/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 Use the Wilcoxon matched-pairs s... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 6

Use the Wilcoxon matched-pairs signedranks test to test the given hypotheses at the \(\alpha=0.05\) level of significance. The dependent samples were obtained randomly. Hypotheses: \(H_{0}: M_{D}=0\) versus \(H_{1}: M_{D}<0\) with \(n=25\) and \(T_{+}=95\)

Short Answer

Expert verified
Reject \(H_{0}\) since \(T_{+} < T_{c}\). Median difference is significantly less than zero.

Step by step solution

01

- State the Hypotheses

The null hypothesis is given as \(H_{0}: M_{D}=0\) (the median difference is zero). The alternative hypothesis is \(H_{1}: M_{D}<0\) (the median difference is less than zero).
02

- Identify the Significance Level

The significance level is given as \(\alpha=0.05\).
03

- Determine the Test Statistic

For the Wilcoxon matched-pairs signed-ranks test, calculate the test statistic \(T_{+} = 95\).
04

- Find the Critical Value

Refer to the Wilcoxon signed-ranks critical value table for \(n=25\) at \(\alpha=0.05\) for a one-tailed test. The critical value \(T_c\) is 98.
05

- Compare Test Statistic to Critical Value

Since \(T_{+} = 95\) is less than the critical value \(T_c = 98\), we reject the null hypothesis.
06

- Conclude the Test

We have sufficient evidence to reject the null hypothesis \(H_{0}\) at the \(\alpha=0.05\) level of significance. Thus, we conclude that the median difference is significantly less than zero.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Hypothesis Testing
In hypothesis testing, we are essentially making an educated guess about a population parameter. We begin by stating two hypotheses: the null hypothesis (denoted as \(H_0\)) and the alternative hypothesis (denoted as \(H_1\)).
The null hypothesis usually suggests that there is no effect or no difference, while the alternative hypothesis proposes that there is an effect or difference. These hypotheses are mutually exclusive.
For example, in this exercise, the hypotheses are:
  • \(H_0: M_D = 0\) - The median difference is zero.
  • \(H_1: M_D < 0\) - The median difference is less than zero.
The goal is to determine if we have enough evidence to reject the null hypothesis in favor of the alternative hypothesis.
This involves comparing test statistics to critical values, which we'll discuss next.
Significance Level
The significance level, often denoted as \(\alpha\), represents the threshold for determining whether to reject the null hypothesis. This level is chosen before conducting the test.
A common choice for \(\alpha\) is 0.05, which means that we are willing to accept a 5% chance of incorrectly rejecting the null hypothesis.
If our test results show a probability (p-value) less than \(\alpha\), we reject \(H_0\); otherwise, we do not reject \(H_0\).
In our exercise, the significance level is \(\alpha = 0.05\). This small value signifies our requirement for strong evidence before we reject the null hypothesis.
Test Statistic
The test statistic is a standardized value that is calculated from sample data during a hypothesis test.
It is used to determine whether to reject the null hypothesis. In the case of the Wilcoxon matched-pairs signed-ranks test, the test statistic is denoted as \(T_+\).
This value is derived from the sum of the ranks of differences that are positive.
For our problem, we are given that \(T_+ = 95\). This statistic will be compared to a critical value to conclude the test.
Critical Value
The critical value is a point on the test's distribution that is compared to the test statistic to decide whether to reject the null hypothesis.
It is determined based on the significance level \(\alpha\), the sample size, and the type of test (one-tailed or two-tailed).
For the Wilcoxon matched-pairs signed-ranks test, we refer to a table that provides critical values.
In our scenario, for \(n=25\) and \(\alpha=0.05\) in a one-tailed test, the critical value \(T_c\) is 98. This means if our test statistic is less than this value, we reject the null hypothesis.
Reject the Null Hypothesis
The decision to reject the null hypothesis is the main goal of hypothesis testing.
After calculating the test statistic and finding the critical value, we compare the two.
If the test statistic falls into the critical region (beyond the critical value), we reject the null hypothesis. Otherwise, we fail to reject it.
In our problem, since \(T_+ = 95\) is less than the critical value \(T_c = 98\), we reject the null hypothesis \(H_0\).
This tells us that there is sufficient evidence to conclude that the median difference is significantly less than zero at the \(\alpha=0.05\) level.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A researcher wants to know if the median weight of NFL offensive tackles is higher than the median weight of NFL defensive tackles. He randomly selected 10 offensive tackles and 8 defensive tackles and obtained the following data: $$ \begin{array}{lllll} {\text { Offensive Linemen }} \\ \hline 323 & 295 & 305 & 380 & 309 \\ \hline 320 & 328 & 313 & 318 & 305 \\ \hline \end{array} $$ $$ \begin{array}{llll} {\text { Defensive Linemen }} \\ \hline 289 & 250 & 305 & 310 \\ \hline 295 & 278 & 300 & 339 \\ \hline \end{array} $$ Do the data indicate that offensive tackles are heavier? Use the \(\alpha=0.05\) level of significance.

Describe the difference between parametric statistical procedures and nonparametric statistical procedures.

Use the sign test to test the given alternative hypothesis at the \(\alpha=0.05\) level of significance. The median is different from 68. An analysis of the data reveals that there are 45 plus signs and 27 minus signs.

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.

Use the Wilcoxon matched-pairs signedranks test to test the given hypotheses at the \(\alpha=0.05\) level of significance. The dependent samples were obtained randomly. Hypotheses: \(H_{0}: M_{D}=0\) versus \(H_{1}: M_{D} \neq 0\) with \(n=14, T_{-}=-45,\) and \(T_{+}=60\)
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.