question archive Suppose a sample of 0-rings was obtained and the wall thickness (in inches) of each was recorded
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Suppose a sample of 0-rings was obtained and the wall thickness (in inches) of each was recorded. Use a normal probably plot to asses whether the sample data could have come from a population that is normally distributed. Using the correlation coefficient of the normal probability plot, is it reasonable to conclude that the population is normally distributed? Select the correct choice below and fill in the answer boxes within your choice. (Round to 3 decimal places as needed) A) yes. The correlation between the expected z-scores and the observation data, ?, exceeds the critical value, ?. Therefore, it is reasonable to conclude that the data come from a normal population.
Yes. The correlation between the expected z-scores and the observation data, 0.994, exceeds the critical value, 0.941. Therefore, it is reasonable to conclude that the data come from a normal population.
I hope this helps you.
I have provided the excel formulas below.
Step-by-step explanation
Excel Procedure
A B C D E F G H K M N Rank , i Proportion Z score Data 0.0588 -1.5647 0. 165 Sample Size, n | Critical Value 0.1176 -1.1868 0. 191 5 0.880 0.1765 -0.9289 0.194 0.2353 -0.7215 0.219 6 0.888 0.2941 -0.5414 0.216 Correlation between z score and observed data = 0.994 7 0.898 0.3529 -0.3774 0.236 critical correlation value for n=16 is 0.941 8 0.906 0.4118 -0.2230 0.233 0.4706 -0.0738 0.240 9 0.912 9 0.5294 0.0738 0.248 10 0.918 10 0.5882 0.2230 0.261 11 11 0.923 0.6471 0.3774 0.272 12 0.7059 0.5414 0.283 12 0.928 14 13 0.7647 0.7215 0.282 13 0.932 15 14 0.8235 0.9289 0.306 14 16 15 0.8824 0.935 1.1868 0.310 17 16 0.9412 1.5647 0.334 15 0.939 18 16 0.941 19
Formula on Excel
A B C D E F G H Rank , i Proportion Z score Data =A2/(16+1) -1.5647 0.165 =A3/(16+1) -1.1868 0. 191 =A4/(16+1) -0.9289 0. 194 =A5/(16+1) -0.7215 0.219 =A6/(16+1) -0.5414 0.216 Correlation between z score and observed data = =CORREL(C2:C17,D2:D17) =A7/(16+1) -0.3774 0.236 critical correlation value for n=16 is 0.941 =A8/(16+1) -0.223 0.233 =A9/(16+1) -0.0738 0.24 =A10/(16+1) 0.0738 0.248 =A11/(16+1) 0.223 0.261 =A12/(16+1) 0.3774 0.272 =A13/(16+1) 0.5414 0.283 14 =A14/(16+1) 0.7215 0.282 15 =A15/(16+1) 0.9289 0.306 16 =A16/(16+1) 1.1868 0.31 17 16 =A17/(16+1) 1.5647 0.334 18
Please see the attached file