AksiomatikMetodologi dan Statistika

Uji Validitas & Reliabilitas

Evaluasi butir instrumen menggunakan korelasi Pearson Product Moment dan uji reliabilitas Cronbach Alpha dengan rincian perhitungan lengkap.

Apa yang dihitung?

Validitas mengukur sejauh mana setiap butir berkorelasi dengan total skor keseluruhan responden. Reliabilitas Cronbach Alpha menilai konsistensi internal keseluruhan instrumen.

Konfigurasi

Input Dataset

Memuat...
Tips cepat
  • Usahakan jumlah responden ≥ 5 × jumlah item.
  • Nilai harus berskala interval atau rasio agar korelasi bermakna.
  • Pastikan tidak ada baris kosong atau jumlah kolom berbeda.

Status data

11 responden • 4 item • Memuat jStat…

Hasil Analisis

Responden: 11
RespondenP1P2P3P4Total
R11419111357
R21113151756
R31812171764
R41016121149
R51616161866
R61211191254
R71312181255
R81914141562
R91314111452
R101912111961
R111913161058

Validitas Pearson

Item valid: 0/4
n = 11
Periksa item dengan p ≥ 0.05
Itemr hitungtpKeputusan
P10.77393.6661-Tidak valid
P2-0.0467-0.1402-Tidak valid
P30.21010.6447-Tidak valid
P40.67282.7280-Tidak valid

Langkah Perhitungan

P1

n = 11
ΣX = 164 • ΣY = 634
ΣXY = 9590 • ΣX² = 2562 • ΣY² = 36812
r=nXYXY(nX2(X)2)(nY2(Y)2)\begin{alignedat}{2} r &= \frac{n\sum XY - \sum X \sum Y}{\sqrt{\left(n\sum X^2 - (\sum X)^2\right)\left(n\sum Y^2 - (\sum Y)^2\right)}} \end{alignedat}
r=11×9590164×634(11×25621642)(11×368126342)=0.7739\begin{alignedat}{2} r &= \frac{11\times 9590 - 164\times 634}{\sqrt{\left(11\times 2562 - 164^2\right)\left(11\times 36812 - 634^2\right)}} \\ &= 0.7739 \end{alignedat}
t=rn21r2\begin{alignedat}{2} t &= \frac{r\sqrt{n-2}}{\sqrt{1-r^2}} \end{alignedat}
t=0.773911210.77392=3.6661\begin{alignedat}{2} t &= \frac{0.7739\sqrt{11-2}}{\sqrt{1-0.7739^2}} \\ &= 3.6661 \end{alignedat}

Nilai p (two-tailed) = - (menunggu jStat)

P2

n = 11
ΣX = 152 • ΣY = 634
ΣXY = 8755 • ΣX² = 2156 • ΣY² = 36812
r=nXYXY(nX2(X)2)(nY2(Y)2)\begin{alignedat}{2} r &= \frac{n\sum XY - \sum X \sum Y}{\sqrt{\left(n\sum X^2 - (\sum X)^2\right)\left(n\sum Y^2 - (\sum Y)^2\right)}} \end{alignedat}
r=11×8755152×634(11×21561522)(11×368126342)=0.0467\begin{alignedat}{2} r &= \frac{11\times 8755 - 152\times 634}{\sqrt{\left(11\times 2156 - 152^2\right)\left(11\times 36812 - 634^2\right)}} \\ &= -0.0467 \end{alignedat}
t=rn21r2\begin{alignedat}{2} t &= \frac{r\sqrt{n-2}}{\sqrt{1-r^2}} \end{alignedat}
t=0.046711210.04672=0.1402\begin{alignedat}{2} t &= \frac{-0.0467\sqrt{11-2}}{\sqrt{1--0.0467^2}} \\ &= -0.1402 \end{alignedat}

Nilai p (two-tailed) = - (menunggu jStat)

P3

n = 11
ΣX = 160 • ΣY = 634
ΣXY = 9254 • ΣX² = 2414 • ΣY² = 36812
r=nXYXY(nX2(X)2)(nY2(Y)2)\begin{alignedat}{2} r &= \frac{n\sum XY - \sum X \sum Y}{\sqrt{\left(n\sum X^2 - (\sum X)^2\right)\left(n\sum Y^2 - (\sum Y)^2\right)}} \end{alignedat}
r=11×9254160×634(11×24141602)(11×368126342)=0.2101\begin{alignedat}{2} r &= \frac{11\times 9254 - 160\times 634}{\sqrt{\left(11\times 2414 - 160^2\right)\left(11\times 36812 - 634^2\right)}} \\ &= 0.2101 \end{alignedat}
t=rn21r2\begin{alignedat}{2} t &= \frac{r\sqrt{n-2}}{\sqrt{1-r^2}} \end{alignedat}
t=0.210111210.21012=0.6447\begin{alignedat}{2} t &= \frac{0.2101\sqrt{11-2}}{\sqrt{1-0.2101^2}} \\ &= 0.6447 \end{alignedat}

Nilai p (two-tailed) = - (menunggu jStat)

P4

n = 11
ΣX = 158 • ΣY = 634
ΣXY = 9213 • ΣX² = 2362 • ΣY² = 36812
r=nXYXY(nX2(X)2)(nY2(Y)2)\begin{alignedat}{2} r &= \frac{n\sum XY - \sum X \sum Y}{\sqrt{\left(n\sum X^2 - (\sum X)^2\right)\left(n\sum Y^2 - (\sum Y)^2\right)}} \end{alignedat}
r=11×9213158×634(11×23621582)(11×368126342)=0.6728\begin{alignedat}{2} r &= \frac{11\times 9213 - 158\times 634}{\sqrt{\left(11\times 2362 - 158^2\right)\left(11\times 36812 - 634^2\right)}} \\ &= 0.6728 \end{alignedat}
t=rn21r2\begin{alignedat}{2} t &= \frac{r\sqrt{n-2}}{\sqrt{1-r^2}} \end{alignedat}
t=0.672811210.67282=2.7280\begin{alignedat}{2} t &= \frac{0.6728\sqrt{11-2}}{\sqrt{1-0.6728^2}} \\ &= 2.7280 \end{alignedat}

Nilai p (two-tailed) = - (menunggu jStat)

Histogram distribusi

349.0 - 54.7454.7 - 60.3460.3 - 66.0

Skor vs. Total

Item skorTotal skor