Question

Suppose a sample of O-rings was obtained and the wall thickness (in inches) of each was recorded Use a normal probability plot to 0.165 0.191 0.194 0 2195 assess whether the sample data could have come from a population that is normally distributed 0.216 0236 0.233 0.240 0.248 0 261 0 272 0.283 0.282 0.308 0.310 0334 Click here to view the table of critical values. Click here to view page 1 of the standard normal distribution table Click here to view page 2 of the standard normal distribution table 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 three decimal places as needed.) OA. Yes. The correlation between the expected 2-scores and the observed data, exceeds the critical value Therefore, it is not reasonable to conclude that the data come from a normal population OB. No. The correlation between the expected 2-scores and the observed data, does not exceed the critical value, Therefore, it is reasonable to conclude that the data come from a normal population. O C. No. The correlation between the expected z-scores and the observed data, does not exceed the critical value, Therefore, it is not reasonable to conclude that the data come from a normal population OD. Yes. The correlation between the expected z-scores and the observed data, exceeds the critical value, Therefore, it is reasonable to conclude that the data come from a normal population
also if you know how to do it through excel, can you please include steps? thank you !

Answer 1

yes, the correlation between expected z score and observed data, 0.994, exceed the critical value 0.941,
therefore, it is reasonable to conclude that data come from a normal population.

Excel procedure:

E F G H J K M N 1 Correlation between z score and observed data = critical correlation value for n=16 is 0.994 0.941 A 1 Rank , i 2 3 2 4 3 5 4 6 5 7 6 8 7 9 8 10 9 11 10 12 11 13 12 14 13 15 14 16 15 17 16 18 19 20 B с D Proportion Z score Data 0.0588 -1.5647 0.165 0.1176 -1.1868 0.191 0.1765 -0.9289 0.194 0.2353 -0.7215 0.219 0.2941 -0.5414 0.216 0.3529 -0.3774 0.236 0.4118 -0.2230 0.233 0.4706 -0.0738 0.240 0.5294 0.0738 0.248 0.5882 0.2230 0.261 0.6471 0.3774 0.272 0.7059 0.5414 0.283 0.7647 0.7215 0.282 0.8235 0.9289 0.306 0.8824 1.1868 0.310 0.9412 1.5647 0.334 Sample Size, n 5 6 7 8 9 10 11 Critical Value 0.880 0.888 0.898 0.906 0.912 0.918 0.923 0.928 0.932 0.935 0.939 0.941 O 13 14 15 16

Formula Reference:

I A с D E F Correlation between z score and observed data = critical correlation value for n=16 is =CORREL(C2:C17,D2:D17) 0.941 1. Rank, i 2 1 3 2 4 3 5 4 6 5 7 6 8 7 9 8 10 9 11 10 12 11 13 12 14 13 15 14 16 15 17 16 18 B Proportion =A2/(16+1) =A3/(16+1) =A4/(16+1) =A5/(16+1) =A6/(16+1) =A7/(16+1) =A8/(16+1) =A9/(16+1) =A10/(16+1) =A11/(16+1) =A12/(16+1) =A13/(16+1) =A14/(16+1) =A15/(16+1) =A16/(16+1) =A17/(16+1) Z score -1.5647 -1.1868 -0.9289 -0.7215 -0.5414 -0.3774 -0.223 -0.0738 0.0738 0.223 0.3774 0.5414 0.7215 0.9289 1.1868 1.5647 Data 0.165 0.191 0.194 0.219 0.216 0.236 0.233 0.24 0.248 0.261 0.272 0.283 0.282 0.306 0.31 0.334 19

Find a z score for each proportion and calculate the correlation between data and the zscore.