Question 13.3: Ear Infections in Swimmers A medical researcher believed the......
Ear Infections in Swimmers
A medical researcher believed the number of ear infections in swimmers can be reduced if the swimmers use earplugs. A sample of 10 people was selected, and the number of infections for a four-month period was recorded. During the first two months, the swimmers did not use the earplugs; during the second two months, they did. At the beginning of the second two-month period, each swimmer was examined to make sure that no infection was present. The data are shown here. At α = 0.05, can the researcher conclude that using earplugs reduced the number of ear infections?
| Number of ear infections | ||
| Swimmer | Before, X_B | After, X_A |
| A | 3 | 2 |
| B | 0 | 1 |
| C | 5 | 4 |
| D | 4 | 0 |
| E | 2 | 1 |
| F | 4 | 3 |
| G | 3 | 1 |
| H | 5 | 3 |
| I | 2 | 2 |
| J | 1 | 3 |
Step 1 State the hypotheses and identify the claim.
H_{0} : The number of ear infections will not be reduced.
H_{1} : The number of ear infections will be reduced (claim).
Step 2 Find the critical value. Subtract the after values X_{A} from the before values X_{B}, and indicate the difference by a positive or negative sign or 0 , according to the value, as shown in the table.
\begin{array}{|c|c|c|c|} \hline \text{Swimmer }& \text{Before,} \boldsymbol{X}_{\boldsymbol{B}} & \text{After}, \boldsymbol{X}_{\boldsymbol{A}} & \text{Sign of difference} \\ \hline A & 3 & 2 & + \\ B & 0 & 1 & – \\ C & 5 & 4 & + \\ D & 4 & 0 & + \\ E & 2 & 1 & + \\ F & 4 & 3 & + \\ G & 3 & 1 & +\\ H & 5 & 3 & + \\ I & 2 & 2 & 0 \\ J & 1 & 3 & – \\ \hline \end{array}
From Table J, with n=9 (the total number of positive and negative signs; the 0 is not counted) and \alpha=0.05 (one-tailed), at most 1 negative sign is needed to reject the null hypothesis because 1 is the smallest entry in the \alpha=0.05 column of Table J.
Step 3 Compute the test value. Since n \leq 25, we will count the number of positive and negative signs found in step 2 and use the smaller value as the test value. There are 7 positive signs and 2 negative signs, so the test value is 2.
Step 4 Make the decision. Compare the test value 2 with the critical value 1 . If the test value is less than or equal to the critical value, the null hypothesis is rejected. In this case, 2>1, so the decision is not to reject the null hypothesis.
Step 5 Summarize the results. There is not enough evidence to support the claim that the use of earplugs reduced the number of ear infections.
| 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 |
||||
| 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.