The mean annual tuition and fees for a sample of 15 private colleges was with a standard deviation of . A dotplot shows that it is reasonabl

Question

The mean annual tuition and fees for a sample of 15 private colleges was with a standard deviation of . 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.

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1 week 2021-10-06T05:58:47+00:00 1 Answer 0

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    2021-10-06T06:00:28+00:00

    Answer:

    The test statistic value is 1.79.

    The degrees of freedom is 14.

    Step-by-step explanation:

    The complete question is:

    The mean annual tuition and fees for a sample of 15 private colleges was 35,500 with a standard deviation of 6500. A dot plot 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 collages is different from 32,500.  

    Compute the value of the test statistic and state the number of degrees of freedom.

    Solution:

    In this case we need to test whether the mean tuition and fees for private collages is different from 32,500.

    The information provided is:

    n = 15

    \bar x = 35,500

    s = 6500

    As the population standard deviation is not known we will use a t-test for single mean.

    Compute the test statistic value as follows:

     t=\frac{\bar x-\mu}{s/\sqrt{n}}

       =\frac{35500-32500}{6500/\sqrt{15}}\\\\=1.79

    Thus, the test statistic value is 1.79.

    The degrees of freedom of the test is,

    Degrees of freedom = n – 1

                                      = 15 – 1

                                      = 14

    Thus, the degrees of freedom is 14.

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45:7+7-4:2-5:5*4+35:2 =? ( )