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The mean annual tuition and fees for a sample of 11 private colleges was $26,500 with a standard deviation of $6,000. 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 $31,000.
i). State the null and alternate hypotheses.
ii). Compute the value of the test statistic and state the number of degrees of freedom.
iii). State a conclusion regarding H. Use the a = 0.05 level of significance.

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Answer:

Step-by-step explanation:

Given that:

Sample size n = 11

Sample Mean X = 26500

standard deviation = 6000

Population mean [tex]\mu[/tex] = 31000

the null and alternate hypotheses are being stated as follows:

[tex]H_o : \mu = 31000[/tex]

[tex]H_1 : \mu \neq 31000[/tex]

The value of the test statistic can be computed as:

[tex]Z = \dfrac{\bar x - \mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

[tex]Z = \dfrac{26500 - 31000}{\dfrac{6000}{\sqrt{11}}}[/tex]

[tex]Z = \dfrac{-4500}{\dfrac{6000}{3.3166}}[/tex]

Z = −2.4875

Z = −2.49

The degree of freedom df = n- 1

The degree of freedom df = 11 - 1

The degree of freedom df = 10

At the level of significance ∝ = 0.05

[tex]t_{\alpha/2}[/tex] = 0.025

From the t distribution table at [tex]t_{\alpha/2, 10}[/tex] and critical value = -2.49;

The p-value =  0.0320

Decision Rule: Reject null hypothesis if p -value is lesser than the level of significance

Conclusion:We reject the null hypothesis , therefore, we conclude that there is no sufficient information to that the mean tuition and fees for private colleges is different from $31,000

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