Question

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.

Answer 1

Explanation:

Given that:

Sample size n = 11

Sample Mean X = 26500

standard deviation = 6000

Population mean
`\mu` = 31000

the null and alternate hypotheses are being stated as follows:

`H_o : \mu = 31000`

`H_1 : \mu \\eq 31000`

The value of the test statistic can be computed as:

`Z = (\bar x - \mu)/((\sigma)/(√(n)))`

`Z = (26500 - 31000)/((6000)/(√(11)))`

`Z = (-4500)/((6000)/(3.3166))`

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

`t_(\alpha/2)` = 0.025

From the t distribution table at
`t_(\alpha/2, 10)` 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