Question

The mean annual tuition and fees for a sample of 12 private colleges was $36,800 with a standard deviation of $5000. 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 $33,700 Compute the value of the test statistic and state the number of degrees of freedom. A) 0.620; 12 degrees of freedom B) 0.620; 11 degrees of freedom C) 2.148; 12 degrees of freedom D) 2.148: 11 degrees of freedom

Answer 1

Answer:

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

Step-by-step explanation:

The mean annual tuition and fees for a sample of 12 private colleges was $36,800 with a standard deviation of $5000

$x= 36800\\s = 5000$

n = 12

Claim: You wish to test whether the mean tuition and fees for private colleges is different from $33,700

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

Since n < 30 and sample standard deviation is given so, we will use t test

Formula : $t = \frac{x-\mu}{\frac{s}{\sqrt{n}}}$

Substitute the values in the formula :

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

$t = 2.148$

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

So,. Option D is true

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