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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