Question

In: Statistics and Probability

Calcium Carbonate has been assayed from 50 patients. The mean of the lnCaCO2 should be -1.25....

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

Expert Solution

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

Please give me a thumbs-up if this helps you out. Thank you! :)

orchestra answered 3 years ago

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