Why is this margin of error problem wrong?

Question

$M.OE = (2.75)*(2.7)/\sqrt{n}\;$?

Answer 1

$t = \frac {\mu - \bar x}{s/\sqrt n}$

$\mu - \bar x$ is the margin of error. (true mean - sample mean)

$s$ is the sample standard deviation.

Then you look up on a table or use a statistical calculator to find the range of t for your confidence interval, and the "degrees of freedom" $(n-1)$ and as either a $1$ tail or $2$ tailed test.

$99$% and $9$ degrees of freedom and $2$-tailed test give $|t|<3.25$

$|\mu - \bar x| < \frac {3.25 \cdot 2.7}{\sqrt {10}} = 2.77$

Answer 2

As a rule of thumb, when $n\leq30$ we should use a $t$-distribution rather than a normal distribution.

With a $t$ distribution our confidence interval is given by

$$\bar{X}\pm t \frac{s}{\sqrt{n}}$$

with degrees of freedom $$df=n-1$$

I leave it to you to find the necessary $t$ value using a $t$-table.