$\mathrm{restart}\:$$\mathrm{with}\left(\mathrm{Statistics}\right)\:$$\mathrm{infolevel}\left[\mathrm{Statistics}\right]≔1\:$

$X:=\mathrm{Array}\left(\left[9\,10\,8\,4\,8\,3\,0\,10\,15\,9\right]\right)\:$$Y≔\mathrm{Array}\left(\left[6\,3\,10\,11\,9\,8\,13\,4\,4\,4\right]\right)\:$$Z≔\mathrm{Array}\left(\left[10\,11\,7\,3\,10\,5\,2\,12\,14\,10\right]\right)\:$

$\mathrm{XProp}:=\mathrm{table}\left(\left[\'\mathrm{\μ}\'\=\mathrm{Mean}\left(X\right)\,\'\mathrm{\σ}\'\=\mathrm{StandardDeviation}\left(X\right)\right]\right)$

$\mathrm{XProp}:=\mathrm{table}\left(\left[\mathrm{\μ}\=7.600000000\,\mathrm{\σ}\=4.247875286\right]\right)$

$\mathrm{YProp}:=\mathrm{table}\left(\left[\'\mathrm{\μ}\'\=\mathrm{Mean}\left(Y\right)\,\'\mathrm{\σ}\'\=\mathrm{StandardDeviation}\left(Y\right)\right]\right)$

$\mathrm{YProp}:=\mathrm{table}\left(\left[\mathrm{\μ}\=7.200000000\,\mathrm{\σ}\=3.489667288\right]\right)$

$\mathrm{ZProp}:=\mathrm{table}\left(\left[\'\mathrm{\μ}\'\=\mathrm{Mean}\left(Z\right)\,\'\mathrm{\σ}\'\=\mathrm{StandardDeviation}\left(Z\right)\right]\right)$

$\mathrm{ZProp}:=\mathrm{table}\left(\left[\mathrm{\μ}\=8.400000000\,\mathrm{\σ}\=3.977715704\right]\right)$

$\mathrm{TwoSampleZTest}\left(X\,Y\,3\,4\,3\right)\:$

Standard Z-Test on Two Samples
------------------------------
Null Hypothesis:
Sample drawn from populations with difference of means equal to 3

Alt. Hypothesis:
Sample drawn from population with difference of means not equal to 3

Sample sizes:            10, 10
Sample means:            7.6, 7.2
Difference in means:     0.4
Distribution:            Normal(0,1)
Computed statistic:      -1.64438
Computed pvalue:         0.100097
Confidence interval:     -2.698975162 .. 3.498975162
                         (difference of population means)

Result: [Accepted]
There is no statistical evidence against the null hypothesis

$\mathrm{TwoSampleTTest}\left(X\,Z\,1\right)\:$

Standard T-Test on Two Samples (Unequal Variances)
--------------------------------------------------
Null Hypothesis:
Sample drawn from populations with difference of means equal to 1
Alt. Hypothesis:
Sample drawn from population with difference of means not equal to 1

Sample sizes:            10, 10
Sample means:            7.6, 8.4
Sample standard devs.:   4.24788, 3.97772
Difference in means:     -0.8
Distribution:            StudentT(17.92283210)
Computed statistic:      -0.978107
Computed pvalue:         0.34104
Confidence interval:     -4.667499017 .. 3.067499017
                         (difference of population means)

Result: [Accepted]
There is no statistical evidence against the null hypothesis

$\mathrm{TwoSamplePairedTTest}\left(X\,Z\,1\right)\:$

Standard T-Test with Paired Samples
-----------------------------------
Null Hypothesis:
Sample drawn from populations with difference of means equal to 1
Alt. Hypothesis:
Sample drawn from population with difference of means not equal to 1

Sample size:             10
Difference in means:     -0.8
Difference std. dev.:    1.31656
Distribution:            StudentT(9)
Computed statistic:      -4.32346

Computed pvalue:         0.00192341
Confidence interval:     -1.741810891 .. .1418108907
                         (difference of population means)

Result: [Rejected]
There exists statistical evidence against the null hypothesis

$\mathrm{restart}\:$$\mathrm{with}\left(\mathrm{Statistics}\right)\:$$\mathrm{infolevel}\left[\mathrm{Statistics}\right]≔1\:$

$S:=\mathrm{Sample}\left(\mathrm{Maxwell}\left(3\right)\,100\right)\:$$T:=\mathrm{Sample}\left(\mathrm{Exponential}\left(2\right)\,100\right)\:$

$\mathrm{S\_sigma}:=\mathrm{evalf}\left(\mathrm{StandardDeviation}\left(\mathrm{Maxwell}\left(3\right)\right)\right)$

$\mathrm{S\_sigma}:=2.020318836$

$\mathrm{T\_sigma}:=\mathrm{evalf}\left(\mathrm{StandardDeviation}\left(\mathrm{Exponential}\left(2\right)\right)\right)$

$\mathrm{T\_sigma}:=2.$

$\mathrm{OneSampleChiSquareTest}\left(S\,2\right)\:$

Chi-Square Test on One Sample
-----------------------------
Null Hypothesis:
Sample drawn from population with standard deviation equal to 2
Alt. Hypothesis:
Sample drawn from population with standard deviation not equal to 2

Sample size:             100
Sample standard dev.:    1.83342

Distribution:            ChiSquare(99)
Computed statistic:      83.1952
Computed pvalue:         0.253798
Confidence interval:     1.609754032 .. 2.129836954
                         (population standard deviation)

Result: [Accepted]
There is no statistical evidence against the null hypothesis

$\mathrm{TwoSampleFTest}\left(S\,T\,2\right)\:$

F-Ratio Test on Two Samples
---------------------------
Null Hypothesis:
Sample drawn from populations with ratio of variances equal to 2
Alt. Hypothesis:
Sample drawn from population with ratio of variances not equal to 2

Sample sizes:            100, 100
Sample variances:        3.36142, 4.08274
Ratio of variances:      0.823326
Distribution:            FRatio(99,99)
Computed statistic:      0.411663
Computed pvalue:         1.45561e-05
Confidence interval:     .5539687377 .. 1.223654982
                         (ratio of population variances)

Result: [Rejected]
There exists statistical evidence against the null hypothesis

$\mathrm{restart}\:$$\mathrm{with}\left(\mathrm{Statistics}\right)\:$$\mathrm{infolevel}\left[\mathrm{Statistics}\right]≔1\:$

$S:=\mathrm{Sample}\left(\mathrm{Normal}\left(5\,2\right)\,20\right)\:$$T:=\mathrm{Sample}\left(\mathrm{Uniform}\left(3\,7\right)\,20\right)\:$

$\mathrm{ShapiroWilkWTest}\left(S\right)\:$

Shapiro and Wilk's W-Test for Normality
---------------------------------------
Null Hypothesis:
Sample drawn from a population that follows a normal distribution
Alt. Hypothesis:
Sample drawn from population that does not follow a normal distribution

Sample size:             20
Computed statistic:      0.972002
Computed pvalue:         0.784909

Result: [Accepted]
There is no statistical evidence against the null hypothesis

$\mathrm{ShapiroWilkWTest}\left(T\right)\:$

Shapiro and Wilk's W-Test for Normality
---------------------------------------
Null Hypothesis:
Sample drawn from a population that follows a normal distribution
Alt. Hypothesis:
Sample drawn from population that does not follow a normal distribution

Sample size:             20
Computed statistic:      0.889151
Computed pvalue:         0.0259262

Result: [Rejected]
There exists statistical evidence against the null hypothesis

$\mathrm{restart}\:$$\mathrm{with}\left(\mathrm{Statistics}\right)\:$$\mathrm{infolevel}\left[\mathrm{Statistics}\right]≔1\:$

$\mathrm{Ob}:=\mathrm{Array}\left(\left[25\,17\,15\,23\,24\,16\right]\right)$

$\mathrm{SalesPerDay}:=\frac{\sum _{i\=1}^6{\'\mathrm{Ob}\'}_i}{6}$

$\mathrm{SalesPerDay}:=20$

$\mathrm{Ex}:=\mathrm{Array}\left(\left[\mathrm{SalesPerDay}\$6\right]\right)$

$\mathrm{ChiSquareGoodnessOfFitTest}\left(\mathrm{Ob}\,\mathrm{Ex}\,\mathrm{level}\=0.05\right)\:$

Chi-Square Test for Goodness-of-Fit
-----------------------------------
Null Hypothesis:
Observed sample does not differ from expected sample
Alt. Hypothesis:
Observed sample differs from expected sample

Categories:              6
Distribution:            ChiSquare(5)
Computed statistic:      5
Computed pvalue:         0.41588
Critical value:          11.07049741

Result: [Accepted]
There is no statistical evidence against the null hypothesis

$\mathrm{SaleTimes}:=\left[1.4\,1.8\,2.2\,2.9\,3.0\,3.4\,3.4\,3.5\,3.6\,3.7\,3.8\,4.0\,4.4\,4.6\,5.3\,7.5\right]\:$

$\mathrm{ChiSquareSuitableModelTest}\left(\mathrm{SaleTimes}\,\mathrm{Uniform}\left(0\,8\right)\right)\:$

Chi-Square Test for Suitable Probability Model

----------------------------------------------
Null Hypothesis:
Sample was drawn from specified probability distribution
Alt. Hypothesis:
Sample was not drawn from specified probability distribution

Bins:                    4
Distribution:            ChiSquare(3)
Computed statistic:      9.5191
Computed pvalue:         0.023129
Critical value:          7.814728288

Result: [Rejected]
There exists statistical evidence against the null hypothesis

$\mathrm{DrugGroup}≔\mathrm{Vector}\left[\mathrm{column}\right]\left(\left[64\,176\right]\right)\:$$$\begin{gathered} \#\mathrm{Recovered,\; Not\; Recovered} \\ \mathrm{PlaceboGroup}≔\mathrm{Vector}\left[\mathrm{column}\right]\left(\left[86\,150\right]\right)\: \#\mathrm{Recovered,\; Not\; Recovered} \end{gathered}$$

$\mathrm{Output} ≔ \mathrm{Matrix}\left(\left[\mathrm{DrugGroup}\, \mathrm{PlaceboGroup}\right]\right)\:$

$\mathrm{ChiSquareIndependenceTest}\left(\mathrm{Output}\right)\:$

Chi-Square Test for Independence
--------------------------------
Null Hypothesis:
Two attributes within a population are independent of one another
Alt. Hypothesis:
Two attributes within a population are not independent of one another
Dimensions:              2
Total Elements:          476

Distribution:            ChiSquare(1)
Computed statistic:      5.26704
Computed pvalue:         0.0217328
Critical value:          3.84145606580278
Result: [Rejected]
There exists statistical evidence against the null hypothesis

$\mathrm{restart}\:$$\mathrm{with}\left(\mathrm{Statistics}\right)\:$

$X:=\mathrm{Array}\left(\left[9\,10\,8\,4\,8\,3\,0\,10\,15\,9\right]\right)\:$

$\mathrm{OneSampleTTest}\left(X\,5\right)$

$\mathrm{hypothesis}\=\mathrm{true}\,\mathrm{confidenceinterval}\=4.561253851..10.63874615\,\mathrm{distribution}\=\mathrm{StudentT}\left(9\right)\,\mathrm{pvalue}\=0.08491508130\,\mathrm{statistic}\=1.935537501$

$\mathrm{OneSampleTTest}\left(X\,5\,\mathrm{output}\=\mathrm{confidenceinterval}\right)$

$4.561253851..10.63874615$