Question

The mean annual tuition and fees for a sample of 12 private colleges was 36,800 with a standard deviation of 5,000 . A dotplot shows that it is reasonable to assume that the population is approximately normal. You wish to test whether the mean tuition and fees for private colleges is different from . Compute the value of the test statistic and state the number of degrees of freedom.

Answer 1

Answer:

The value of the test statistic and degrees of freedom is 2.148 and 11 respectively.

Step-by-step explanation:

Consider the provided information.

The mean annual tuition and fees for a sample of 12 private colleges was 36,800 with a standard deviation of 5,000 .

Thus, n = 12, $\bar x=36800$ σ = 5000

degrees of freedom = n-1 = 12-1 = 11

$H_0: \mu = 33700\ and\ H_a: \mu \neq 33700$

Formula to find the value of z is: $z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}$

Where $\bar x$ is mean of sample, μ is the mean of population, σ is the standard deviation of population and n is number of observations.

$z=\frac{36800-33700}{\frac{5000}{\sqrt{12}}}$

$z=2.148$

Hence, the value of the test statistic and degrees of freedom is 2.148 and 11 respectively.