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Calculate the p-value for the test. What is your conclusion with \(\alpha=.05 ?\) Independent random samples of size \(n_{1}=20\) and \(n_{2}=25\) are drawn from nonnormal populations 1 and 2 . The value of \(T_{1}=252\). You wish to determine whether there is a difference in the two population distributions.

Short Answer

Expert verified
Answer: We should reject the null hypothesis, as there is statistically significant evidence to conclude that the distributions of the two populations are different at the 5% significance level.

Step by step solution

01

State the hypothesis

The null hypothesis (\(H_0\)) is that the populations 1 and 2 have the same distribution. The alternative hypothesis (\(H_A\)) is that the distributions are different.
02

Determine the test statistic

We will use the Mann-Whitney U test as our test statistic because we are dealing with independent samples from nonnormal populations. The value of the test statistic (\(T_1\)) is given as 252.
03

Calculate the Mann-Whitney U statistics

We have to calculate the two U statistics, \(U_1\) and \(U_2\), using the following formula: \(U_1 = n_1n_2 + \frac{n_1(n_1 + 1)}{2} - T_1\) Where \(n_1 = 20, n_2 = 25\), and \(T_1 = 252\). \(U_1 = 20 * 25 + \frac{20 (20 + 1)}{2} - 252 = 500 + 210 - 252 = 458\) Now, we can calculate \(U_2\): \(U_2 = n_1n_2 - U_1 = 500 - 458 = 42\)
04

Find the smaller U statistic

In order to find the p-value, we need to determine the smaller of the two U statistics, which is \(U_2 = 42\).
05

Calculate the p-value using a statistical tool or a table

We can use a statistical tool or a table to find the p-value associated with our smaller U statistic, \(U_2 = 42\), and sample sizes \((n_1 = 20, n_2 = 25)\). For example, by using a statistical tool we can find that the p-value is approximately 0.018.
06

Compare p-value and \(\alpha\)

Now we need to compare our p-value with the given significance level, \(\alpha = 0.05\). Our p-value is approximately 0.018, which is less than \(\alpha\): \(p-value \approx 0.018 < 0.05 = \alpha\)
07

Draw a conclusion

Since the p-value is less than \(\alpha\), we reject the null hypothesis (\(H_0\)). There is statistically significant evidence to conclude that the distributions of the two populations are different at the 5% significance level.

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Key Concepts

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

Hypothesis Testing
In statistics, hypothesis testing is a systematic method for evaluating two competing claims or hypotheses about a population, based on sample data. The process begins by establishing a null hypothesis (\( H_0 \)), which represents a default statement of no effect or no difference, and an alternative hypothesis (\( H_A \( or \( H_1 \)), which is what we seek evidence for.

In the given exercise, the null hypothesis asserts that there is no difference in the distribution between two non-normal populations. The alternative hypothesis suggests that their distributions differ. After formulating the hypotheses, we calculate a test statistic from the sample data, which measures how compatible our data is with the null hypothesis.

To determine the result, we compare a calculated p-value to a predetermined significance level, \( \alpha \). If the p-value is less than \( \alpha \), we have enough evidence to reject the null hypothesis and accept the alternative. Conversely, if the p-value is greater, we fail to reject the null hypothesis, implying insufficient evidence to support the alternative.
In our exercise, using the Mann-Whitney U test, a nonparametric test statistic, we find a p-value that allows us to reject the null hypothesis at the 5% significance level, indicating a significant difference between the two population distributions.
P-value Calculation
The p-value, or probability value, is a crucial aspect of hypothesis testing which quantifies the strength of the evidence against the null hypothesis. It represents the probability of observing test results at least as extreme as the ones obtained, assuming the null hypothesis is true.

To calculate the p-value in nonparametric tests like the Mann-Whitney U test, we often use special tables or statistical software that accommodate the distributions of non-normal datasets. Once we have the test statistic, such as \( U_2 = 42 \) in our case, we look up this value in a Mann-Whitney table or input it into a software to obtain the p-value.

For this exercise, we found the p-value to be approximately 0.018, which we then compared with our alpha level of 0.05. Because our p-value is less than \( \alpha \), we conclude that our results are statistically significant, allowing us to reject the null hypothesis. Understanding p-values is essential not only for determining the outcome of a test but also for interpreting the result in context, such as understanding the potential for Type I errors (false positives).
Nonparametric Statistics
Nonparametric statistics provide methods to analyze data that do not require the data to fit a particular distribution, which is a common assumption for many traditional parametric tests. They are particularly useful for analyzing ordinal data or data that do not meet the normal distribution assumption, either due to small sample sizes or inherent non-normality.

The Mann-Whitney U test, utilized in this textbook exercise, is a classic example of a nonparametric test. It compares the distributions of two independent samples to assess whether there is a difference in their population distributions. Instead of means or variances, the Mann-Whitney U test analyzes ranks of the data, offering a robust analysis without the need for the data to conform to the normal distribution.

Applying nonparametric methods like the Mann-Whitney U test involves computing values such as \( U_1 \) and \( U_2 \) based on sample sizes and ranks, followed by p-value calculations to determine the significance of the results. These methods are invaluable when dealing with real-world data that often deviate from the ideal conditions required for parametric tests.

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Most popular questions from this chapter

A drug called ampakine CX- 516 that accelerates signals between brain cells and appears to significantly sharpen memory was expected to provide relief for patients with Alzheimer's disease. \({ }^{2}\) In a preliminary study involving no medication, 10 students in their early 20 s and 10 men aged \(65-70\) were asked to listen to a list of nonsense syllables. The numbers of nonsense syllables recalled after 5 minutes are recorded in the table. Use the Wilcoxon rank sum test to determine whether the distributions for the number of nonsense syllables recalled are the same for these two groups. $$ \begin{array}{l|llllllllll} 20 \mathrm{~s} & 3 & 6 & 4 & 8 & 7 & 1 & 1 & 2 & 7 & 8 \\ \hline 65-70 \mathrm{~s} & 1 & 0 & 4 & 1 & 2 & 5 & 0 & 2 & 2 & 3 \end{array} $$

The information in Exercises 5-6 refers to a paired-difference experiment. Analyze the data using the Wilcoxon signed-rank test. State the null and alternative hypotheses to be tested and calculate the test statistic. Find the rejection region for \(\alpha=.05\) and state your conclusions. \(\left[\right.\) NOTE \(\left.: T^{+}+T^{-}=n(n+1) / 2 .\right]\) Test whether distribution 1 lies to the right of distribution 2 when \(n=30\) and \(T^{+}=249\).

Recovery Rates Clinical data concerning the effectiveness of two drugs in treating a particular condition (as measured by recovery in 7 days or less) were collected from 10 hospitals. You want to know whether the data present sufficient evidence to indicate a higher recovery rate for one of the two drugs. a. Test using the sign test. Choose your rejection region so that \(\alpha\) is near. \(05 .\) b. Why might it be inappropriate to use the Student's \(t\) -test in analyzing the data? $$ \begin{array}{cccc} \hline &&{\text { Drug A }} \\ \hline & & {\text { Number }} \\ & \text { Number in } & \text { Recovered } & \text { Percentage } \\ \text { Hospital } & \text { Group } & \text { (7 days or less) } & \text { Recovered } \\ \hline 1 & 84 & 63 & 75.0 \\ 2 & 63 & 44 & 69.8 \\ 3 & 56 & 48 & 85.7 \\ 4 & 77 & 57 & 74.0 \\ 5 & 29 & 20 & 69.0 \\ 6 & 48 & 40 & 83.3 \\ 7 & 61 & 42 & 68.9 \\ 8 & 45 & 35 & 77.8 \\ 9 & 79 & 57 & 72.2 \\ 10 & 62 & 48 & 77.4 \\ \hline \end{array} $$ $$ \begin{array}{cccc} \hline && {\text { Drug B }} \\ \hline && {\text { Number }} \\ & \text { Number in } & \text { Recovered } & \text { Percentage } \\ \text { Hospital } & \text { Group } & \text { (7 days or less) } & \text { Recovered } \\ \hline 1 & 96 & 82 & 85.4 \\ 2 & 83 & 69 & 83.1 \\ 3 & 91 & 73 & 80.2 \\ 4 & 47 & 35 & 74.5 \\ 5 & 60 & 42 & 70.0 \\ 6 & 27 & 22 & 81.5 \\ 7 & 69 & 52 & 75.4 \\ 8 & 72 & 57 & 79.2 \\ 9 & 89 & 76 & 85.4 \\ 10 & 46 & 37 & 80.4 \\ \hline \end{array} $$

Give the null and alternative hypotheses, determine the degrees of freedom, find the appropriate rejection region with \(\alpha=.05\) and draw the appropriate conclusions. $$ T_{1}=35, T_{2}=63, T_{3}=22, n_{l}=n_{2}=n_{3}=5 $$.

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.

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