Question

In: Statistics and Probability

Calcium Carbonate Has been assayed from 50 patients. To make the data more “Bell-Shaped” it has...

Calcium Carbonate Has been assayed from 50 patients. To make the data more “Bell-Shaped” it has been log-transformed into the following datafile LnCaCO2 -0.116533816 -0.314710745 -1.771956842 -0.478035801 -0.634878272 -0.342490309 -0.314710745 -0.634878272 -1.514127733 -1.078809661 -1.272965676 -0.867500568 -1.560647748 -2.525728644 -2.525728644 -0.867500568 -2.207274913 -1.386294361 -2.659260037 -1.966112856 -1.660731207 -1.514127733 -1.30933332 -1.203972804 -0.967584026 -1.897119985 -1.171182982 -1.771956842 -1.660731207 -2.302585093 -1.832581464 -0.916290732 -1.897119985 -2.207274913 -1.609437912 -2.659260037 -1.966112856 -1.897119985 -2.120263536 -2.407945609 -1.966112856 -3.506557897 -2.302585093 -1.139434283 -2.120263536 -3.506557897 -3.218875825 -2.120263536 -1.386294361 -1.771956842

The mean of the lnCaCO2 should be -1.25.

1. (20) Test to see if the mean is -1.25. Use two tests, one must be nonparametric.

2. (10) Is the data normally distributed? Calculate:

Expert Solution

1)

sample mean = -1.661

sd = 0.7946

parametric test

one-sample t-test

non-parametric test

data when sorted

-3.506557897   -3.506557897   -3.218875825   -2.659260037   -2.659260037   -2.525728644   -2.525728644   -2.407945609   -2.302585093   -2.302585093
-2.207274913   -2.207274913   -2.120263536   -2.120263536   -2.120263536   -1.966112856   -1.966112856   -1.966112856   -1.897119985   -1.897119985
-1.897119985   -1.832581464   -1.771956842   -1.771956842   -1.771956842   -1.660731207   -1.660731207   -1.609437912   -1.560647748   -1.514127733
-1.514127733   -1.386294361   -1.386294361   -1.30933332   -1.272965676   -1.203972804   -1.171182982   -1.139434283   -1.078809661   -0.967584026
-0.916290732   -0.867500568   -0.867500568   -0.634878272   -0.634878272   -0.478035801   -0.342490309   -0.314710745   -0.314710745   -0.116533816

There are 15 points which are more than -1.25

Sign and binomial test

Number of "successes": 15
Number of trials (or subjects) per experiment: 50

The two-tail P value is 0.0066
This is the chance of observing either 15 or fewer successes, or 35 or more successes, in 50 trials.

since p-value < alpha

we reject the null hypothesis

we conclude that the mean is different from -1.25

orchestra answered 2 years ago

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