Imibuzo Yezibonelo kanye Nengxoxo Yokuhlobana Kwesikhathi Somkhiqizo
Ukuhlanganiswa Kwesikhathi Somkhiqizo, okwaziwa nangokuthi i-Pearson Correlation, kuyindlela yezibalo esetshenziselwa ukukala amandla kanye nesiqondiso sobudlelwano obuqondile phakathi kweziguquguquko ezimbili. Le ndlela iwusizo emikhakheni ehlukahlukene, kusukela ocwaningweni lwezemfundo kanye nokuhlaziywa kwebhizinisi kuya ekuhlolweni kokuhlolwa kwesayensi yemvelo. Lesi sihloko sizoxoxa ngezinkinga eziningana zezibonelo kanye nezixazululo zazo zokubala Ukuhlanganiswa Kwesikhathi Somkhiqizo.
I-Pendahuluan
Ngaphambi kokuthi singene emibuzweni eyisibonelo, kungumqondo omuhle ukuqonda umqondo oyisisekelo woHlelo Lokuhlobana Kwesikhathi Somkhiqizo. Ifomula ejwayelekile esetshenziswa ukubala i-Pearson correlation coefficient (\(r\)) ithi:
\[ r = \frac{n(\sum{XY}) – (\sum{X})(\sum{Y})}{\sqrt{[n\sum{X^2} – (\sum{X})^2][n\sum{Y^2} – (\sum{Y})^2]}} \]
Kuphi:
– \( n \) inani lama-data pairs.
– \( \sum{XY} \) yisamba semikhiqizo ka-\( X \) kanye no-\( Y \).
– \( \sum{X} \) yisamba seziguquguquko \( X \).
– \( \sum{Y} \) yisamba seziguquguquko \( Y \).
– \( \sum{X^2} \) yisamba sezikwele ze-variable \( X \).
– \( \sum{Y^2} \) yisamba sezikwele ze-variable \( Y \).
I-coefficient yokuxhumana kukaPearson (\( r \)) ihlala iphakathi kuka--1 no-1. I-correlation enhle ikhombisa ukuthi zombili iziguquguquko zihamba ngendlela efanayo, kuyilapho i-correlation engemihle ikhombisa ukuthi njengoba i-variable eyodwa ikhula, enye iyancipha. Uma \( r = 0 \), khona-ke akukho ukuhlangana okuqondile phakathi kweziguquguquko ezimbili.
Isibonelo Umbuzo 1
Idatha
Okulandelayo yidatha yamaphuzu okuhlolwa kwezibalo nefiziksi yabafundi aba-5:
| Umfundi | Izibalo (X) | Ifiziksi (Y) |
|——-|——————-|———-|
| 1 | 85 | 90 |
| 2 | 78 | 85 |
| 3 | 85 | 80 |
| 4 | 70 | 70 |
| 5 | 80 | 88 |
Izinyathelo Zokuxazulula
1. Ukubala Izingxenye Ezibalulekile:
– \( \sum{X} \) = 85 + 78 + 85 + 70 + 80 = 398
– \( \sum{Y} \) = 90 + 85 + 80 + 70 + 88 = 413
– \( \sum{XY} \) = (85\ 90) + (78\ 85) + (85\ 80) + (70\ 70) + (80\ 88) = 7650 + 6630 + 6800 + 4900 + 7040 = 33020
– \( \sum{X^2} \) = (85^2) + (78^2) + (85^2) + (70^2) + (80^2) = 7225 + 6084 + 7225 + 4900 + 6400 = 31834
– \( \sum{Y^2} \) = (90^2) + (85^2) + (80^2) + (70^2) + (88^2) = 8100 + 7225 + 6400 + 4900 + 7744 = 34369
2. Faka ifomula:
\[ r = \frac{n(\sum{XY}) – (\sum{X})(\sum{Y})}{\sqrt{[n\sum{X^2} – (\sum{X})^2][n\sum{Y^2} – (\sum{Y})^2]}} \]
\[ r = \frac{5(33020) – (398)(413)}{\sqrt{[5(31834) – (398)^2][5(34369) – (413)^2]}} \]
3. Ukubala Imiphumela:
– Inombolo: \( 5(33020) – (398)(413) = 165100 – 164474 = 626 \)
– I-Denominator:
– \( n\sum{X^2} – (\sum{X})^2 = 5(31834) – (398)^2 = 159170 – 158404 = 766 \)
– \( n\sum{Y^2} – (\sum{Y})^2 = 5(34369) – (413)^2 = 171845 – 170569 = 1276 \)
– \( \sqrt{766 \times 1276} \cishe \sqrt{976856} \cishe 989.36 \)
\[r = \frac{626}{989.36} \cishe kube ngu-0.633 \]
Ngakho-ke, i-Pearson correlation coefficient phakathi kwamaphuzu okuhlolwa kwezibalo kanye nefiziksi ingu-0.633, okubonisa ukuthi kukhona ubudlelwano obuhle obuphakathi kwalezi ziguquguquko ezimbili.
Isibonelo Umbuzo 2
Idatha
Okulandelayo idatha ngenani lokuthengisa kanye nezindleko zokukhangisa kusukela ezinyangeni eziyi-6 enkampanini:
| Inyanga | Ukukhangisa (X) | Ukuthengisa (Y) |
|——-|————–|———————|
| 1 | 2000 | 2500 |
| 2 | 1800 | 2100 |
| 3 | 2200 | 2700 |
| 4 | 2400 | 2900 |
| 5 | 2300 | 3000 |
| 6 | 2500 | 3200 |
Izinyathelo Zokuxazulula
1. Ukubala Izingxenye Ezibalulekile:
– \( \sum{X} \) = 2000 + 1800 + 2200 + 2400 + 2300 + 2500 = 13200
– \( \sum{Y} \) = 2500 + 2100 + 2700 + 2900 + 3000 + 3200 = 16400
– \( \sum{XY} \) = (2000\ 2500) + (1800\ 2100) + (2200\ 2700) + (2400\ 2900) + (2300\ 3000) + (2500\ 3200) = 5000000 + 3780000 + 5940000 + 6960000 + 6900000 + 8000000 = 36580000
– \( \sum{X^2} \) = (2000^2) + (1800^2) + (2200^2) + (2400^2) + (2300^2) + (2500^2) = 4000000 + 3240000 + 4840000 + 5760000 + 5290000 + 6250000 = 29380000
– \( \sum{Y^2} \) = (2500^2) + (2100^2) + (2700^2) + (2900^2) + (3000^2) + (3200^2) = 6250000 + 4410000 + 7290000 + 8410000 + 9000000 + 10240000 = 45590000
2. Faka ifomula:
\[ r = \frac{n(\sum{XY}) – (\sum{X})(\sum{Y})}{\sqrt{[n\sum{X^2} – (\sum{X})^2][n\sum{Y^2} – (\sum{Y})^2]}} \]
\[ r = \frac{6(36580000) – (13200)(16400)}{\sqrt{[6(29380000) – (13200)^2][6(45590000) – (16400)^2]}} \]
3. Ukubala Imiphumela:
– Inombolo: \( 6(36580000) – (13200)(16400) = 219480000 – 216480000 = 3000000 \)
– I-Denominator:
– \( n\sum{X^2} – (\sum{X})^2 = 6(29380000) – (13200)^2 = 176280000 – 174240000 = 2040000 \)
– \( n\sum{Y^2} – (\sum{Y})^2 = 6(45590000) – (16400)^2 = 273540000 – 268960000 = 4580000 \)
– \( \sqrt{2040000 \times 4580000} \cishe \sqrt{9343200000000} \cishe 3056246.20 \)
\[r = \frac{3000000}{3056246.20} \cishe kube ngu-0.981 \]
Ngakho-ke, i-coefficient yokuxhumana kwe-Pearson phakathi kwezindleko zokukhangisa kanye nenani lokuthengisa ingu-0.981, okubonisa ukuthi kukhona ubudlelwano obuhle kakhulu phakathi kwalezi ziguquguquko ezimbili.
Isiphetho
I-Pearson correlation coefficient (\(r\)) iyithuluzi eliwusizo kakhulu lokuqonda ubudlelwano obuqondile phakathi kweziguquguquko ezimbili. Ezibonelweni ezinikeziwe, sibona indlela yokubala inani le-\(r\) bese silihumusha. Ubudlelwano obuphezulu (obuseduze no-1 noma -1) bubonisa ubudlelwano obuqinile, kanti ubudlelwano obuphansi (obuseduze no-0) bubonisa ubudlelwano obubuthakathaka. Kubalulekile ukuqaphela ukuthi ubudlelwano abusho imbangela; bumane bubonisa ukuthi kukhona ubudlelwano phakathi kweziguquguquko ezimbili.
