Question 13.1: Patients at a Medical Center The manager of Green Valley Med......
Patients at a Medical Center
The manager of Green Valley Medical Center claims that the median number of patients seen by doctors who work at the center is 80 per day. To test this claim, 20 days are randomly selected and the number of patients seen is recorded and shown. At \alpha=0.05, test the claim.
\begin{array}{rrrrr}82 & 85 & 93 & 81 & 80 \\ 86 & 95 & 89 & 74 & 62 \\ 72 & 84 & 88 & 81 & 83 \\ 105 & 80 & 86 & 81 & 87\end{array}
Step 1 State the hypotheses and identify the claim.
H_{0} : Median =80 (claim).
H_{1} : Median \neq 80.
Step 2 Find the critical value.
Subtract the hypothesized median, 80, from each data value. If the data value falls above the hypothesized median, assign the value \mathrm{a}+ sign. If the data value falls below the hypothesized median, assign the data value a – sign. If the data value is equal to the median, assign it a 0.
82-80=+2, so 82 is assigned a + sign.
86-80=+6, so 86 is assigned a + sign .
72-80=-8, so 72 is assigned a – sign.
etc.
The completed table is shown.
\begin{array}{lllll}+ & + & + & + & 0 \\ + & + & + & – & – \\ – & + & + & + & + \\ + & 0 & + & + & +\end{array}
Since n \leq 25, refer to Table J in Appendix A. In this case, n=20-2=18 (There are two zeros) and \alpha=0.05. The critical value for a two-tailed test is 4. See Figure 13-1.
Step 3 Compute the test value. Count the number of + and – signs in step 2, and use the smaller value as the test value. In this case, there are 15 plus signs and 3 minus signs, so the test value is 3 .
Step 4 Make the decision. Compare the test value 3 with the critical value 4 . If the test value is less than or equal to the critical value, the null hypothesis is rejected. In this case, the null hypothesis is rejected since 3<4.
Step 5 Summarize the results. There is enough evidence to reject the null hypothesis that the median of the number of patients seen per day is 80 .
| TABLE J Critical Values for the Sign Test | ||||
| Reject the null hypothesis if the smaller number of positive or negative signs is less than or equal to the value in the table |
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| n | One-tailed, 𝜶 = 0.005 |
𝜶 =0.01 | 𝜶 = 0.025 | 𝜶 = 0.05 |
| Two-tailed, 𝜶 = 0.01 |
𝜶 = 0.02 | 𝜶 = 0.05 | 𝜶 = 0.10 | |
| 8 | 0 | 0 | 0 | 1 |
| 9 | 0 | 0 | 1 | 1 |
| 10 | 0 | 0 | 1 | 1 |
| 11 | 0 | 1 | 1 | 2 |
| 12 | 1 | 1 | 2 | 2 |
| 13 | 1 | 1 | 2 | 3 |
| 14 | 1 | 2 | 3 | 3 |
| 15 | 2 | 2 | 3 | 3 |
| 16 | 2 | 2 | 3 | 4 |
| 17 | 2 | 3 | 4 | 4 |
| 18 | 3 | 3 | 4 | 5 |
| 19 | 3 | 4 | 4 | 5 |
| 20 | 3 | 4 | 5 | 5 |
| 21 | 4 | 4 | 5 | 6 |
| 22 | 4 | 5 | 5 | 6 |
| 23 | 4 | 5 | 6 | 7 |
| 24 | 5 | 5 | 6 | 7 |
| 25 | 5 | 6 | 6 | 7 |
Note: Table J is for one-tailed or two-tailed tests. The term n represents the total number of positive and negative signs. The test value is the number of less frequent signs.
