Question

86-8. A manufacturer of powdered soap checks the weights of soap in the company's 6-pound boxes periodically throughout the day. At each of 22 times, four boxes are selected at random, and the average of the weights of the soap in the boxes is recorded. Use the following 22 average weights to test the hypothesis of randomness against the alternative hypothesis of a trend at an approximate significance level of a = 0.025: 6,050 6.038 6.003 6.015 6.025 0.063 6.033 6.010 5.995 6,020 6.060 6.060 6.0656.050 6,043 6.040 6.045 6.065 6.055 6.060 6.060 6.070

Answer 1

The problem states that manufacturer of powdered soap checks the weights of soap in the company's 6-pound boxes periodically throughout the day. At each of 22 times, four boxes are selected at random, and the average of the weights of the soap in the boxes is recorded. The hypothesis to be tested is stated below: H: The selected sample is random. H: The sample is not random. To test the null hypothesis, Run test will be used. To conduct the Run test in Minitab, the steps are given as follows: 1. Enter the given data column-wise in the Minitab spreadsheet. 2. Go to STATS >> NONPARAMETRICS >> RUNS TEST. 3. The following dialog box will appear: Runs Test Variables: Above and below the mean Above and below: Select Help OK Cancel

4. Click "OK" and the following output will appear on the screen. Runs Test: C2 Descriptive Statistics Number of Observations N K SK > K 22 6.042059 13 K = sample mean Test Ho: The order of the data is random H.: The order of the data is not random Null hypothesis Alternative hypothesis Number of Runs Observed Expected 7 11.64 P-value 0.036 The p-value obtained from the Minitab output is 0.036. The level of significance is given to be 0.025. The decision rule is to reject the null hypothesis if the p-value is less than the level of significance. Here, the p-value is greater than the level of significance, that is, 0.036 > 0.025. Thus, we fail to reject the null hypothesis and conclude that the sample is random.