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Two Simple Examples Use the sign test to compare two populations for significant differences for the paired data. State the null and alternative hypotheses to be tested. Determine an appropriate rejection region with \(\alpha \leq .10 .\) Calculate the observed value of the test statistic and present your conclusion. $$ \begin{array}{cccccccc} & &&& {\text { Pair }} \\ \hline \text { Population } & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline 1 & 8.9 & 8.1 & 9.3 & 7.7 & 10.4 & 8.3 & 7.4 \\ 2 & 8.8 & 7.4 & 9.0 & 7.8 & 9.9 & 8.1 & 6.9 \\ \hline \end{array} $$

Short Answer

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#tag_title# Short answer The null hypothesis states that there is no significant difference between the medians of the two populations, while the alternative hypothesis states that there is a significant difference. After applying the sign test and comparing the observed test statistic (T = 1) with the critical value (Tα = 0), we fail to reject the null hypothesis. This means we do not have enough evidence to conclude that there is a significant difference between the medians of the two populations.

Step by step solution

01

Null hypothesis (H0)

The null hypothesis states that there is no significant difference between the medians of the two populations: \(H_0: M_1 - M_2 = 0\)
02

Alternative hypothesis (H1)

The alternative hypothesis states that there is a significant difference between the medians of the two populations: \(H_1: M_1 - M_2 \neq 0\) #Step 2: Calculate the differences for each pair#
03

Calculating the differences

For each pair, we need to calculate the difference between the values of population 1 and population 2: Pair 1: \(8.9 - 8.8 = 0.1\) Pair 2: \(8.1 - 7.4 = 0.7\) Pair 3: \(9.3 - 9.0 = 0.3\) Pair 4: \(7.7 - 7.8 = -0.1\) Pair 5: \(10.4 - 9.9 = 0.5\) Pair 6: \(8.3 - 8.1 = 0.2\) Pair 7: \(7.4 - 6.9 = 0.5\) #Step 3: Count the positive and negative differences#
04

Counting positive and negative differences

We will count the number of positive and negative differences: Positive differences: 6 Negative differences: 1 #Step 4: Determine the observed value of the test statistic#
05

Observed value of the test statistic (T)

In the sign test, the observed value of the test statistic (T) is the smaller count of positive or negative differences: \(T = \min (6, 1)\) \(T = 1\) #Step 5: Determine the critical value and rejection region#
06

Determine the critical value (Tα)

Since it is a two-tailed test with a significance level of α = 0.10, we should look up in the table/appropriate resource for the critical value (Tα) corresponding to our sample size (n = 7) and α value: \(T_{(1-\alpha/2)} = T_{0.95} = 0\) (This value can be found in tables or using calculators or statistical software)
07

Set the rejection region

Our rejection region is: \(T \leq T_{0.95}\) #Step 6: Compare the test statistic and make a conclusion#
08

Comparing test statistic and critical value

We need to compare the observed value of the test statistic (T = 1) with the critical value (Tα = 0): \(T = 1 > T_{0.95} = 0\)
09

Conclusion

Since the observed value of the test statistic (T = 1) is not in the rejection region, we fail to reject the null hypothesis. This means we do not have enough evidence to conclude that there is a significant difference between the medians of the two populations.

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Key Concepts

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

Null Hypothesis
When performing a sign test, one of the primary steps is to define the null hypothesis, which provides a starting point for statistical analysis. The null hypothesis (\( H_0 \) ) is a statement of no effect or no difference. It is essentially a skeptical position, which assumes that any observed differences in the data are due to chance rather than a real effect.

For instance, in the given exercise, the null hypothesis is that there is no significant difference between the medians of the two populations being compared, mathematically stated as \( H_0: M_1 - M_2 = 0 \). This hypothesis is what we presuppose to be true before collecting any data, and our test aims to determine whether the observed data provide enough evidence to reject this assumption.
Alternative Hypothesis
The alternative hypothesis (\( H_1 \) or \( H_a \)) provides contrast to the null hypothesis. It suggests that there is an effect or a difference, and in the context of the sign test, it posits that the medians of the two populations are not equal. This hypothesis reflects the research question or the suspicion that led to the investigation.

In our example, the alternative hypothesis is defined as \( H_1: M_1 - M_2 eq 0 \), which means we suspect that there is a significant difference between the population medians. The alternative hypothesis is what we seek to support by showing that the null hypothesis is unlikely given the data.
Test Statistic
A test statistic is a calculated value used to determine whether to reject the null hypothesis. It is derived from sample data and measures the degree of agreement between the null hypothesis and the observed data. For the sign test, the test statistic (\( T \) ) is determined by the smaller number of positive or negative differences in paired observations.

In the exercise provided, it was necessary to calculate the differences for each pair and then count the number of positive and negative differences. The test statistic was the minimum of these counts (\( T = \min(6, 1) = 1 \)). This value is then compared to a critical value to help decide if there is enough evidence to reject the null hypothesis.
Rejection Region
The rejection region defines the values of the test statistic that would lead to rejection of the null hypothesis. It is determined based on the significance level of the test, denoted by \( \alpha \). A common significance level used in hypothesis testing is 0.05, but in the exercise, the significance level was set as \( \alpha \leq 0.10 \).

By consulting statistical tables or software with a given \( \alpha \) value and the number of pairs (\( n \) ), we find the critical value (\( T_\alpha \) ) that delineates the rejection region. In this case, the rejection region is \( T \leq T_{0.95} \). Since the observed test statistic (\( T = 1 \) ) is greater than the critical value (\( T_{0.95} = 0 \)), it falls outside the rejection region, meaning we do not reject the null hypothesis.

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Most popular questions from this chapter

Dissolved oxygen content is a measure of the ability of a river, lake, or stream to support aquatic life, with high levels being better. A pollution- control inspector who suspected that a river community was releasing semitreated sewage into a river, randomly selected five specimens of river water at a location above the town and another five below. These are the dissolved oxygen readings (in parts per million): $$ \begin{array}{l|lllll} \text { Above Town } & 4.8 & 5.2 & 5.0 & 4.9 & 5.1 \\ \hline \text { Below Town } & 5.0 & 4.7 & 4.9 & 4.8 & 4.9 \end{array} $$ a. Use a one-tailed Wilcoxon rank sum test with \(\alpha=.05\) to confirm or refute the theory. b. Use a Student's \(t\) -test (with \(\alpha=.05\) ) to analyze the data. Compare the conclusion reached in part a.

The time required for kindergarten children to assemble a specific Lego creation was measured for children who had been instructed for four different lengths of time. Five children were randomly assigned to each instructional group. The length of time (in minutes) to assemble the Lego creation was recorded for each child in the experiment. $$ \begin{array}{rlll} \hline & {\text { Training Period (hours) }} \\ \hline .5 & 1.0 & 1.5 & 2.0 \\ \hline 8 & 9 & 4 & 4 \\ 14 & 7 & 6 & 7 \\ 9 & 5 & 7 & 5 \\ 12 & 8 & 8 & 5 \\ 10 & 9 & 6 & 4 \\ \hline \end{array} $$ Use the Kruskal-Wallis \(H\) -Test to determine whether there is a difference in the distribution of times for the four different lengths of instructional time. Use \(\alpha=01\).

Use the Wilcoxon rank sum test to determine whether population 1 lies to the left of population 2 by (1) stating the null and alternative hypotheses to be tested, (2) calculating the values of \(T_{1}\) and \(T_{l}^{*},(3)\) finding the rejection region for \(\alpha=.05,\) and (4) stating your conclusions. $$ \begin{array}{l|lllll} \text { Sample } 1 & 1 & 3 & 2 & 3 & 5 \\ \hline \text { Sample } 2 & 4 & 7 & 6 & 8 & 6 \end{array} $$

The data given result from experiments run in completely randomized designs. Use the Kruskal-Wallis H statistic to determine whether there are significant differences between at least two of the treatment groups at the \(5 \%\) level of significance. You can use a computer program if one is available. Summarize your results. $$ \begin{array}{lcc} \hline & \text { Treatment } & \\ \hline 1 & 2 & 3 \\ \hline 26 & 27 & 25 \\ 29 & 31 & 24 \\ 23 & 30 & 27 \\ 24 & 28 & 22 \\ 28 & 29 & 24 \\ 26 & 32 & 20 \\ & 30 & 21 \\ & 33 & \\ \hline \end{array} $$

A psychology class performed an experiment to determine whether a recall score in which instructions to form images of 25 words were given differs from an initial recall score for which no imagery instructions were given. Twenty students participated in the experiment with the results listed in the table. $$ \begin{array}{ccc|ccc} \hline & \text { With } & \text { Without } & & \text { With } & \text { Without } \\ \text { Student } & \text { Imagery } & \text { Imagery } & \text { Student } & \text { Imagery } & \text { Imagery } \\ \hline 1 & 20 & 5 & 11 & 17 & 8 \\ 2 & 24 & 9 & 12 & 20 & 16 \\ 3 & 20 & 5 & 13 & 20 & 10 \\ 4 & 18 & 9 & 14 & 16 & 12 \\ 5 & 22 & 6 & 15 & 24 & 7 \\ 6 & 19 & 11 & 16 & 22 & 9 \\ 7 & 20 & 8 & 17 & 25 & 21 \\ 8 & 19 & 11 & 18 & 21 & 14 \\ 9 & 17 & 7 & 19 & 19 & 12 \\ 10 & 21 & 9 & 20 & 23 & 13 \\ \hline \end{array} $$ a. What three testing procedures can be used to test for differences in the distribution of recall scores with and without imagery? What assumptions are required for the parametric procedure? Do these data satisfy these assumptions? b. Use both the sign test and the Wilcoxon signed-rank test to test for differences in the distributions of recall scores under these two conditions. c. Compare the results of the tests in part b. Are the conclusions the same? If not, why not?

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