tag:blogger.com,1999:blog-3568555666281177719.post8856480015846339024..comments2020-08-15T04:18:55.835-07:00Comments on Excel Master Series Blog: A Quick Normality Test Easily Done In ExcelAnonymoushttp://www.blogger.com/profile/02423338645515400885noreply@blogger.comBlogger12125tag:blogger.com,1999:blog-3568555666281177719.post-54348106763328224162014-08-10T03:28:47.987-07:002014-08-10T03:28:47.987-07:00A formula for Uniform Order Statistic Medians is w...A formula for Uniform Order Statistic Medians is wrong. m(i) = 0.5 ^ (1/n) for i = n. That is m(6) = m(i) = 0.5 ^ (1/n) for i = n = m(i) = 0.5 ^ (1/6) = 0.8909. It results in a smoother line for your example. Source: http://www.itl.nist.gov/div898/handbook/eda/section3/normprpl.htmAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-3568555666281177719.post-44677369443860604132013-05-25T06:52:19.859-07:002013-05-25T06:52:19.859-07:00In your first method you can quantify what you see...In your first method you can quantify what you see by using the R² function --> =rsq(actual values;z-scores at each sample point if sample is normal distributed). The more closer to 1 the more the actual values are normal distributed.Anonymoushttps://www.blogger.com/profile/00372871320913341086noreply@blogger.comtag:blogger.com,1999:blog-3568555666281177719.post-76584465554879403052012-11-06T09:58:15.600-08:002012-11-06T09:58:15.600-08:00I found the first method of creating the normal pr...I found the first method of creating the normal probability plot helpful. But when I use the data from the second method for the first, the corresponding correlations do not agree. It looks like it is because of an error in the expression for m(n). You use 0.5(1/n) when it should be 0.5 to the 1/n power. When I make this change the correlations are close.Steve Eastmannoreply@blogger.comtag:blogger.com,1999:blog-3568555666281177719.post-38808343571163348802012-10-25T22:41:01.290-07:002012-10-25T22:41:01.290-07:00This was of very timely help. Thanks muchThis was of very timely help. Thanks muchAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-3568555666281177719.post-28364584957616482012-08-15T03:57:48.847-07:002012-08-15T03:57:48.847-07:00This is extremely usefulThis is extremely usefulLongolongohttps://www.blogger.com/profile/03133288874153785255noreply@blogger.comtag:blogger.com,1999:blog-3568555666281177719.post-52700763304615837342012-05-12T02:22:25.198-07:002012-05-12T02:22:25.198-07:00Excel is also very useful as SPSS.Excel is also very useful as SPSS.Sarinahnoreply@blogger.comtag:blogger.com,1999:blog-3568555666281177719.post-83816242627431530902012-03-30T07:25:23.909-07:002012-03-30T07:25:23.909-07:00Just fixed it. Thanks for catching that.Just fixed it. Thanks for catching that.Mark Harmonhttp://excelmasterseries.comnoreply@blogger.comtag:blogger.com,1999:blog-3568555666281177719.post-75937010025844630432012-03-30T01:07:58.262-07:002012-03-30T01:07:58.262-07:00How does m(6) = 0.5(1/6) = 0.8909? By my math, ...How does m(6) = 0.5(1/6) = 0.8909? By my math, 0.5*(1/6) = 0.0833Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3568555666281177719.post-32939356801788485282012-01-04T19:07:50.755-08:002012-01-04T19:07:50.755-08:00Quite helpful.Quite helpful.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3568555666281177719.post-55634991313697814012011-10-27T21:33:55.614-07:002011-10-27T21:33:55.614-07:00quite helpful to young statisticians and can do it...quite helpful to young statisticians and can do it alonestanhttps://www.blogger.com/profile/09637705132550179012noreply@blogger.comtag:blogger.com,1999:blog-3568555666281177719.post-71058034145498637572011-02-24T10:34:24.089-08:002011-02-24T10:34:24.089-08:00The values of 1/14 etc. are the probability interv...The values of 1/14 etc. are the probability intervals. You get them as follows: start with 1/(2*n), so 1/(2*7), then add 2/2n to the previous value. You'll get 1/14, 3/14, 5/14, etc.<br /><br />As the author said, there's a 1/7th distance between each probability value, or 1/n in general.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-3568555666281177719.post-57412970051994234202011-02-16T09:30:31.246-08:002011-02-16T09:30:31.246-08:00Where are you getting 1/14, 3/14, 5/14 ... for CDF...Where are you getting 1/14, 3/14, 5/14 ... for CDF at each sample point in the first Normal Plat example?Anonymousnoreply@blogger.com