suppl_mtab_1.pdfCh. 7 Addendum: Using Minitab Stat 218, Dr. Jimmy A. DoiAdditional Notes for Minitab – Please add to the current “Guide to Minitab”Exercise (A) from Course Notes:Researchers suspect that the exposure to toluene will increase the levels of NE in the brain.Let group 1 be the toluene group, and let group 2 be the control.Hypotheses of Interest: H0: µ1= µ2versus Ha: µ1> µ2.The summary statistics are shown here:n mean SDToluene 16 550.2 64.3Control 18 441.5 70.2Using Minitab – If the raw data values are NOT available, but summary statistics are given such as thoseshown above, then we use Minitab in the following way:• Click on Stat -> Basic Statistics -> Two Sample t. In the dialogue box which appears, type inthe appropriate information for ‘Summarized data’ (see below left)• Under ‘Options’, if testing the hypothesis H0: µ1= µ2, ensure you input 0 into ‘Test difference’.Select the appropriate setting for ‘Alternative’ depending upon the form of Ha(greater than or lessthan) (see below right)• Click on ‘OK’ to generate the result of the hypothesis test. Note that the value of the test statisticand the corresponding p-value will be reported in the Session window.Ch. 7 Addendum – Using Minitab: Page 1Ch. 7 Addendum: Using Minitab Stat 218, Dr. Jimmy A. DoiMinitab OutputTwo-Sample T-Test and CISESample N Mean StDev Mean1 16 550.2 64.3 162 18 441.5 70.2 17Difference = mu (1) - mu (2)Estimate for difference: 108.795% lower bound for difference: 69.6T-Test of difference = 0 (vs >): T-Value = 4.71 P-Value = 0.000 DF = 31Minitab OutputIn the exercise we computed the following:1. test statistic: ts= 4.712. DF = 15 (using our conservative method)3. p-value < 0.0005Notes:1. When Minitab reports the P-value = 0.000, this really means the actual p-value is less than0.001.2. The degrees of freedom is not equal to the conservative value we use in class. The DF Minitabuses is based on the complicated formula discussed in the book (Equation 7.1 on Page 227).Exercise (B) from Course Notes:The National Center for Health Statistics has been tracking the mean systolic blood pressure for males 35to 44 years of age. Let us denote this as µ1. Do the executives of a large company have a different meanblood pressure?The summary statistics are shown here:n mean SDRegular 47 122.26 15.0Executives 49 126.07 14.2Using Minitab – Following the same steps we have:• Click on Stat -> Basic Statistics -> Two Sample t. In the dialogue box which appears, type inthe appropriate information for ‘Summarized data’ (see below left)• Under ‘Options’, if testing the hypothesis H0: µ1= µ2, ensure you input 0 into ‘Test difference’.Select the appropriate setting for ‘Alternative’ depending upon the form of Ha(not equal) (see belowright)Ch. 7 Addendum – Using Minitab: Page 2Ch. 7 Addendum: Using Minitab Stat 218, Dr. Jimmy A. Doi• Click on ‘OK’ to generate the result of the hypothesis test. Note that the value of the test statisticand the corresponding p-value will be reported in the Session window.Minitab OutputTwo-Sample T-Test and CISample N Mean StDev SE Mean1 47 122.3 15.0 2.22 49 126.1 14.2 2.0Difference = mu (1) - mu (2)Estimate for difference: -3.8195% CI for difference : (-9.7 3, 2.11)T-Test of difference = 0 (vs not =): T-Value = -1.28 P-Value = 0.205 DF = 93Minitab OutputIn the exercise we computed the following:1. test statistic: ts= -1.272. DF = 46 (using our conservative method)(a) Table 4 requires that we use DF=403. 0.20 < p-value < 0.40Notes:1. Minitab reports T-Value = -1.28 which is not exactly what we obtained by hand. If you roundthe denominator of the test s tatistic to 2.98, then the value of tsis -1.28. The exact value,however, should be -1.27. No big difference here ... our conclusions would be the same.2. Once again, note that the degrees of freedom is not equal to the conservative value we use inclass. The DF Minitab uses is based on the complicated formula discussed in the book (Equation7.1 on Page 227).Confidence IntervalUnder ‘Options’, you can determine the two-sided corresponding 95% confidence interval by setting the‘Confidence level’ to 95% and by setting ‘Alternative’ to ‘not equal’.Based on our output ab ove, the 95% CI for µ1− µ2is (-9.73, 2.11).Ch. 7 Addendum – Using Minitab: Page