Questions: A researcher wanted to determine if carpeted rooms contain more bacteria than uncarpeted rooms. The table shows the results for the number of bacteria per cubic foot for both types of rooms. Full data set Carpeted Carpeted Carpeted Uncarpeted Uncarpeted Uncarpeted --- --- --- --- --- --- 10.9 10.5 9.4 6.7 8.2 5.6 12.5 12.4 9.3 7.5 9.9 8.8 11.2 10.5 5.5 8.3 Determine whether carpeted rooms have more bacteria than uncarpeted rooms at the α=0.01 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. H₀: μ₁=μ₂ H₁: μ₁<μ₂ B. H₀: μ₁=μ₂ H₁: μ₁ ≠ μ₂ C. H₀: μ₁=μ₂ H₁: μ₁>μ₂ D. H₀: μ₁<μ₂ H₁: μ₁>μ₂

A researcher wanted to determine if carpeted rooms contain more bacteria than uncarpeted rooms. The table shows the results for the number of bacteria per cubic foot for both types of rooms. Full data set Carpeted Carpeted Carpeted Uncarpeted Uncarpeted Uncarpeted --- --- --- --- --- --- 10.9 10.5 9.4 6.7 8.2 5.6 12.5 12.4 9.3 7.5 9.9 8.8 11.2 10.5 5.5 8.3 Determine whether carpeted rooms have more bacteria than uncarpeted rooms at the α=0.01 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. H₀: μ₁=μ₂ H₁: μ₁<μ₂ B. H₀: μ₁=μ₂ H₁: μ₁ ≠ μ₂ C. H₀: μ₁=μ₂ H₁: μ₁>μ₂ D. H₀: μ₁<μ₂ H₁: μ₁>μ₂
Transcript text: A researcher wanted to determine if carpeted rooms contain more bacteria than uncarpeted rooms. The table shows the results for the number of bacteria per cubic foot for both types of rooms. Full data set \begin{tabular}{|c|c|c|c|c|c|} \hline \multicolumn{3}{|c|}{Carpeted} & \multicolumn{3}{|c|}{Uncarpeted} \\ \hline 10.9 & 10.5 & 9.4 & 6.7 & 8.2 & 5.6 \\ \hline 12.5 & 12.4 & 9.3 & 7.5 & 9.9 & 8.8 \\ \hline 11.2 & 10.5 & & 5.5 & 8.3 & \\ \hline \end{tabular} Determine whether carpeted rooms have more bacteria than uncarpeted rooms at the $\alpha=0.01$ 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. $\mathrm{H}_{0}: \mu_{1}=\mu_{2}$ $H_{1}: \mu_{1}<\mu_{2}$ B. $H_{0}: \mu_{1}=\mu_{2}$ $H_{1}: \mu_{1} \neq \mu_{2}$ C. $H_{0}: \mu_{1}=\mu_{2}$ $H_{1}: \mu_{1}>\mu_{2}$ D. $H_{0}: \mu_{1}<\mu_{2}$ $H_{1}: \mu_{1}>\mu_{2}$
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Solution

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Solution Steps

Step 1: State the Hypotheses

We are testing whether carpeted rooms contain more bacteria than uncarpeted rooms. The hypotheses are stated as follows:

  • Null Hypothesis (\(H_0\)): \( \mu_1 = \mu_2 \) (The mean number of bacteria in carpeted rooms is equal to that in uncarpeted rooms)
  • Alternative Hypothesis (\(H_1\)): \( \mu_1 > \mu_2 \) (The mean number of bacteria in carpeted rooms is greater than that in uncarpeted rooms)
Step 2: Significance Level

The significance level is set at \( \alpha = 0.01 \).

Step 3: Calculate the Standard Error

The standard error (\(SE\)) is calculated using the formula:

\[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{1.4284}{8} + \frac{2.3998}{8}} = 0.6918 \]

Step 4: Calculate the Test Statistic

The test statistic (\(t\)) is calculated as follows:

\[ t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{10.8375 - 7.5625}{0.6918} = 4.7343 \]

Step 5: Calculate Degrees of Freedom

The degrees of freedom (\(df\)) are calculated using the formula:

\[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} = \frac{0.229}{0.0174} = 13.1531 \]

Step 6: Calculate the p-value

The p-value is calculated as:

\[ P = 1 - T(t) = 1 - T(4.7343) = 0.0002 \]

Step 7: Compare p-value with Significance Level

Since the p-value \(0.0002\) is less than the significance level \(0.01\), we reject the null hypothesis.

Final Answer

Based on the analysis, we conclude that carpeted rooms have more bacteria than uncarpeted rooms. The answer is:

\(\boxed{C}\)

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