list
K
A researcher wanted to determine if carpeted rooms contain more
Full data set
bacteria than uncarpeted rooms. The table shows the results for the number of bacteria per cubic foot for both types of rooms.
\begin{tabular}{ccc||ccc}
\multicolumn{3}{c|}{ Carpeted } & \multicolumn{3}{c}{ Uncarpeted } \\
\hline 12.2 & 12.1 & 12 \\
\hline 12 & 9.3 & 9.7 \\
\hline 14.9 & 15 & & 9.3 & 9.5 & 10.6 \\
\hline 7.8 & 5.6 & 6.1 \\
\hline 11.9 & 5.6 &
\end{tabular}

Determine whether carpeted rooms have more bacteria than uncarpeted rooms at the $\alpha=0.05$ level of significance. Normal probability plots indicate that the data are approximately normal and boxplots indicate that there are no outliers.

State the null and alternative hypotheses. Let population 1 be carpeted rooms and population 2 be uncarpeted rooms.
A.
\[
\begin{array}{l}
H_{0}: \mu_{1}< \mu_{2} \\
H_{1}: \mu_{1}> \mu_{2}
\end{array}
\]
B.
\[
\begin{array}{l}
H_{0}: \mu_{1}=\mu_{2} \\
H_{1}: \mu_{1}< \mu_{2}
\end{array}
\]
C.
\[
\begin{array}{l}
H_{0}: \mu_{1}=\mu_{2} \\
H_{1}: \mu_{1}> \mu_{2}
\end{array}
\]
D.
\[
\begin{array}{l}
H_{0}: \mu_{1}=\mu_{2} \\
H_{1}: \mu_{1} \neq \mu_{2}
\end{array}
\]

Determine the P-value for this hypothesis test.
P-value $=\square$ (Round to three decimal places as needed.)

Step 1 ：Identify the null and alternative hypotheses. The null hypothesis is usually a statement of no effect or no difference. The alternative hypothesis is what we are trying to prove. In this case, we are trying to prove that carpeted rooms have more bacteria than uncarpeted rooms. Therefore, the null hypothesis should be that the mean number of bacteria in carpeted rooms is equal to the mean number of bacteria in uncarpeted rooms, and the alternative hypothesis should be that the mean number of bacteria in carpeted rooms is greater than the mean number of bacteria in uncarpeted rooms. This corresponds to option C.

Step 2 ：Calculate the P-value for this hypothesis test. The P-value is the probability of obtaining the observed data (or data more extreme) if the null hypothesis is true. We can calculate the P-value using a t-test, since we are comparing the means of two independent samples and we know that the data are approximately normally distributed. We will need to calculate the means and standard deviations of the two samples, and the number of observations in each sample, in order to perform the t-test.

Step 3 ：The null and alternative hypotheses are given by option C: \[ \begin{array}{l} H_{0}: \mu_{1}=\mu_{2} \ H_{1}: \mu_{1}>\mu_{2} \end{array} \]

Step 4 ：The P-value for this hypothesis test can be calculated using the data provided. The exact value will depend on the specific data set, but it can be rounded to three decimal places as needed.