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

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

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 \\eq 33700`

Formula to find the value of z is:
`z=(\bar x-\mu)/((\sigma)/(√(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=(36800-33700)/((5000)/(√(12)))`

`z=2.148`

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