In: Statistics and Probability
Calcium Carbonate has been assayed from 50 patients. The mean of the lnCaCO2 should be -1.25.
1) Test to see if the mean is -1.25. Use two tests, one must be nonparametric (Wilcoxon)
2) Is the data normally distributed? Use Shapiro-Wilk
| LnCaCO2 | 
| -0.11653 | 
| -0.31471 | 
| -1.77196 | 
| -0.47804 | 
| -0.63488 | 
| -0.34249 | 
| -0.31471 | 
| -0.63488 | 
| -1.51413 | 
| -1.07881 | 
| -1.27297 | 
| -0.8675 | 
| -1.56065 | 
| -2.52573 | 
| -2.52573 | 
| -0.8675 | 
| -2.20727 | 
| -1.38629 | 
| -2.65926 | 
| -1.96611 | 
| -1.66073 | 
| -1.51413 | 
| -1.30933 | 
| -1.20397 | 
| -0.96758 | 
| -1.89712 | 
| -1.17118 | 
| -1.77196 | 
| -1.66073 | 
| -2.30259 | 
| -1.83258 | 
| -0.91629 | 
| -1.89712 | 
| -2.20727 | 
| -1.60944 | 
| -2.65926 | 
| -1.96611 | 
| -1.89712 | 
| -2.12026 | 
| -2.40795 | 
| -1.96611 | 
| -3.50656 | 
| -2.30259 | 
| -1.13943 | 
| -2.12026 | 
| -3.50656 | 
| -3.21888 | 
| -2.12026 | 
| -1.38629 | 
| -1.77196 | 
1)
a) one-sample t-test
stat-> basic tatistics -> 1-sample t

since p-value = 0.001 < alpha
we reject the null hypothesis
we conclude that mean is different from -1.25
ii)
stat -> non-parametrric -> 1-sample wilcoxon

again p-value = 0.001 < alpha
we reject the null hypothesis
2)
stat -> basic stat - > normality

since p-value = 0.837 > alpha
we fail to reject the null
normality asssumption is met
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