Question

Keith sampled 10 private universities in Colorado and recorded the tuition cost. Keith wants to test the hypothesis that the tuition of these 10 universities is more expensive than the national average tuition, which is $29,056 with a population standard deviation of $3,339.
z equals fraction numerator x with bar on top minus mu over denominator begin display style fraction numerator sigma over denominator square root of n end fraction end style end fraction

If the sample mean is $31,650, what is the z-score? Answers are rounded to the hundredths place.

Answer 1

Answer: The z-score = 2.46

Step-by-step explanation:

Given:  sample size : n= 10

Population mean : $\mu= 29056$

population standard deviation : $\sigma= 3339$

Sample mean : $\overline{x} = 31650$

Formula to calculate z :

$z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

Substitute all values, we get

$z=\dfrac{31650-29056}{\dfrac{3339}{\sqrt{10}}}\\\\=\dfrac{2594}{\dfrac{3339}{3.16227766}}\\\\=\dfrac{2594}{3339}\times3.16227766\approx2.46$

Hence, the z-score = 2.46