Minitab assignment 6

Minitab assignment 6 - MINITAB Assignment 6 1. a. Copy and...

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Unformatted text preview: MINITAB Assignment 6 1. a. Copy and Paste all Responses in the Sessions windows as you opened your MINITAB 6 Data. Be sure to include the date. 11/21/2009 10:03:20 PM Welcome to Minitab, press F1 for help. 4/28/2010 9:45:59 AM Welcome to Minitab, press F1 for help. Retrieving project from file: 'C:\USERS\EVAN\APPDATA\LOCAL\MICROSOFT\WINDOWS\TEMPORARY INTERNET FILES\CONTENT.IE5\YW3DGE2L\MINITAB_6_DATA[1].MPJ' b. Make a probability plot and a boxplot of each data set and see if there are any major deviations from Normality. Is it reasonable to use the t procedures? Why or why not? (Copy and paste the graphs that result with your answer. Make sure you have meaningful titles for your graph.) In both the probability plots and boxplots of the active and passive methods number of correct identifications of blissymbols, there are no major deviations from Normality. Therefore, it is reasonable to use the t procedures, because the data is approximately normal. Probability Plots of Active and Passive Methods MTB > pplot c1. Probability Plot of Active MTB > pplot c2. Probability Plot of Passive Boxplots of Active and Passive Methods MTB > boxplot c1; SUBC> transpose. Boxplot of Active MTB > boxplot c2; SUBC> transpose. Boxplot of Passive c. If your conclusion in part (a) is yes , give a 90% confidence interval for the difference in mean number of Blissymbols identified correctly by children after active and passive lessons. (Copy and paste the MINITAB commands you used and the results found in the Session window after your answer. Make sure you interpretation of the 90% confidence interval is in a complete sentence in the context of the problem. ) MTB > twosample c1 c2; SUBC> conf 90. Two-Sample T-Test and CI: Active, Passive Two-sample T for Active vs Passive N Mean StDev SE Mean Active 22 23.05 4.25 0.91 Passive 22 17.14 3.30 0.70 Difference = mu (Active) - mu (Passive) Estimate for difference: 5.91 90% CI for difference: (3.98, 7.84) T-Test of difference = 0 (vs not =): T-Value = 5.15 P-Value = 0.000 DF = 39 The 90% Confidence Interval for the mean difference of number of Blissymbols correctly identified by children after active and passive lessons is (3.98, 7.84). In the context of this problem, this means that in 90% of all samples of the mean difference between the two groups, the population mean difference of correct identifications will be between 3.98 and 7.84. d. If your conclusion in part (a) is yes , give a 99% confidence interval for the difference in mean number of Blissymbols identified correctly by children after active and passive lessons. (Copy and paste the MINITAB commands you used and the results found in the Session window after your answer. Make sure your interpretation of the 99% confidence interval is in a complete sentence in the context of the problem....
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Minitab assignment 6 - MINITAB Assignment 6 1. a. Copy and...

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