Population normal distribution

Question

Assuming the population has an approximate normal distribution, if a sample size n=21 has a sample mean x-bar=37 with a sample standard deviation s=6, find the margin of error at an 80% confidence level. Round the answer to two decimal places.

Answer 1

Hi Bailey,

No population standard deviation is given, so we have to use a t confidence interval:

CI=x-bar +/- t*SE

Let's break this down:

x-bar=sample mean

t*=t critical value, obtained from t table

SE=standard error, which is computed as:

SE=s/sqrt(n)

s=sample standard deviation

n=sample size

Let's compute our standard error right now:

s=6

n=21

SE=6/sqrt(21)

SE=1.31

Now, we also must find t*. To do this, we first need our degrees of freedom, computed as:

df=n-1

n=21

df=21-1

df=20

Now, go to the t-table, search for 80% confidence in column, 20 degrees of freedom in row:

t*=1.325

Returning to our initial equation:

CI=x-bar +/- t*SE

x-bar=37

t*=1.325

SE=1.31

CI=37 +/- (1.325*1.31)

CI=37 +/- 1.74

CI=(35.26, 38.74)

I hope this helps.