Confidence level margin of error

Question

Assuming the population has an approximate normal distribution, if a sample size n = 12 has a sample mean x-bar = 47 with a sample standard deviation s = 6, find the margin of error at a 98% confidence level.

Answer 1

Hi Bailey,

We do not know population standard deviation and our sample size is small, so we will use a t-confidence interval to compute margin of error. Formula for t-confidence interval:

CI=x-bar +/- t*SE

Margin of error is simply the last part of that equation. For the record, margin of error is always standard error times critical value.

MoE=t*SE

Let's break this down:

t*=t critical value, obtained from t-table

SE=standard error, which has its own formula:

SE=s/sqrt(n) where:

s=sample standard deviation

n=sample size

So, for our problem:

s=6

n=12

SE=6/sqrt(12)

SE=1.73

Now, we still need t*, the t-critical value.. To get this, we first compute degrees of freedom:

df=n-1

n=sample size, so:

df=12-1

df=11

Now, we have a 98% confidence level here, so go to that confidence level's column, and go to the row for 11 degrees of freedom:

t*=2.718

Now, we have all we need to compute margin of error:

MoE=t*SE

t*=2.718

SE=1.73

MoE=(2.718*1.73)

MoE=4.70

I hope this helps.