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Researchers wanted to discover whether the median amount of bacteria in carpeted rooms was greater than that in uncarpeted rooms. To determine the amount of bacteria in a room, researchers pumped the air from the room over a Petri dish at the rate of 1 cubic foot per minute for 8 carpeted rooms and 8 uncarpeted rooms. Colonies of bacteria were allowed to form in the 16 Petri dishes. The results (bacteria/cubic foot) are presented in the following table: $$ \begin{array}{|rc|cc|} \hline {\text { Carpeted Rooms }} & {\text { Uncarpeted Rooms }} \\ \hline 11.8 & 10.8 & 12.1 & 12.0 \\ \hline 8.2 & 10.1 & 8.3 & 11.1 \\ \hline 7.1 & 14.6 & 3.8 & 10.1 \\ \hline 13.0 & 14.0 & 7.2 & 13.7 \\ \hline \end{array} $$ Is the median amount of bacteria in carpeted rooms greater than the median amount of bacteria in uncarpeted rooms? Use the \(\alpha=0.05\) level of significance.

Short Answer

Expert verified
The median amount of bacteria in carpeted rooms is not greater than in uncarpeted rooms.

Step by step solution

01

State the Hypotheses

Formulate the null hypothesis and the alternative hypothesis. \( H_0: \text{The median bacteria count in carpeted rooms} \leq \text{the median bacteria count in uncarpeted rooms} \). \( H_a: \text{The median bacteria count in carpeted rooms} > \text{the median bacteria count in uncarpeted rooms} \).
02

Choose the Significance Level

Given in the problem, the significance level \( \alpha \) is 0.05.
03

Organize the Data

List the bacteria counts from the carpeted rooms and uncarpeted rooms. Carpeted: [11.8, 12.1, 8.2, 8.3, 7.1, 3.8, 13.0, 7.2]. Uncarpeted: [10.8, 12.0, 10.1, 11.1, 14.6, 10.1, 14.0, 13.7].
04

Compute the Medians

Find the median of the carpeted and uncarpeted room data sets. For carpeted rooms, reorder the data: [3.8, 7.1, 7.2, 8.2, 8.3, 11.8, 12.1, 13.0]. The median is \( (8.2 + 8.3)/2 = 8.25 \). For uncarpeted rooms, reorder the data: [10.1, 10.1, 10.8, 11.1, 12.0, 13.7, 14.0, 14.6]. The median is \( (11.1 + 12.0)/2 = 11.55 \).
05

Determine the Test Statistic

Use the Wilcoxon Rank-Sum test to compare the medians. Calculate the rank sums for the two groups.
06

Find Critical Value

Refer to a table of critical values for the Wilcoxon Rank-Sum test at \( \alpha = 0.05 \).
07

Make a Decision

Compare the test statistic to the critical value to determine whether to reject the null hypothesis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hypothesis Testing
Hypothesis testing is a fundamental method in statistics used to make decisions based on data. When a researcher wants to know if a certain condition or difference exists, they set up hypotheses.
There are two types of hypotheses: the null hypothesis (ull hypothesis\text{null hypothesis}\(H_0\)), and the alternative hypothesis (\text{alternative hypothesis}Hi\(\text{alternative hypothesis}H_a\)).
The null hypothesis represents the idea that there is no effect or difference, often implying a status quo or baseline scenario. In contrast, the alternative hypothesis suggests that there is an effect or difference.
In our exercise, the null hypothesis (\(H_0\)) is that the median bacteria count in carpeted rooms is less than or equal to that in uncarpeted rooms. On the other hand, the alternative hypothesis (\text{alternative hypothesis}Hi\(\text{alternative hypothesis}H_a\)) suggests that the median bacteria count in carpeted rooms is greater.
Formulating these hypotheses helps us focus our analysis and determine if the data provides enough evidence to reject the null hypothesis in favor of the alternative.

Significance Level
The significance level, denoted as \text{significance level} alphaα (\text{alphaα}\(α\)), is a threshold set by the researcher to decide whether to reject the null hypothesis. Typically, it is set at 0.05, which corresponds to a 5% risk of concluding a difference exists when there is none.
This is a crucial component of hypothesis testing, as it quantifies the probability of making a Type I error—rejecting a true null hypothesis.
In our exercise, the significance level is 0.05. This means we are willing to accept a 5% chance of incorrectly concluding that the median bacteria count in carpeted rooms is greater than that in uncarpeted rooms.
By keeping the significance level in mind, we can later determine whether the evidence from our data is strong enough to reject the null hypothesis. If our test statistic exceeds a critical value corresponding to the 0.05 significance level, we will reject the null hypothesis.

Median Comparison
Comparing medians is a useful method to understand the central tendency of two different groups, in this case, carpeted and uncarpeted rooms.
The median is the middle value in a data set when it is ordered from smallest to largest. If there is an even number of observations, the median is the average of the two middle numbers.
For our exercise, after ordering the bacteria counts, we found the medians for both groups:
  • Carpeted rooms: 8.25
  • Uncarpeted rooms: 11.55

This initial calculation shows that the median bacteria count is lower in carpeted rooms. However, to determine statistical significance, we must use a formal test.
The Wilcoxon Rank-Sum test is ideal for this because it does not assume normal distribution. It allows us to compare the two independent samples to see if their population medians significantly differ. By using the rank sums and comparing them to critical values, we can determine if the observed difference is statistically significant.
Understanding these key steps helps demystify the testing process and shows students how to make data-driven decisions in research.

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Most popular questions from this chapter

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