Question

The mean age of De Anza College students in a previous term was $26.6$ years old. An instructor thinks the mean age for online students is older than $26.6.$ She randomly surveys$56$online students and finds that the sample mean is$29.4$with a standard deviation of $2.1.$ Conduct a hypothesis test.

Short answer

The average age of online students is older than $26.6$and the null hypothesis is rejected.

Step-by-step solution

Given information

Standard deviation $s=2.1$

Sample size, $n=56$

The mean age for online student$=29.4$

Explanation

Hypothesis: The null hypothesis shows that online students have an average age of about or equal to $26.6$while the alternative hypothesis claims that online students are older than $26.6$years old. The hypothesis can be represented as

$H_0:\mu \le 26.6$

$H_a:\mu >26.6$

The degree of freedom is:

$$\begin{gathered} df=n-1 \\ =56-1 \\ =55 \end{gathered}$$

From the t distribution table, for a right-tailed test with $a=0.05\text{for}df=55$, the critical value of t values for the critical region are $t=1.673$.

The test statistic is:

localid="1649829543725" role="math" $$\begin{gathered} t=\frac{M-\mu }{{\displaystyle \frac{S}{\sqrt{n}}}} \\ \text{whereMisthemeanage,nisthesamplesize} \\ M=29.4,n=56,\mu =26.6,s=2.1 \\ t=\frac{29.4-26.6}{{\displaystyle \frac{2.1}{\sqrt{56}}}} \\ =\frac{2.8}{0.2806} \\ =9.98 \end{gathered}$$

We can conclude that the mean age of online students is older than$26.6$ since the null hypothesis was rejected.