Answer
$t_0\lt t_α$: null hypothesis is not rejected.
There is not enough evidence to conclude that carpeted rooms have more bacteria than uncarpeted rooms.
Work Step by Step
$x ̅_1,n_1~and~s_1$ refer to carpeted rooms and $x ̅_2,n_2~and~s_2$ refer to uncarpeted rooms.
$x ̅_1=\frac{∑x_{1_i}}{n_1}=11.2$
$s_1=\sqrt {\frac{∑(x_{1_i}-x ̅_1)^2}{n_1}}=2.6774$
$x ̅_2=\frac{∑x_{2_i}}{n_2}=9.7875$
$s_2=\sqrt {\frac{∑(x_{2_i}-x ̅_2)^2}{n_2}}=3.2100$
$H_0:~µ_1=µ_2$ versus $H_1:~µ_1\gt µ_2$
$t_0=\frac{(x ̅_1-x ̅_2)-(µ_1-µ_2)}{\sqrt {\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}}=\frac{(11.2-9.7875)-0}{\sqrt {\frac{2.6774^2}{8}+\frac{3.2100^2}{8}}}=0.956$
$n=8$, so:
$d.f.=n-1=7$
Right-tailed test:
$t_α=t_{0.05}=1.895$
(According to Table VI, for d.f. = 7 and area in right tail = 0.05)
Since $t_0\lt t_α$, we do not reject the null hypothesis.
