# Confidence Interval for a Mean

$15$ credit cards were randomly sampled and analyzed with the following results:

(Sample)

Mean $= 50.50$

Sample Variance $= 400$

$\sqrt{400} = 20 =$ Standard deviation right?

$95\%$ confidence interval $= z = 1.96$

"A $95\%$ confidence interval for the average amount the credit card customers spend on their first visit to the chain's new store in the mall is:"

So my understanding was the confidence interval is

$\text{Mean} \pm Z\cdot \text{stdev} / \sqrt{\text{sample size}}$.

I do $\frac{20\cdot 1.96}{\sqrt{15}}$ in the calculator and I get $10.12$.

The book says the answer is $11.08$. What am I doing wrong?

• For a sample size this small, perhaps use the Student T distribution instead? – barrycarter Nov 3 '14 at 19:54
• That worked! If you could put that as and answer I'll select you as correct answer. – Jazz Man Nov 3 '14 at 19:57