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Chapter 15: Problem 21

Outpatient Treatment The median length of stay for substance-abuse outpatient treatment completers is 107 days for those referred by the criminal justice system. The following data represent the length of stays for a random sample of substanceabuse outpatient treatment completers who were referred by an employer. $$ \begin{array}{cccccc} \hline 80 & 108 & 95 & 107 & 89 & 100 \\ \hline 85 & 102 & 115 & 109 & 99 & 94 \\ \hline \end{array} $$ Using the Wilcoxon signed-ranks test at the \(\alpha=0.05\) level of significance, does the median length of stay seem different for employer referrals than for those referred by the criminal justice system?

Short Answer

Expert verified
Reject the null hypothesis. The median length of stay for employer referrals is different from 107 days.

Step by step solution

01

- Formulate the Hypotheses

State the null and alternative hypotheses. The null hypothesis ( H_0) is that the median length of stay for employer referrals is equal to 107 days. The alternative hypothesis ( H_a) is that the median length of stay for employer referrals is different from 107 days. $$ H_0: \text{Median} = 107 $$ $$ H_a: \text{Median} eq 107 $$
02

- Organize the Data

List the differences between each sample data point and the median 107. Calculate the differences: $$ 80-107 = -27, \ 108-107 = 1, \ 95-107 = -12, \ 107-107 = 0, \ 89-107 = -18, \ 100-107 = -7, \ 85-107 = -22, \ 102-107 = -5, \ 115-107 = 8, \ 109-107 = 2, \ 99-107 = -8, \ 94-107 = -13 $$
03

- Calculate Absolute Differences and Rank Them

Ignore the differences that are zero and rank the absolute values of the remaining differences. $$ \begin{array}{cccccc} |27| = 27 & |1| = 1 & |12| = 12 & |0|= 0 & |18| = 18 & |7| = 7 \ |22| = 22 & |5| = 5 & |8| = 8 & |2| = 2 & |8| = 8 & |13| = 13 \ \ Ranks: & 1 & 2 & & 4.5 & 6 & & 9 & 4.5 & 8 & 3 & 4.5 & 7 \ \ \text{Tied ranks averaged} & 3 & 4.5 & 4.5 \ \end{array}$$
04

- Assign Signs to Ranks

Assign the signs of the original differences to the ranks. For negative differences: -27 (12), -12 (7), -18 (9), -7 (4.5), -22 (8), -5 (3), -8 (4.5), -13 (6). For positive differences: 1 (1), 8 (4.5), 2 (2), 8 (4.5).
05

- Calculate Test Statistic

Calculate the Wilcoxon signed-rank test statistic: W = Sum of positive ranks or sum of negative ranks, whichever is smaller $$ W = 1 + 4.5 + 2 + 4.5 = 12 $$ (positive ranks total = 12), $$ W = 12 + 6 + 9 + 4.5 + 8 + 3 + 4.5 + 7 = 54 $$ (negative ranks total = 54).
06

- Determine Critical Value

Determine the critical value for Wilcoxon signed-rank test at 伪 = 0.05 level for n = 11 non-zero differences. From the Wilcoxon signed-rank test table, the critical value is 14.
07

- Make Decision

Compare the test statistic to the critical value. Since W = 12 is less than 14, we reject the null hypothesis.
08

- Conclusion

At 伪 = 0.05 significance level, there is enough evidence to conclude that the median length of stay for employer referrals is significantly different from 107 days.

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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 method used to decide whether there is enough evidence to reject a null hypothesis. In this context, we want to know if the median length of stay for employer referrals differs from the established median of 107 days for criminal justice referrals. We start with two hypotheses: the null hypothesis (H0), which states there is no difference (Median = 107), and the alternative hypothesis (Ha), which suggests there is a difference (Median 鈮� 107). This process helps us make an objective decision based on sample data.
statistical significance
Statistical significance helps us understand if our findings are likely to be real or if they could have happened by chance. We use a significance level, often denoted as 伪 (alpha), to determine this. In our case, the 伪 is set at 0.05. This means there is a 5% risk of concluding that a difference exists when there is none. If our test statistic (W) is less than the critical value (14, from the table), we say our results are statistically significant, indicating enough evidence to reject the null hypothesis. In this exercise, W = 12, which is less than 14, leading us to reject H0 and conclude a significant difference.
non-parametric tests
Non-parametric tests, like the Wilcoxon signed-rank test used here, do not assume a normal distribution of the data. This makes them useful when dealing with small sample sizes or data that do not meet normality assumptions. The Wilcoxon signed-rank test specifically compares the median of paired samples or differences against a hypothesized median. It ranks the absolute differences, considers their signs, and sum these ranks, making it robust against outliers and non-normal data distributions.
median comparison
Comparing medians is crucial when dealing with skewed data or outliers, as the median is less affected by extreme values compared to the mean. In our exercise, we compare the sample median length of stay with a hypothetical median of 107 days. We rank the absolute differences between each data point and 107, then analyze these ranks to judge if the observed differences are significant. By rejecting H0, we conclude that the median length of stay for employer referrals is statistically different, showing the strength of median comparison in capturing true central tendencies in non-normal data.

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