8-5 Resampling: Using Technology for Hypothesis Testing 427 Original Data (minutes) x = 6.0 Modified Data (minutes) x = 6.5 Resample #1 (Resample Modified Data With Replacement) x1 = 7.1 2 2.5 3.5 3 3.5 3.5 6 6.5 8.5 8 8.5 8.5 11 11.5 11.5 If we proceed to obtain many (such as 1000) resamples and find the sample mean for each of them, we can use the 1000 sample means to determine whether the original sample mean of x = 6.0 minutes can easily occur with a population having a mean of m = 6.5 minutes, or whether such a sample mean is very unlikely, suggesting that the population mean is not 6.5 minutes. This is the key question: Do the resamples provide means suggesting that the original sample mean is significantly different from the claimed mean? YOUR TURN. Do Exercise 9 “Lead in Medicine.” SOLUTION The following procedure is based on 1000 resamples. While any number of resamples could be used instead, 1000 is a recommended minimum. The steps in the following procedure are vastly simplified by using technology. See the Tech Center at the end of this section. Step 1: Find the difference d between the claimed mean (from the null hypothesis) and the sample mean. Example: The null hypothesis H0: m = 6.5 minutes shows that the claimed mean is 6.5 minutes. The sample mean is x = 6.0 minutes. The difference is d = 6.5 - 6.0 = 0.5. Step 2: Modify the Sample Data to Conform to the Null Hypothesis. Example: Add d = 0.5 to each sample value so that the modified sample has a mean equal to the claimed mean of 6.5 minutes. Step 3: Resample with replacement 1000 times or more. Find the Value of the Sample Mean for Each Resampling. Example: Use technology to resample the modified data set 1000 times (or more). (You really need technology for this step.) Step 4: Find the number of sample means (found in Step 3) that are “at least as extreme” as the mean of the original sample. Example: The author used randomization to find that there were 743 samples with means at least as extreme as x = 6.0 from the original sample. For more details about the correct interpretation of “at least as extreme,” see the comments following this example. Step 5: Divide the value from Step 4 by the number of randomizations to get the estimated P-value. Example: P-value is estimated to be 743>1000 = 0.743. INTERPRETATION Using a significance level of 0.05, we fail to reject the null hypothesis of H0: m = 6.5 because the P-value of 0.743 is greater than the significance level of 0.05. Using the randomization method, we find that there is not sufficient evidence to warrant rejection of the claim that the mean service time is equal to 6.5 minutes.
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