90% Confidence Interval or use technology to compute it (you know the population's StdDev here, which is why it's a z-score). Accuracy can be important here, so I'll use 1.645. Using this with the given information, you'll compute:

(1.890*1.645/1.08)^2 = (2.8788....)^2 = 8.2872...

However, the sample size must be a whole number (and you rarely get one when using this formula). PAY ATTENTION TO ROUNDING HERE...although you would usually round the above result to 8, this would actually give a larger Margin of Error (compute it if you're skeptical; 1.890*1.645/sqrt(8) = 1.0992, which is too big). To make a long story short, ALWAYS ROUND UP when using this formula, so your minimum sample size would be 9 in this case.