Normal distribution

Question

Given a normal distribution with 𝜎 = 0.340. Find the required sample size for a 99% confidence level (estimating the mean) with a 0.14 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 99% 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 2.576. Using this with the given information, you'll compute:

(0.340*2.576/0.14)^2 = (6.256....)^2 = 39.1375...

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 39, this would actually give a larger Margin of Error (compute it if you're skeptical; 0.340*2.576 = 0.1402, which is slightly too big). To make a long story short, ALWAYS ROUND UP when using this formula, so your minimum sample size would be 40 in this case.