Sebaran normal Nilai Mean Margin of Error dan Uji-

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Susila IN, Kartasasmita B, Ruwuh, Terjemahan; Jakarta: Erlangga. Terjemahan dari: Calculus With Analytic Geometry, 2 nd Edition. Pritsker AA, O’Reilly JJ. 1999. Simulation with Visual Slam and Awesim. John Wiley Sons. USA Ross SM. 2000. Stochastic Process. New York: Macmillan Publishing Company. Triola MF. 2006. Elementary Statistics. Tenth Edition. Pearson Education, Inc. Walpole RE. 1993. Pengantar Statistika. Edisi ke-3. PT. Gramedia Pustaka Utama. Jakarta. L A M P I R A N Lampiran 1. Program Simulasi kategorimaks=15;q=1000; 1=50;2=50;1=15;2=15;=0.05; talphaperdua=2.101; Membangkitkan bilangan acak berdistribusi normal dengan nilai tengah  dan simpangan baku . acak[mu_,sigma_,n_,a_,b_]:=Module[{lis,nd},lis={}; While[Length[lis]n,nd=RandomVariate[NormalDistribution[mu,sigma]]; If[a ndb,lis=Append[lis,nd]]];lis] Kasus 1. ukuran contoh 10 , menyebar normal. Membangkitkan populasi kelompok 1 sebanyak 1000 contoh bernilai 0 sampai 100 berdistribusi normal dengan nilai tengah 50 dan simpangan baku 15. pt1=q;pt2=q;n1=10;n2=n1;For[i=1,i pt1,i++,contoh1[i]=acak[1,1,n1,0,100]] For[i=1,i q,i++,{contoh[1,i]=contoh1[i];xmean[1,i]=Mean[contoh[1,i]]; ragam[1,i] = Variance[contoh[1,i]]}] rerata1=Mean[Flatten[Array[xmean,{1,q},{1,1}]]]; sb1=Sqrt[Mean[Flatten[Array[ragam,{1,q},{1,1}]]]]; sbnol1=Count[Flatten[Sqrt[Array[ragam,{1,q},{1,1}]],1],0]; Membangkitkan populasi kelompok 2 sebanyak 1000 contoh bernilai 0 sampai 100 berdistribusi normal dengan nilai tengah 50 dan simpangan baku 15. For[i=1,i pt2,i++,contoh2[i]=acak[2,2,n2,0,100]] For[i=1,i q,i++,{contoh[2,i]=contoh2[i];xmean[2,i]=Mean[contoh[2,i]]; ragam[2,i]=Variance[contoh[2,i]]}]; rerata2=N[Mean[Flatten[Array[xmean,{1,q},{2,1}]]]]; sb2=N[Sqrt[Mean[Flatten[Array[ragam,{1,q},{2,1}]]]]]; sbnol2=Count[Flatten[Sqrt[Array[ragam,{1,q},{2,1}]],1],0]; Peluang Nilai P Value data awal  Taraf Nyata For[i=1,i q,i++,PValue[i]=TTest[{contoh[1,i],contoh[2,i]},1-2, PValue,SignificanceLevel 0.05,VerifyTestAssumptionsNone, AlternativeHypothesis Unequal]] reratapv=Mean[Array[PValue,q]]; For[i=1,i q,i++,PValue[i]=TTest[{contoh[1,i],contoh[2,i]},1-2, ShortTestConclusion,SignificanceLevel 0.05,VerifyTestAssumptionsNone, AlternativeHypothesis Unequal]] tolakk=Count[Array[PValue,q],Reject]; terimak=Count[Array[PValue,q],Do not reject]; Melakukan uji nilai tengah mencari true value For[i = 1, i = q, i++, {spkontinu[i] = Sqrt[n1 - 1ragam[1, i] + n2 - 1ragam[2, i]n1 + n2 -2]; deltakontinu[i] = N[talphaperduaspkontinu[i] Sqrt[1n1 + 1n2]]; thkontinu[i] = N[xmean[1, i] - xmean[2, i] - Subscript[ , 1] - Subscript[ , 2]spkontinu[i]Sqrt[1n1 + 1n2]]; deltarerata[i] = Abs[xmean[1, i] - xmean[2, i]]; batasatas[i] = deltarerata[i] + deltakontinu[i]; batasbawah[i] = deltarerata[i] - deltakontinu[i]; keputusankontinuTrue[i] = If[batasbawah[i] Subscript[ , 1] - Subscript[, 2] batasatas[i], True, False]; keputusankontinuFalse[i] = If[batasbawah[i] Subscript[ , 1] - Subscript[, 2] || Subscript[ , 1] - Subscript[, 2] batasatas[i], False, True]}] reratagalat=Mean[Array[deltakontinu,q]]; dalamkontinu=Count[Array[keputusankontinuTrue,q],True]; luarkontinu=Count[Array[keputusankontinuFalse,q],False]; reratathkontinu=Mean[Array[thkontinu,q]]; reratabatasatas=Mean[Array[batasatas,q]]; reratabatasbawah=Mean[Array[batasbawah,q]]; Masing-masing 1000 pasangan contoh data kontinu dikonversi ke data kategori berukuran 2 sampai 15. Kemudian masing-masing pasangan data kategori dilakukan uji nilai tengah. For[j=1,j 2,j++,For[l=1,lq,l++,For[kategorimin=2,kategoriminkategorimaks,kategorimin++,For [k=1,k kategorimin,k++,bbkelas[j,l,kategorimin,1]=0]]]] For[j = 1, j = 2, j++, For[l = 1, l = q, l++, For[kategorimin = 2, kategorimin = kategorimaks, kategorimin++, For[k = 1, k = kategorimin, k++, selisih[j, l, kategorimin, k] = N[100kategorimin, 4]]]]]; For[j=1,j 2,j++,For[l=1,lq,l++,For[kategorimin=2,kategoriminkategorimaks,kategorimin++,For [k=1,k kategorimin,k++,{bakelas[j,l,kategorimin,k]=bbkelas[j,l,kategorimin,k]+selisih[j,l,kategori min,k]+0.0005,bbkelas[j,l,kategorimin,k+1]=bakelas[j,l,kategorimin,k]}]]]]; For[j=1,j 2,j++,For[l=1,lq,l++,For[kategorimin=2,kategoriminkategorimaks,kategorimin++,For [k=1,k kategorimin,k++,fkelas[j,l,kategorimin,k]=Length[Select[contoh[j,l],bbkelas[j,l,kategorimi n,k]  bakelas[j,l,kategorimin,k]]]]]]] For[j = 1, j = 2, j++, For[l = 1, l = q, l++, For[kategorimin = 2, kategorimin = kategorimaks, kategorimin++, jumlahfkelas[j, l, kategorimin] = Sum[fkelas[j, l, kategorimin, k], {k, 1, kategorimin}]]]] For[j=1,j 2,j++,For[l=1,lq,l++,For[kategorimin=2,kategoriminkategorimaks,kategorimin++,For [k=1,k kategorimin,k++,xtengah[j,l,kategorimin,k]=12 bbkelas[j,l,kategorimin,k]+bakelas[j,l,kategorimin,k]]]]] For[j = 1, j = 2, j++, For[l = 1, l = q, l++, For[kategorimin = 2, kategorimin = kategorimaks, kategorimin++, For[k = 1, k = kategorimin, k++,xmean[j, l, kategorimin] = N[Sum[xtengah[j, l, kategorimin, p]fkelas[j, l, kategorimin, p], {p, 1, kategorimin}]Sum[fkelas[j, l, kategorimin, p], {p, 1, kategorimin}]]]]]] For[j = 1, j = 2, j++, For[l = 1, l = q, l++, For[kategorimin = 2, kategorimin = kategorimaks, kategorimin++, For[k = 1, k = kategorimin, k++,{ragam[j, l, kategorimin] = N[1Sum[fkelas[j, l, kategorimin, p], {p, 1, kategorimin}] - 1 Sum[fkelas[j, l, kategorimin, p]xtengah[j, l, kategorimin, p] - xmean[j, l, kategorimin]2, {p, 1, kategorimin}]]}]]]] For[j=1,j 2,j++,For[l=1,lq,l++,For[kategorimin=2,kategoriminkategorimaks,