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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 \(n=18, T_{-}=-121,\) and \(T_{+}=50\)

Short Answer

Expert verified
Fail to reject \( \ H_{0} \). Median difference is not significantly different from zero.

Step by step solution

01

- State the hypotheses

Each hypothesis needs to be clearly defined before conducting the test. Here, the null hypothesis is given as: \(H_{0}: M_{D}=0 \) This implies that the median of the differences between the pairs is zero. The alternative hypothesis is: \(H_{1}: M_{D} eq 0 \)
02

- Determine the test statistic

In the Wilcoxon matched-pairs signed-ranks test, the test statistic is the smaller of the two sums of the signed ranks. In this case, we have \(T_{-}=-121\) and \(T_{+}=50\). Therefore, the test statistic (T) is: \[ T = \text{min}( \lvert T_{-} \rvert, T_{+} ) = \text{min}( \lvert -121 \rvert, 50 ) = 50 \]
03

- Obtain the critical value

The critical value for the Wilcoxon signed-ranks test at \( \alpha = 0.05 \) for \( n = 18 \) pairs can be found in statistical tables. For a two-tailed test with \( n = 18 \), the critical value (T_c) at \( \alpha = 0.05 \) is 46.
04

- Compare the test statistic and critical value

Compare the calculated test statistic (T) to the critical value (T_c). \[ T = 50 \] \[ T_c = 46 \]
05

- Make a decision

Since \(T \) (50) is greater than the critical value \[ T_c \] (46), we fail to reject the null hypothesis \ (H_{0}) \. There is not enough evidence to suggest that the median difference is different from zero.

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Key Concepts

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

Wilcoxon test
The Wilcoxon matched-pairs signed-ranks test is a non-parametric statistical test. It is used to compare two related samples, matched samples, or repeated measurements on a single sample. This test is often referred to simply as the Wilcoxon test. It evaluates whether the median difference between pairs of observations is zero.
It considers the magnitude and direction of differences between matched pairs. In our example, we have dependent samples and we are testing at a significance level of \( \alpha = 0.05 \).
Remember, the Wilcoxon signed-ranks test is similar to the paired t-test, but it does not assume a normal distribution of differences. Therefore, it is a good alternative when the normality assumption is in question or when the sample size is small.
Non-parametric tests
Non-parametric tests are a set of statistical tests that do not assume a specific distribution for the data. They are particularly useful when we cannot make assumptions about the population from which our sample is drawn.
Some key points about non-parametric tests:
  • They do not assume the data follow a normal distribution.
  • They are less powerful than parametric tests when the assumptions of the parametric tests are met.
  • Useful for small sample sizes and ordinal data.
  • Examples include the Wilcoxon signed-rank test, Mann-Whitney U test, and Kruskal-Wallis test.
In our context, the Wilcoxon signed-ranks test is a non-parametric test that helps us to compare the medians of two related samples without making assumptions about their distribution.
Hypothesis testing
Hypothesis testing is a statistical method used to make decisions about a population based on sample data. It involves making a claim or hypothesis and then using sample data to test whether this claim is likely to be true.
Steps in Hypothesis Testing include:
1. **Formulate Hypotheses**: The null hypothesis (\(H_0\)) and the alternative hypothesis (\(H_1\)). In our exercise, \( H_{0}: M_{D}=0 \) and \( H_{1}: M_{D} eq 0 \)
2. **Select Significance Level**: Commonly used levels are 0.05 or 0.01. Our exercise uses \( \alpha = 0.05 \). This represents a 5% risk of concluding that a difference exists when there is no actual difference.
3. **Calculate Test Statistic**: Derived from the sample data. For the Wilcoxon test, the test statistic is the smaller of the sums of the signed ranks.
4. **Determine Critical Value**: From statistical tables or distributions corresponding to the test used.
5. **Compare Statistic and Critical Value**: Decide whether to reject the null hypothesis based on this comparison.
If the test statistic is greater than the critical value, as in our example, we fail to reject the null hypothesis, meaning there is not enough evidence to support a difference. This methodical approach helps ensure our conclusions are based on data and not random chance.

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Most popular questions from this chapter

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 \(n=35\) and \(T_{+}=210 .\)

Reaction-Time Experiment Researchers at the University of Mississippi wanted to learn the reaction times of students to different stimuli. In the following data, the reaction times for subjects were measured after they received a simple stimulus and a go/no-go stimulus. The simple stimulus was an auditory cue, and the time from when the cue was given to when the student reacted was measured. The go/no-go stimulus required the student to respond to a particular stimulus and not respond to other stimuli. Again, the reaction time was measured. The following data were obtained: $$ \begin{array}{ccc|ccc} \begin{array}{l} \text { Subject } \\ \text { Number } \end{array} & \text { Simple } & \text { Go/No-Go } & \begin{array}{l} \text { Subject } \\ \text { Number } \end{array} & \text { Simple } & \text { Go/No-Go } \\ \hline 1 & 0.220 & 0.375 & 16 & 0.498 & 0.565 \\ \hline 2 & 0.430 & 1.759 & 17 & 0.262 & 0.402 \\ \hline 3 & 0.338 & 0.652 & 18 & 0.620 & 0.643 \\ \hline 4 & 0.266 & 0.467 & 19 & 0.300 & 0.351 \\ \hline 5 & 0.381 & 0.651 & 20 & 0.424 & 0.380 \\ \hline 6 & 0.738 & 0.442 & 21 & 0.478 & 0.434 \\ \hline 7 & 0.885 & 1.246 & 22 & 0.305 & 0.452 \\ \hline 8 & 0.683 & 0.224 & 23 & 0.281 & 0.745 \\ \hline 9 & 0.250 & 0.654 & 24 & 0.291 & 0.290 \\ \hline 10 & 0.255 & 0.442 & 25 & 0.453 & 0.790 \\ \hline 11 & 0.198 & 0.347 & 26 & 0.376 & 0.792 \\ \hline 12 & 0.352 & 0.698 & 27 & 0.328 & 0.613 \\ \hline 13 & 0.285 & 0.803 & 28 & 0.952 & 1.179 \\ \hline 14 & 0.259 & 0.488 & 29 & 0.355 & 0.636 \\ \hline 15 & 0.200 & 0.281 & 30 & 0.368 & 0.391 \\ \hline \end{array} $$ The researchers used Minitab to test whether the simple stimulus had a lower reaction time than the go/no-go stimulus. The results of the analysis are as follows: (a) State the null and alternative hypotheses. (b) Is the median reaction time for the go/no-go stimulus higher than the median reaction time for the simple stimulus? Use the \(\alpha=0.05\) level of significance. Why?

Problems 17 and 18 illustrate the use of the sign test to test hypotheses regarding a population proportion. The only requirement for the sign test is that our sample be obtained randomly. When dealing with nominal data, we can identify a characteristic of interest and then determine whether each individual in the sample possesses this characteristic. Under the null hypothesis in the sign test, we expect that half of the data will result in minus signs and half in plus signs. If we let a plus sign indicate the presence of the characteristic (and a minus sign indicate the absence), we expect half of our sample to possess the characteristic while the other half will not. Letting \(p\) represent the proportion of the population that possesses the characteristic, our null hypothesis will be \(H_{0}: p=0.5 .\) Use the sign test for Problems 17 and 18 , following the sign convention indicated previously. Trusting the Press In a study of 2302 U.S. adults surveyed online by Harris Interactive 1243 respondents indicated that they tend to not trust the press. Using an \(\alpha=0.05\) level of significance, does this indicate that more than half of U.S. adults tend to not trust the press?

A quality-control inspector tracks the compressive strength of parts created by an injection molding process. If the process is in control, the strengths should fluctuate randomly around the target value of 75 psi. The inspector measures the strength of 20 parts as they come off the production line and obtains the following strengths, in order: 82.0,78.3,73.5,74.4,72.6,79.8,77.0,83.4,76.2,75.2 81.5,69.8,71.3,69.4,82.1,77.6,76.9,77.1,72.7,73.6 Conduct a runs test for randomness of the compression strengths at the \(\alpha=0.05\) level of significance. Should the quality-control inspector be concerned about the process?

Write a paragraph that describes the logic of the test statistic in a two- tailed sign test.

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