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اﻟﻔﺼﻞ اﻻول
1- اﻟﻤﻘﺪﻣﺔ
ان اﻻھﻤﯿﺔ اﻟﻜﺒﯿﺮة ﻟﺘﻘﺪﯾﺮ ﻗﯿﻤﺔ 05 LDﻓﻲ اﻟﻌﻠﻮم اﻟﺒﺎﯾﻮﻟﻮﺟﯿﺔ اﻧﻌﻜﺲ ﻋﻠﻰ اﯾﺠﺎد
اﻟﻌﺪﯾ ﺪ ﻣ ﻦ اﻟﺒ ﺮاﻣﺞ اﻻﺣ ﺼﺎﺋﯿﺔ اﻟﺨﺎﺻ ﺔ ﺑﺘﻘ ﺪﯾﺮھﺎ ﻣﺜ ﻞ ﺑﺮﻧ ﺎﻣﺞ
Graphpad Prismو AOT 425 StatPgmو TOPKAT package
و BMDPﻛﻤﺎ ﺗﻮﻓﺮت ﺑﻌﺾ ﻃﺮق ﺗﻘﺪﯾﺮھﺎ ﻓﻲ اھﻢ اﻟﺒﺮاﻣﺢ اﻻﺣﺼﺎﺋﯿﺔ اﻟﻌﺎﻣﺔ
واﻛﺜﺮھ ﺎ رواﺟ ﺎ ﻣﺜ ﻞ ، STATISTICA ، Minitab، SPSS ، SAS
.STATGRAPHاﻻ ان ﺗﻌ ﺪد ﻃ ﺮق ﺗﻘ ﺪﯾﺮ ﻗﯿﻤ ﺔ 05 LDادى اﻟ ﻰ وﺟ ﻮد
اﺧﺘﻼﻓ ﺎت ﺑ ﯿﻦ ﺗﻠ ﻚ اﻟﺒ ﺮاﻣﺞ ﻓ ﻲ ﻋ ﺪد ﻃ ﺮق اﻟﺘﻘ ﺪﯾﺮاﻟﺘﻲ ﯾﻤﻜ ﻦ ان ﺗﺘ ﻮﻓﺮ ﻓﯿﮭ ﺎ ،
اذ ﯾﻤﻜ ﻦ اﺳ ﺘﻌﻤﺎل ﺟﻤﯿ ﻊ اﻟﺒ ﺮاﻣﺞ اﻟﻤ ﺬﻛﻮرة ﻟﺘﻘ ﺪﯾﺮ ﻗﯿﻤ ﺔ 05 LDﺑﺎﺳ ﺘﻌﻤﺎل
اﻻﻧﺤﺪار اﻟﺒﺴﯿﻂ او ﻃﺮﯾﻘﺔ Probitﻓﻀﻼ ﻋﻦ رﺳﻢ اﻟﻌﻼﻗ ﺔ ﺑﯿﻨﮭﻤ ﺎ ﻓﯿﻤ ﺎ ﻧﺠ ﺪ ان
اﻟﺒ ﺮاﻣﺞ اﻟﺜﻼﺛ ﺔ ) SASو SPSSو ( Minitabﺗﺘ ﻮﻓﺮ ﻓﯿﮭ ﺎ اﻛﺜ ﺮ ﻣ ﻦ ﻃﺮﯾﻘ ﺔ
ﻟﻠﺤﻞ وﺑﺬﻟﻚ ﻓﮭﻲ ﺗﺘﻔﻮق ﻋﻠﻰ ﺑﻘﯿﺔ اﻟﺒﺮاﻣﺞ اﻟﻌﺎﻣﺔ.
ان اﻟﻜ ﻼم ﻋ ﻦ اھﻤﯿ ﺔ ﺗﻘ ﺪﯾﺮ ﻗﯿﻤ ﺔ 05 LDواﺳ ﺘﻌﻤﺎﻟﮭﺎ ﻓ ﻲ ﺗﻘ ﺪﯾﺮ دﻟﯿ ﻞ
اﻟﻌ ﻼج TI
)Index
(Therapeuticوﻋﺎﻣ ﻞ اﻻﻣ ﺎن اﻟﻤﺤ ﺪد CSF
) (Certain Safety Factorﻻﺑ ﺪ ان ﯾﻘﺘ ﺮن ﺑﺘﻮﺿ ﯿﺢ ﻣﻨﺤﻨ ﻰ اﻻﺳ ﺘﺠﺎﺑﺔ
ﻟﻠﺠﺮﻋ ﺔ ﻟ ﯿﺲ ﺑ ﺴﺒﺐ ان 05 LDﺗﻤﺜ ﻞ اﺣ ﺪى ﻧﻘﺎﻃ ﮫ ﻓﺤ ﺴﺐ واﻧﻤ ﺎ ﻻن ﻣﻨﺤﻨﯿ ﺎت
اﻻﺳﺘﺠﺎﺑﺔ ﯾﻤﻜﻦ ان ﺗﺎﺧﺬ اﺷﻜﺎﻻ ﻣﺘﻌﺪدة وان ﻛﺎن اﻟﺸﻜﻞ اﻟﺸﺎﺋﻊ ﻟﮭ ﺎ ھ ﻮ اﻟﻤﻨﺤﻨ ﻰ
Sواﻟﺒﺎﺣ ﺚ ﯾﺤﺘ ﺎج اﻟ ﻰ اﻟﺘﻌ ﺮف ﻋﻠ ﻰ ﺷ ﻜﻞ اﻟﻤﻨﺤﻨ ﻰ ﻟﻠﻤﻘﺎرﻧ ﺔ ﺑ ﯿﻦ اﻟﻤ ﻮاد ﻣ ﻦ
ﺣﯿﺚ اﻟﻘﻮة و اﻟﻔﻌﺎﻟﯿﺔ ) .(Efficacy ، Potency
ﻟﻘﺪ ﺣﺎوﻟﺖ ﻓﻲ ھﺬا اﻟﻜﺘﺎب ﺟﺎھﺪا ﺷﻤﻮل ﺟﻤﯿﻊ اﻟﻄﻠﺒﺔ واﻟﺒ ﺎﺣﺜﯿﻦ ﻓ ﻲ اﻻﻓ ﺎدة ﻣﻨ ﮫ
ﺗﻮﺧﯿﺎ ﻟﺰﯾﺎدة اﻟﻤﻌﻠﻮﻣﺎت اﻟﻌﻠﻤﯿﺔ وﺗﻌﻤ ﯿﻢ اﻟﻤﻨﻔﻌ ﺔ اﺧ ﺬا ﺑﻨﻈﺮاﻻﻋﺘﺒﺎراﻟﺘﻔ ﺎوت ﺑ ﯿﻦ
اﻟﻤﺴﺘﻮﯾﺎت ﻓﻲ ﻣﺪى اﻟﻤﻌﺮﻓﺔ ﺑﻌﻠﻤﻲ اﻻﺣ ﺼﺎء واﻟﺤﺎﺳ ﻮب ﻣﻤ ﺎ دﻋ ﺎﻧﻲ ذﻟ ﻚ اﻟ ﻰ
اﺳﺘﻌﻤﺎل اﻛﺜﺮ ﻣﻦ وﺳﯿﻠﺔ ﻟﻠﺤﻞ واﻛﺜﺮ ﻣ ﻦ ﺑﺮﻧ ﺎﻣﺞ وﻟﻜ ﻦ ذﻟ ﻚ ﻟ ﻦ ﯾﺤ ﻮل دون ان
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2. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
ﯾﻌﺘﺮي ﻣﮭﻤﺘﻲ ھﺬه ﻧﻘﺼﺎ ھﻨﺎ وﻧﻘﺼﺎ ھﻨﺎك وھ ﻮ اﻣ ﺮا وارد ﻓ ﻲ ﺗﻐﻄﯿ ﺔ ﻣﺜ ﻞ ھ ﺬا
اﻟﻤﻮﺿﻮع اﻟﺬي ﯾﻤﺜﻞ ﻣﺤ ﺼﻠﺔ ﻟﻌﻠ ﻮم اﻻدوﯾ ﺔ واﻟ ﺴﻤﻮم واﻻﺣ ﺼﺎء واﻟﺤﺎﺳ ﻮب.
وﯾﻤﻜﻦ ﺗﻘﺴﯿﻢ ﻣﺤﺎوراﻟﻜﺘﺎب ﻓﻲ ﺗﻮﺿﯿﺢ ﻃﺮق ﺗﻘﺪﯾﺮ ﻗﯿﻤﺔ 05 LDاﻟﻰ ﻣﺎﯾﻠﻲ:
1- اﻟﺤ ﻞ اﻟﯿ ﺪوي اﻟ ﺬي ﯾﻤﻜ ﻦ ﺗﻄﺒﯿﻘ ﮫ ﻣ ﻦ ﻗﺒ ﻞ اﻟ ﺬﯾﻦ ﻟﯿ ﺴﺖ ﻟ ﺪﯾﮭﻢ ﻣﻌﺮﻓ ﺔ
ﺑﺎﻟﺒﺮاﻣﺞ اﻻﺣﺼﺎﺋﯿﺔ.
2- اﻟﺤ ﻞ ﺑﺎﺳ ﺘﻌﻤﺎل ﺑﺮﻧ ﺎﻣﺞ ﻋﻠ ﻰ اﺣ ﺪ اﻟﻤﻮاﻗ ﻊ ﻋﻠ ﻰ اﻻﻧﺘﺮﻧﯿ ﺖ اذ ﯾﻤﻜ ﻦ
اﺳﺘﻌﻤﺎﻟﮫ ﻣﻦ ﻗﺒ ﻞ اﻟ ﺬﯾﻦ ﻟ ﺪﯾﮭﻢ ﻣﻌﺮﻓ ﺔ ﺑ ﺴﯿﻄﺔ ﺑﺎﻟﺤﺎﺳ ﻮب ، وﯾﺘﻤﯿ ﺰ ھ ﺬا
اﻟﺒﺮﻧﺎﻣﺞ ﺑﺎﺳﻠﻮﺑﮫ اﻟﺒﺴﯿﻂ ﺟﺪا ،اذ ﻻ ﯾﺤﺘﺎج اﻟﻰ ﺗﻨﺼﯿﺐ ﻋﻠﻰ اﻟﺤﺎﺳ ﻮب
واﻧﻤﺎ ﯾﻤﻜﻦ ﺧﺰﻧﮫ ﻣﺒﺎﺷﺮة ، وﻻﯾﺘﻄﻠﺐ اﺟﺮاء اﻟﺘﺤﻠﯿﻞ اﻻ وﺿ ﻊ اﻻرﻗ ﺎم
وﻣ ﻦ ﺛ ﻢ اﻟﺘﻨﻔﯿ ﺬ ، وﯾﻤﻜ ﻦ اﺳ ﺘﻌﻤﺎل ھ ﺬا اﻟﺒﺮﻧ ﺎﻣﺞ ﻟﺘﻘﺪﯾﺮﻗﯿﻤ ﺔ 05LD
ﺑﺜﻼﺛ ﺔ ﻃ ﺮق ) Logisticو Probitو (Linear Regressionﻋﻠﻤ ﺎ
ﺑﺎن ﻣﻮﻗﻊ اﻟﺒﺮﻧﺎﻣﺞ ھﻮ:
.http://faculty.vassar.edu/lowry/VassarStats.html
وﺳ ﯿﺠﺪ اﻟﻘ ﺎرىء ﻓ ﻲ ھ ﺬا اﻟﻜﺘ ﺎب ﻣﺜ ﺎل ﯾﺘ ﻀﻤﻦ ﻃﺮﯾﻘ ﺔ ﺗﻘ ﺪﯾﺮ 05LD
ﺑﺎﻋﺘﻤﺎد ﺑﺮﻧﺎﻣﺞ اﻟﻤﻮﻗﻊ اﻟﻤﺸﺎراﻟﯿﮫ ﺳﻠﻔﺎ ﻟﯿﺴﮭﻞ ﻋﻠﯿﮫ اﺳﺘﻌﻤﺎﻟﮫ ﻣ ﺴﺘﻘﺒﻼ،
وﻗ ﺪ اﺳ ﺘﻌﻤﻠﻨﺎ ﺑﺮﻧ ﺎﻣﺞ SASﻟﺤ ﻞ اﻟﻤﺜ ﺎل ﻧﻔ ﺴﮫ ﻟﻐ ﺮض ﺗﺄﻛﯿ ﺪ ﺗﻄ ﺎﺑﻖ
اﻟﻨﺘﺎﺋﺞ.
3- اﻟﺤ ﻞ ﺑﺎﺳ ﺘﻌﻤﺎل اﻛﺜ ﺮ ﻣ ﻦ ﺑﺮﻧ ﺎﻣﺞ اﺣ ﺼﺎﺋﻲ ﻟﻐ ﺮض اﺗﺎﺣ ﺔ ﻓﺮﺻ ﺔ
اﻛﺒﺮاﻣﺎم اﻟﺒﺎﺣﺜﯿﻦ ﻟﻤﻌﺮﻓﺔ ﻃﺮﯾﻘﺔ اﻟﺘﻘﺪﯾﺮ وھﻮ ﯾﺨﺺ اﻟﺬﯾﻦ ﻟﺪﯾﮭﻢ ﻣﻌﺮﻓﺔ
ﺑﺎﺣ ﺪ اﻟﺒ ﺮاﻣﺞ اﻻﺣ ﺼﺎﺋﯿﺔ اﻟﻌﺎﻣ ﺔ ) SASو SPSSو، Minitab
(STATGRAPH ، STATISTICAوھ ﺬه اﻟﺒ ﺮاﻣﺞ ﺗﻌ ﺪ ﻣ ﻦ اﻛﺜ ﺮ
اﻟﺒ ﺮاﻣﺞ ﺷ ﯿﻮﻋﺎ واﻧﺘ ﺸﺎرا ، ﻟ ﺬا ﻓﺎﻧﻨ ﺎ ﺳ ﻨﺤﺎول ﺷ ﺮح اﻟﻄﺮﯾﻘ ﺔ او ﺗﺤﺪﯾ ﺪ
ﻗﻄﻌ ﺔ اﻟﺒﺮﻧ ﺎﻣﺞ دون ذﻛ ﺮ ﺗﻔﺎﺻ ﯿﻞ دﻗﯿﻘ ﺔ ﻋ ﻦ ﺗﻨ ﺼﯿﺐ او ﻋﻤ ﻞ ﻛ ﻞ
ﺑﺮﻧ ﺎﻣﺞ اذ ان ﻣﺎﺳ ﻨﺬﻛﺮه ﺳ ﯿﻜﻮن ﻣﻔﮭﻮﻣ ﺎ ﻟﻠ ﺬﯾﻦ ﻟ ﺪﯾﮭﻢ ﻣﻌﺮﻓ ﺔ ﺑﮭ ﺬه
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3. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
اﻟﺒﺮاﻣﺞ ﻻن ﺗﻮﺿﯿﺢ ﻋﻤﻞ ﻛﻞ ﺑﺮﻧﺎﻣﺞ ﻟﯿﺲ ﻋﻤﻼ ﺳ ﮭﻼ ﻛﻤ ﺎ اﻧ ﮫ ﻻﯾﻤﺜ ﻞ
اﻟﮭﺪف اﻟﺮﺋﯿﺴﻲ ﻟﮭﺬا اﻟﻜﺘﺎب.
4- ﻟﻘ ﺪ ﻗﺎدﻧ ﺎ ﺑﺤﺜﻨ ﺎ ﻓ ﻲ 05 LDاﻟ ﻰ اﻟﺘﻄ ﺮق اﻟ ﻰ ﻣﻮﺿ ﻮﻋﯿﻦ ﻏﺎﯾ ﺔ ﻓ ﻲ
اﻻھﻤﯿ ﺔ ھﻤ ﺎ اﻟﺘﻔﺎﻋ ﻞ اﻟ ﺪواﺋﻲ ودﻟﯿ ﻞ اﻟﺘﻮﻟﯿﻔ ﺔ ﻟ ﺴﺒﺒﯿﻦ اﻻول ﻋﻼﻗﺘﮭﻤ ﺎ
ﺑﻘﯿﻤﺔ 05 LDواﻟﺜﺎﻧﻲ ﻻھﻤﯿﺘﮭﻤﺎ اﻟﻜﺒﯿﺮة ﻓﻲ ﺑﺤﻮث اﻟﺼﯿﺪﻟﺔ.
5- ان ﻋﻤﻠﯿ ﺔ وﺻ ﻒ ﻣﻨﺤﻨ ﻰ اﻻﺳ ﺘﺠﺎﺑﺔ ﺑﺎﺳ ﺘﻌﻤﺎل ﻋ ﺪة ﻧﻤ ﺎذج رﯾﺎﺿ ﯿﺔ
ﺑﺎﻻﺳ ﺘﻌﺎﻧﺔ ﺑﻌ ﺪة ﺑ ﺮاﻣﺞ ﻟ ﻢ ﯾﻜ ﻦ ﺑﺎﻟ ﺴﮭﻮﻟﺔ اﻟﺘ ﻲ ﻋﻠﯿﮭ ﺎ ﻋﻨ ﺪ ﺗﻘ ﺪﯾﺮ ﻗﯿﻤ ﺔ
05 LDوﺧﻼﺻ ﺔ اﻟﻘ ﻮل ﻓ ﺎن اﻟﻘ ﺎرىء ﺳ ﯿﻼﺣﻆ ان ﺑﺮﻧ ﺎﻣﺞ SASﻛ ﺎن
اﻓﻀﻠﮭﺎ اذ ﺗﺘﻮﻓﺮ ﻓﯿﮫ اﻛﺜ ﺮ ﻣ ﻦ ﻃﺮﯾﻘ ﺔ ﻟﻠﺤ ﻞ ، وﻗ ﺪ ﺣﺎوﻟﻨ ﺎ وﺻ ﻒ ﻋ ﺪة
اﺷﻜﺎل ﻣﻦ اﻟﻤﻨﺤﻨﯿﺎت واﺳﺘﻌﻤﻠﻨﺎ ﺑﺮﻧﺎﻣﺞ Excelﻟﺮﺳﻢ اﻟﻌﻼﻗﺔ ﺑ ﯿﻦ اﻟﻘ ﯿﻢ
اﻟﻤ ﺸﺎھﺪة واﻟﻤﺘﻮﻗﻌ ﺔ ﻻﻧ ﮫ ﯾﻌﻄ ﻲ رﺳ ﻮﻣﺎ اﻓ ﻀﻞ ﻣﻘﺎرﻧ ﺔ ﺑ ﺎﻟﺒﺮاﻣﺞ
اﻻﺧﺮى. وھﻨ ﺎ ﻻﺑ ﺪ ﻣ ﻦ اﻻﺷ ﺎرة اﻟ ﻰ ان ھ ﺬا اﻟﻤﻮﺿ ﻮع ﯾ ﺴﺘﻮﺟﺐ ﻣ ﻦ
اﻟﺒﺎﺣ ﺚ ان ﯾﻜ ﻮن ﻣﻠﻤ ﺎ ﺑﺎﺣ ﺪ اﻟﺒ ﺮاﻣﺞ اﻻﺣ ﺼﺎﺋﯿﺔ ﻟﯿﺘ ﺴﻨﻰ ﻟ ﮫ ﺗﺤﻠﯿ ﻞ
اﻟﺒﯿﺎﻧﺎت.
6- ان اﺿ ﺎﻓﺔ ﻓ ﺼﻼ ﯾﺘ ﻀﻤﻦ ﻣ ﺴﺎﺋﻞ ﺗﺨ ﺺ ﺗﻘ ﺪﯾﺮ ﻗﯿﻤ ﺔ 05 LDﺳﯿ ﺴﺎھﻢ
ﺑﻼﺷﻚ ﻓﻲ زﯾﺎدة ﻗﺪرة اﻟﺒﺎﺣﺚ ﻋﻠﻰ اﻟﺘﻘﺪﯾﺮوﺑﺼﻮرة اﻛﺜﺮ دﻗﺔ.
7- ان اﻟﻤﻌﺎﻟﺠﺔ اﻻﺣﺼﺎﺋﯿﺔ ﻟ ﺒﻌﺾ اﻟﻤﻘ ﺎﯾﯿﺲ واﻟﺘ ﻲ ﺗﻤﺜ ﻞ اﻟﮭ ﺪف اﻻﺳﺎﺳ ﻲ
ﻟﮭ ﺬا اﻟﻜﺘ ﺎب ﻛﺎﻧ ﺖ ﺗ ﺴﺘﻠﺰم اﯾ ﻀﺎح ﺑﻌ ﺾ اﻟﻤﻔ ﺎھﯿﻢ اﻟﺘ ﻲ ﺗﺨ ﺺ ﻋﻠ ﻢ
اﻟﺴﻤﻮم واﻻدوﯾﺔ وﻗﺪ اوﺿﺤﻨﺎھﺎ ﺑﺼﻮرة ﻣﺒﺴﻄﺔ وﺿﻤﻦ ﻣ ﺪى اﻟﺘ ﺪاﺧﻞ
ﺑ ﯿﻦ اﻻﺣ ﺼﺎء وﺗﻠ ﻚ اﻟﻌﻠ ﻮم دوﻧﻤ ﺎ اﻟﺨ ﻮض ﻓ ﻲ ﻏﻤﺎرھ ﺎ ﻻﻧﮭ ﺎ ﻻﺗﻤﺜ ﻞ
اﺧﺘﺼﺎﺻﻨﺎ ﻋﻼوة ﻋﻠﻰ ﻗﻨﺎﻋﺘﻨﺎ ﺑﺎن ﺗﻮﺿﯿﺤﮭﺎ ﺑﺸﻜﻞ ادق واﺷ ﻤﻞ ﯾﻌ ﻮد
ﻟﺬوي اﻻﺧﺘﺼﺎص اوﻻ واﺧﯿﺮا.
3
4. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
اﻟﻔﺼﻞ اﻟﺜﺎﻧﻲ
“All substances are poisons: there is none which
is not a poison. The right dose differentiates a
.)1451-3941( poison and a remedy.” Paracelsus
2- 1 : اﻟﺠﺮﻋﺔ اﻟﻤﻤﯿﺘﺔ ﻟﻨﺼﻒ اﻟﻤﺠﻤﻮﻋﺔ) 05(LD
ﺗﻌ ﺪ ﺗﻘ ﺪﯾﺮات 05 LDذات اھﻤﯿ ﺔ ﻛﺒﯿ ﺮة ﻓ ﻲ اﻟﻌﻠ ﻮم اﻟﻄﺒﯿ ﺔ واﻟﺤﯿﺎﺗﯿ ﺔ اذ ﺗﻤﺜ ﻞ
اﻟﺠﺮﻋﺔ اﻟﻮﺳﻄﯿﺔ ﻟﻤﺎدة ﻣﻌﯿﻨﺔ واﻟﺘﻲ ﺗﺴﺒﺐ اﻟﻤ ﻮت ﻟﻨ ﺼﻒ ﻋ ﺪد اﻓ ﺮاد ﻣﺠﻤﻮﻋ ﺔ
ﻣﻦ ﺣﯿﻮاﻧﺎت اﻟﺘﺠﺮﺑﺔ ﺧﻼل ﻣﺪة ﻣﻌﯿﻨﺔ ﻓﻤﺜﻼ اذا ﻛﺎﻧﺖ اﻟﻔﺘ ﺮة 7 أﯾ ﺎم ﻓﯿﻌﺒ ﺮ ﻋﻨﮭ ﺎ
7/05 LDاو 03 ﯾ ﻮم وﺗﻜ ﻮن 03/05LDوھﻜ ﺬا وھ ﻲ ﺗﻤﺜ ﻞ اﺣ ﺪى ﻧﻘ ﺎط ﻣﻨﺤ ﻰ
اﻻﺳ ﺘﺠﺎﺑﺔ وﺗﻘ ﻊ ﺑ ﯿﻦ ادﻧ ﻰ واﻗ ﺼﻰ ﻣ ﺴﺘﻮى ﻟﻠﺠﺮﻋ ﺔ وﯾﻤﻜ ﻦ اﻟﺘﻌﺒﯿ ﺮ ﻋ ﻦ
اﻟﺘﺄﺛﯿﺮاﻟﻤﻤﯿ ﺖ ﻟﻤ ﺎدة ﻋﻠ ﻰ اﺳ ﺎس اﻟﺘﺮﻛﯿ ﺰ05(Lethal concentration) LC
او ﻓﺎﻋﻠﯿ ﺔ اﻟﻌﻘ ﺎر)05dose) (ED
(Effectiveاو اﻟ ﺴﻤﯿﺔ اﻟﻤ ﻮاد 05TD
) .(Toxic dose
ان ﺗﻘ ﺪﯾﺮات 05 LDﯾ ﺘﻢ ﺑﻤﻮﺟﺒﮭ ﺎ ﺗﺤﺪﯾﺪﻋ ﺪة ﻣ ﺴﺘﻮﯾﺎت ﻟﻠ ﺴﻤﯿﺔ واﻟﺘ ﻲ ﺗ ﺸﻤﻞ
ﺗﻌﺎﻃﻲ ھﺬه اﻟﻤﻮاد ﻋﻦ ﻃﺮق اﻟﻔﻢ او اﻟﺠﻠﺪ او اﻟﮭﻮاء.
6- Super toxic
5- Extremly toxic
4- Very toxic
3- Moderitly toxic
2- Slightly toxic
1- Practically toxic
4
5. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
وﻟﺘﻮﺿﯿﺢ اھﻤﯿﺔ ﺗﻄﺒﯿﻘﺎت 05 LDﻋﻠﻰ اﻻﻧ ﺴﺎن ﺳ ﻨﻔﺘﺮض ان ھﻨ ﺎك ﻣ ﺎدة ﺳ ﻤﯿﺔ
وان ﻗﯿﻤﺔ 05 LDﻟﮭﺎ ﻓﻲ اﻟﻔﺌﺮان ﺑﻠﻐﺖ 003 ﻣﻠﻐ ﻢ/ ﻛﻐ ﻢ ﻣ ﻦ وزن اﻟﺠ ﺴﻢ وﺑ ﺬﻟﻚ
ﻓﺎن اﻟﻜﻤﯿﺔ اﻟﻤﻨﺎﻇﺮة ﻟﮭﺎ واﻟﺘﻲ ﺋﺆدي اﻟﻰ اﻟﻤﻮت ﻟﺸﺨﺺ ﯾﺰن 07 ﻛﻐﻢ ﺗﻜﻮن:
70 kg x 300mg/kg = 21000 mg = 21 g
ﻋﻨﺪ وﺟﻮد ھﺬه اﻟﻤﺎدة ﻓﻲ ﻣﻨﺘﻮج وﺑﺘﺮﻛﯿ ﺰ 521 ﻏ ﺮام/ ﻟﺘ ﺮ ﻓ ﺎن اﻟﻜﻤﯿ ﺔ اﻟﻼزﻣ ﺔ
ﻟﻘﺘﻞ اﻟﺸﺨﺺ:
21g/125g/L = 0.168 L = 168 mL
اذا ﻣﺰﺟﻨ ﺎ اﻟﻤﻨﺘ ﻮج ﻣ ﻊ ﻣﺤﻠ ﻮل وﺑﻮاﻗ ﻊ 001 ﻣ ﻞ/ 01 ﻟﺘ ﺮ ﻓ ﺎن اﻟﻜﻤﯿ ﺔ اﻟﻼزﻣ ﺔ
ﻟﻘﺘﻞ اﻟﺸﺨﺺ:
168mL/ 10mL/L = 16.8 L
ھﻨ ﺎك ﻃ ﺮق ﻋﺪﯾ ﺪة ﻟﺘﻘﺪﯾﺮﻗﯿﻤ ﺔ 05 LDوﻣﻌﻈ ﻢ ھ ﺬه اﻟﻄ ﺮق ﯾﻜ ﻮن ﻓﯿﮭ ﺎ اﻟﺘﻘ ﺪﯾﺮ
رﯾﺎﺿﯿﺎ وﺑﻌﻀﺎ ﻣﻨﮭﺎ ﺑﺎﻋﺘﻤﺎد اﻟﺮﺳﻢ اﻟﺒﯿﺎﻧﻲ وﻣﻦ اھﻢ ﻃﺮق اﻟﺘﻘﺪﯾﺮ ھﻲ :
1- )8491( )Up and Down (Dixon-Mood
2- )1391( Spearman- Karber
3- )4491( Miller and Tainter
4- )3891( Lorke
5- )9291( Dragster-Behrens
6- )8391( Reed-Muench
7- )7491( Thompson Moving Average
8- )5791( Shuster-Yang
9- ) 9491( Litchfield-Wilcoxon
01- Probit
11- )6791( Shuster-Dietrich
21- ٍSimple Linear Regression
5
6. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
Polynomial Probit -13
Polynomial Logistic -14
Molinengo (1979) -15
Nonlinear Regression -16
Linear Interpolation -17
Logistic -18
Robbins and Monro (1964) -19
Litchfield and Fertig (1941) -20
ً
Wilson and Worcester (1948) -21
Knudson and Curtis (1947) -22
Fixed Dose (1984) -23
Sunٍ (1963) -24
Schutz and Fuchs (1982) -25
Deichmann and LeBlanc (1943) -26
Non-Animal Test (2004) -27
Ramsey (1972)-28
Davis (1972)-29
Chmiel (1976) -30
Freeman (1980)-31
Joan and Staniswalis (1988)-32
Hans-Georg Muller (1998)-33
.Bhattacharya and Kong (2007)-34
Berkson (1949) -35
6
7. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
2-2 ﺑﻌﺾ ﻃﺮق ﺗﻘﺪﯾﺮ ﻗﯿﻤﺔ 05LD
ھﻨﺎﻟﻚ اﻟﻌﺪﯾ ﺪ ﻣ ﻦ ﻃ ﺮق اﻟﺘﻘ ﺪﯾﺮ واﻟﺘ ﻲ ﺗﺘﻔ ﺎوت ﻓ ﻲ درﺟ ﺔ اﻟﺪﻗ ﺔ وﺳ ﮭﻮﻟﺔ
اﻟﺘﻄﺒﯿ ﻖ وﺳ ﻌﺔ اﻻﻧﺘ ﺸﺎر وھﻨ ﺎك ﺑﺤﺜ ﺎن ﺑﮭ ﺬا اﻟﺨ ﺼﻮص وﯾﻤﻜ ﻦ ﻟﻠﻘ ﺎرىء
اﻟﺮﺟﻮع اﻟﯿﮭﻤﺎ اذ ﯾﺘﻀﻤﻦ ﻛﻞ ﻣﻨﮭﻤﺎ اﺳﺘﻌﺮاﺿﺎ ﻻھﻢ ﻃﺮق اﻟﺘﻘﺪﯾﺮ ﻧﺸﺮا ﻣﻦ
ﻗﺒﻞ ﻛﻞ ﻣﻦ )7791( Stephanو )5891( ,..Gelber et al
ان اھﻤﯿ ﺔ ﺗﻘ ﺪﯾﺮ ﻗﯿﻤ ﺔ 05 LDﻓ ﻲ اﻟﻌﻠ ﻮم اﻟﺒﺎﯾﻮﻟﻮﺟﯿ ﺔ ﺗ ﺴﺘﻮﺟﺐ ﺗﻮﺿ ﯿﺢ
ﺑﻌﻀﺎ ﻣﻦ اھﻢ ﺗﻠﻚ اﻟﻄﺮق ﻓﻀﻼ ﻋﻦ ﺗﻮﺿﯿﺢ ﻛﯿﻔﯿﺔ ﺗﻘﺪﯾﺮھﺬا اﻟﻤﻘﯿﺎس ﯾ ﺪوﯾﺎ
وﻛﺬﻟﻚ ﺑﺎﺳﺘﻌﻤﺎل اﻛﺜﺮﻣﻦ ﺑﺮﻧﺎﻣﺞ اﺣﺼﺎﺋﻲ.
ﺗﻌﺘﻤﺪ ﻗﯿﻤﺔ 05 LDﻓﻲ ﺗﻘﺪﯾﺮھﺎ ﻋﻠﻰ اﺳﺎس ان ﻧ ﺴﺒﺔ اﻟﺰﯾ ﺎدة ﻓ ﻲ اﻟﺤﯿﻮاﻧ ﺎت
اﻟﮭﺎﻟﻜﺔ ﺗﺘﺒ ﻊ داﻟ ﺔ اﻟﺘﻮزﯾ ﻊ اﻟﺘﺠﻤﯿﻌ ﻲ وﺗﺨ ﻀﻊ ﻟﺨ ﺼﺎﺋﺺ اﻟﺘﻮزﯾ ﻊ اﻟﻄﺒﯿﻌ ﻲ
،ﻓﻠﻮ ﻓﺮﺿﻨﺎ ان ﺗﺠﺮﺑﺔ ﻣﺎ اﺟﺮﯾﺖ ﻋﻠ ﻰ ﻣﺠﻤﻮﻋ ﺔ ﻣ ﻦ اﻟﻔﺌ ﺮان ﻟﺘﺤﺪﯾ ﺪ ﺳ ﻤﯿﺔ
ﻣﺎدة ﻣﻌﯿﻨﺔ وﻗﺪ ﻗﻤﻨﺎ ﺑﺮﺳﻢ ﻧﺘﺎﺋﺞ اﻟﺘﺠﺮﺑﺔ ﺑﯿﺎﻧﯿﺎ ﺑﺤﯿ ﺚ ﯾﻜ ﻮن اﻟﻤﺤ ﻮر اﻟ ﺴﯿﻨﻲ
ھ ﻮ اﻟﺠﺮﻋ ﺔ )ﻣﻠﻐ ﻢ/ﻛﻐ ﻢ( واﻟﻤﺤ ﻮر اﻟ ﺼﺎدي ﯾﻤﺜ ﻞ ﻧ ﺴﺒﺔ اﻻﻓ ﺮاد اﻟﮭﺎﻟﻜ ﺔ
ﺑﻮﺣﺪات اﻟﻨﺴﺒﺔ اﻟﻤﺆﯾﺔ اﻟﺘﺠﻤﯿﻌﯿﺔ ﻓﺴﻨﺤﺼﻞ ﻋﻠﻰ اﻟﺮﺳﻢ ادﻧﺎه:
ﺷﻜﻞ 1: اﻟﻌﻼﻗﺔ ﺑﯿﻦ اﻟﺠﺮﻋﺔ واﻻﺳﺘﺠﺎﺑﺔ
7
8. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
ان ﻛﻞ ﻧﻘﻄﺔ ﻓﻲ اﻟﺮﺳﻢ اﻟﺒﯿﺎﻧﻲ ﺗﻤﺜﻞ ﻧﺴﺒﺔ اﻻﻓﺮاد اﻟﮭﺎﻟﻜ ﺔ ﻋﻨ ﺪ ﺟﺮﻋ ﺔ ﻣﻌﯿﻨ ﺔ
وﯾﻼﺣﻆ ﺑﺎن ﻧ ﺴﺒﺔ اﻻﻓ ﺮاد اﻟﮭﺎﻟﻜ ﺔ ﯾ ﺴﺎوي 0 ﻋﻨ ﺪ اول ﺟﺮﻋ ﺔ وﻟﻜ ﻦ ﺑﺰﯾ ﺎدة
اﻟﺠﺮع وﺟﺪ ان اﻟﻨﺴﺒﺔ ﺑﺪأت ﺑﺎﻻرﺗﻔﺎع ، واذا ﺣﺎوﻟﻨﺎ وﺻﻒ ھﺬه اﻟﻨﻘﺎط ﺑﺨﻂ
ﺳﻨﺠﺪ ان اﻟﺨﻂ ﺳﯿﺄﺧﺬ ﺷﻜﻼ ﻣﻠﺘﻮﯾﺎ ) (sigmoid shapeﻛﻤﺎ ﻣﻮﺿﺢ ادﻧﺎه:
ﺷﻜﻞ 2: ﻣﻨﺤﻨﻰ اﻟﻌﻼﻗﺔ ﺑﯿﻦ اﻟﺠﺮﻋﺔ واﻻﺳﺘﺠﺎﺑﺔ
وھﺬا ھﻮ اﻟﺸﻜﻞ اﻟﻤﺜ ﺎﻟﻲ ﻟﻤﻨﺤﻨ ﻰ اﻻﺳ ﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋ ﺔ. وﻋﻨ ﺪﻣﺎ ﻧﻘ ﻮل وﺻ ﻒ
اﻟﻨﻘﺎط ﻧﻘﺼﺪ اﻓﻀﻞ ﺧﻂ ﻟﻮﺻﻒ ﺷﻜﻞ ﺗﻮزﯾﻊ اﻟﻨﻘﺎط اي اﻓﻀﻞ ﺧﻂ ﻟﻮﺻﻒ
ﻣﻨﺤﻨﻰ اﻻﺳﺘﺠﺎﺑﺔ وﻧﻼﺣﻆ اﯾﻀﺎ ان ھﺬا اﻟﺨﻂ ﻻﯾ ﺸﺘﺮط ﺑﺎﻟ ﻀﺮورة ان ﯾﻤ ﺮ
ﺑﺠﻤﯿ ﻊ اﻟﻨﻘ ﺎط . وﻟ ﻮ اﻣﻌﻨ ﺎ اﻟﻨﻈ ﺮ ﻓ ﻲ اﻟﻤﻨﺤﻨ ﻰ اﻋ ﻼه ﺳ ﻨﺠﺪ ان اﻟﺠ ﺰء
اﻟﻮﺳ ﻄﻲ ﻣﻨ ﮫ واﻟ ﺬي ﯾﺘ ﺮاوح ﻣ ﻦ 61 – 48% ﯾﻜ ﻮن ﺧﻄ ﺎ ﻣ ﺴﺘﻘﯿﻤﺎ. وھ ﺬا
ﯾﻌﻨﻲ ان اﻻﺳ ﺘﺠﺎﺑﺔ ﻣ ﺎﺑﯿﻦ اﻟﺤ ﺪﯾﻦ ﻋﻠ ﻰ اﺳ ﺎس اﻟﻤﺤ ﻮر اﻟ ﺴﯿﻨﻲ ) اﻟﺠﺮﻋ ﺔ(
ﺗﻤﺜ ﻞ اﻧﺤﺮاﻓ ﺎ ﻗﯿﺎﺳ ﯿﺎ ﻟﻠﻤﺘﻮﺳ ﻂ ﻗ ﺪره – 1 / + 1 ﻓ ﻲ اﻟﻤﺠﺘﻤ ﻊ ذو اﻟﺘﻮزﯾ ﻊ
اﻟﻄﺒﯿﻌ ﻲ. وﻧﻘ ﺼﺪ ﺑ ﺬﻟﻚ ان ﺗﻮزﯾ ﻊ اﻻﺳ ﺘﺠﺎﺑﺔ ﯾﻜ ﻮن ﺗﻮزﯾ ﻊ ﻃﺒﯿﻌ ﻲ وﯾﺄﺧ ﺬ
ﺷﻜﻞ اﻟﺠﺮس اي ان ﻣﻌﻈﻢ اﻻﺳﺘﺠﺎﺑﺔ ﺳﺘﻜﻮن ﻓﻲ اﻟﻮﺳ ﻂ وﺗﻘ ﻞ ﻛﻠﻤ ﺎ اﺑﺘﻌ ﺪﻧﺎ
8
9. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
ﻋﻦ اﻟﻤﺘﻮﺳﻂ وﻓﻲ ﻛﻼ اﻻﺗﺠﺎھﯿﻦ وﺳ ﺒﺐ ذﻟ ﻚ ﯾﻌ ﻮد اﻟ ﻰ وﺟ ﻮد ﺗﻔ ﺎوت ﺑ ﯿﻦ
اﻟﺤﯿﻮاﻧﺎت ﻓﻲ اﻻﺳﺘﺠﺎﺑﺔ وھﻮ ﻣﺎﯾﺴﻤﻰ .Biological Variability
05
04
03
02
01
51
11
31
7
9
3
5
0
1
ﺷﻜﻞ3: اﻟﺘﻮزﯾﻊ اﻟﻄﺒﯿﻌﻲ ﻟﻼﺳﺘﺠﺎﺑﺔ
وﻋﻨ ﺪﻣﺎ ﻧﺤ ﺎول ﺗﻤﺜﯿ ﻞ اﻟﺘﻮزﯾ ﻊ اﻟﺘﺠﻤﯿﻌ ﻲ ﻟﻠﺘﻮزﯾ ﻊ اﻟﻄﺒﯿﻌ ﻲ ﺳﻨﺤ ﺼﻞ ﻋﻠ ﻰ
اﻟﺸﻜﻞ Sوﻛﻤﺎ ﻣﻮﺿﺢ ﻓﻲ اﻟﺮﺳﻢ اﻟﺒﯿﺎﻧﻲ ادﻧﺎه:
001
08
06
04
02
51
31
11
7
9
5
3
1
0
ﺷﻜﻞ 4: اﻟﺘﻮزﯾﻊ اﻟﺘﺠﻤﯿﻌﻲ ﻟﻤﻨﺤﻨﻰ اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ
9
10. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
ﻟﻐﺮض ﺗﻮﺿﯿﺢ اﻟﻔﻜﺮة ﺑﺼﻮرة ادق ﺳﻨﻔﺮض ان ﺗﺠﺮﺑﺔ اﺟﺮﯾﺖ ﻋﻠﻰ ﻣﺠﻤﻮﻋ ﺔ
ﻣﻦ اﻟﻔﺌﺮان وﻛﺎﻧﺖ اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮع ﻛﻤﺎ ﻣﻮﺿﺢ ﻓﻲ اﻟﺠﺪول ادﻧﺎه:
%Response
%Increasing
in Response
0
5
7
8
51
02
02
51
5
3
2
0
Dose
0
5
21
02
53
55
57
09
59
89
001
001
1
2
3
4
5
6
7
8
9
01
11
21
اﻟﻌﻤ ﻮد اﻻول ﻣ ﻦ اﻟﯿﻤ ﯿﻦ ﯾﻤﺜ ﻞ اﻟﺰﯾ ﺎدة ﻓ ﻲ ﻧ ﺴﺒﺔ اﻻﺳ ﺘﺠﺎﺑﺔ ﻟﻜ ﻞ ﻣ ﺴﺘﻮى ﻣ ﻦ
اﻟﺠﺮﻋ ﺔ ﻋ ﻦ اﻟﻤ ﺴﺘﻮى اﻟ ﺬي ﻗﺒﻠ ﮫ وﻟ ﻮ ﺣﺎوﻟﻨ ﺎ رﺳ ﻢ اﻟﻌﻼﻗ ﺔ ﺑﯿﺎﻧﯿ ﺎ ﺑ ﯿﻦ ﻧ ﺴﺒﺔ
اﻻﺳﺘﺠﺎﺑﺔ واﻟﺰﯾﺎدة ﻓﻲ ﻧﺴﺒﺔ اﻻﺳﺘﺠﺎﺑﺔ ﺳﻨﺠﺪ ان ﻧﺴﺒﺔ اﻻﺳ ﺘﺠﺎﺑﺔ ﺳ ﺘﺄﺧﺬ اﻟ ﺸﻜﻞ
Sﻓﯿﻤﺎ ﺳﻨﺠﺪ ان اﻟﺰﯾﺎدة ﻓﻲ ﻧﺴﺒﺔ اﻻﺳﺘﺠﺎﺑﺔ ﯾﻜﻮن ﺗﻮزﯾﻌﮭﺎ ﻃﺒﯿﻌﯿﺎ.
021
08
06
04
02
0
21 11 01
9
8
7
6
5
4
3
2
1
Dose
Cumulative response
Increasing response
ﺷﻜﻞ 5: اﻟﺘﻮزﯾﻊ اﻟﻄﺒﯿﻌﻲ ﻟﻠﺰﯾﺎدة ﻓﻲ اﻻﺳﺘﺠﺎﺑﺔ
01
Cumulative response
001
11. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
2-2-1 ﻃﺮﯾﻘﺔ Simple Linear Regression
ھﺬه اﻟﻄﺮﯾﻘﺔ ﺗﻤﺜﻞ اﻗﺪم ﻃﺮق اﻟﺘﻘﺪﯾﺮ وﻓﯿﮭﺎ ﯾ ﺘﻢ اﻟﺘﻮﺻ ﻞ اﻟ ﻰ اﯾﺠ ﺎد ﻣﻌﺎدﻟ ﺔ ﺧ ﻂ
ﻣ ﺴﺘﻘﯿﻢ ﻟﻠﻌﻼﻗ ﺔ ﺑ ﯿﻦ ﻧ ﺴﺒﺔ اﻟﺤﯿﻮاﻧ ﺎت اﻟﻤﯿﺘ ﺔ )ﻋ ﺪد اﻟﺤﯿﻮاﻧ ﺎت اﻟﻤﯿﺘ ﺔ / ﻋ ﺪد
اﻟﺤﯿﻮاﻧﺎت اﻟﻜﻠﻲ ( )اﻻﺳﺘﺠﺎﺑﺔ( وﻟﻮﻏﺎرﯾﺘﯿﻢ اﻟﺠﺮﻋﺔ وھﺬه اﻟﻤﻌﺎدﻟﺔ ﺗﻤﺜﻞ ﻣﻌﺎدﻟ ﺔ
ﺗﻨﺒﻮء ) .( Prediction equation
ﻟﺬا ﻓﺄن ﻣﻌﺎدﻟﺔ اﻻﻧﺤﺪار اﻟﺨﻄﻲ اﻟﺒﺴﯿﻂ ھﻲ:
Ŷ= a + bx
اذ ان :
= Ŷﻧﺴﺒﺔ اﻟﺤﯿﻮاﻧﺎت اﻟﻤﯿﺘﺔ اﻟﻤﺘﻮﻗﻌﺔ
=aﻧﻘﻄﺔ اﻟﺘﻘﺎﻃﻊ
= bﻣﻌﺎﻣﻞ اﻟﻤﯿﻞ
= xاﻟﺠﺮﻋﺔ
اﻻﻧﺤ ﺪار ﻓ ﻲ ﻣﻌﻨ ﺎه اﻟﻌ ﺎم ھ ﻮ اﻟﻌﻼﻗ ﺔ ﺑ ﯿﻦ ﻧ ﻮﻋﯿﻦ او اﻛﺜ ﺮ ﻣ ﻦ اﻟﻤﺘﻐﯿ ﺮات ،
واﻟﻤﺘﻐﯿ ﺮات ھ ﻲ اي ﺻ ﻔﺔ ﻛﻤﯿ ﺔ ﺗﺄﺧ ﺬ ﻗ ﯿﻢ ﻣﺨﺘﻠﻔ ﺔ ﻣﺜ ﻞ ﻃ ﻮل اﻟﺠ ﺴﻢ او اﻟ ﻮزن
وﻏﯿﺮھ ﺎ ، اﻟﻨ ﻮع اﻻول ﻣ ﻦ اﻟﻤﺘﻐﯿ ﺮات ﯾ ﺴﻤﻰ اﻟﻤﺘﻐﯿ ﺮات اﻟﺘﺎﺑﻌ ﺔ
) (Dependentوﺗﻤﺜ ﻞ اي ﻣﺘﻐﯿ ﺮ ﯾﺘ ﺄﺛﺮ ﺑﻤﺘﻐﯿ ﺮ اﺧ ﺮ او اﻛﺜ ﺮ واﻟﻨ ﻮع اﻟﺜ ﺎﻧﻲ
ھ ﻮاﻟﻤﺘﻐﯿﺮات اﻟﻤ ﺴﺘﻘﻠﺔ ) (Independentوﺗﻤﺜ ﻞ اي ﻣﺘﻐﯿ ﺮ ﯾ ﺆﺛﺮ ﻋﻠ ﻰ ﻣﺘﻐﯿ ﺮ
اﺧﺮ او اﻛﺜﺮ، ﻓﻤﺜﻼ ﻧﺴﺒﺔ اﻟﻤﻠﻮﺣﺔ ﻓﻲ اﻟﻤ ﺎء ﺗﻌﺘﺒ ﺮ ﻣﺘﻐﯿ ﺮ ﻣ ﺴﺘﻘﻞ وﻋ ﺪد اﻻﺣﯿ ﺎء
اﻟﻤﺎﺋﯿﺔ ﻟﻨﻮع ﻣﻌﯿﻦ ﺗﻤﺜﻞ ﻣﺘﻐﯿﺮ ﺗﺎﺑﻊ ﺑﻤﻌﻨﻰ ان اي زﯾﺎدة ﻓﻲ اﻟﻤﻠﻮﺣﺔ ﺳﺘﺆﺛﺮ ﻋﻠﻰ
ﻋ ﺪد ھ ﺬه اﻻﺣﯿ ﺎء. ﻣﺜ ﺎل اﺧ ﺮ ﻋ ﻦ اﻟﻤﺘﻐﯿ ﺮات ھ ﻮ اﻟﺠﺮﻋ ﺔ وﺗﻌ ﺪ ﻣﺘﻐﯿ ﺮ ﻣ ﺴﺘﻘﻞ
وﻧﺴﺒﺔ اﻟﺤﯿﻮاﻧﺎت اﻟﮭﺎﻟﻜﺔ ﺗﻤﺜﻞ اﻟﻤﺘﻐﯿﺮ اﻟﺘﺎﺑﻊ.
ﯾ ﺴﺘﻌﻤﻞ اﻟﻘ ﺎﻧﻮن اﻟﺘ ﺎﻟﻲ ﻟﺘﻘ ﺪﯾﺮ ﻗﯿﻤ ﺔ اﻻﻧﺤ ﺪار اﻟ ﺬي ﯾﻌﺒ ﺮ ﻋﻨ ﮫ ﻓ ﻲ اﻟﻜﺘ ﺐ
اﻻﺣﺼﺎﺋﯿﺔ ﺑﺎﻟﺤﺮف .b
11
12. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
b= ∑xy – {(∑x)(∑Y)}/n
²)∑x² – (∑ x
n
وﺑﺪﻻﻟﺔ ﻗﯿﻤﺔ bﻧﻘﺪر ﻗﯿﻤﺔ aوﺑﺬﻟﻚ ﻧﺴﺘﻌﻤﻞ ﻗﯿﻤﺔ xﻓﻲ اﻟﻤﻌﺎدﻟﺔ ﻟﺘﻘﺪﯾﺮ ﻗﯿﻤﺔ Ŷ
Ŷ= a + bx
ان ﻣﻌﺎدﻟﺔ اﻻﻧﺤﺪار ﯾﻤﻜﻦ اﻻﺳﺘﻔﺎدة ﻣﻨﮭﺎ ﻟﻐﺮض ﺗﻘﺪﯾﺮاي ﻗﯿﻤﺔ ﻣ ﻦ ﻗ ﯿﻢ اﻟﻤﺘﻐﯿ ﺮ
اﻟﺘ ﺎﺑﻊ ) (Yواﻟﺘ ﻲ ﻻﺗﺘ ﻮﻓﺮ ﻓ ﻲ اﻟﺒﯿﺎﻧ ﺎت اﻟﻤﻨ ﺎﻇﺮة ﻟﻘﯿﻤ ﺔ اﻟﻤﺘﻐﯿ ﺮ اﻟﻤ ﺴﺘﻘﻞ )(X
اﻟﺘﻲ ﻟﺪﯾﻨﺎ ﻛﻤﺎ ﯾﻤﻜﻦ اﺳﺘﻌﻤﺎﻟﮭﺎ ﺑﺼﻮرة ﻣﻌﻜﻮﺳﺔ اي ﺗﻘﺪﯾﺮ ﻗﯿﻤﺔ اﻟﻤﺘﻐﯿ ﺮ اﻟﻤ ﺴﺘﻘﻞ
اﻋﺘﻤﺎدا ﻋﻠﻰ ﻗﯿﻤﺔ اﻟﻤﺘﻐﯿﺮ اﻟﺘﺎﺑﻊ ﻛﻤﺎ ھ ﻮ اﻟﺤ ﺎل ﻋﻨ ﺪ ﺗﻘ ﺪﯾﺮ ﻗﯿﻤ ﺔ 05 ، LDﻓﻤ ﺜﻼ
ﻋﻨ ﺪﻣﺎ ﻧ ﺴﺘﻌﻤﻞ ﺟ ﺮع ﻣﺨﺘﻠﻔ ﺔ ﻣ ﻦ ﻣ ﺎدة ﺳ ﻤﯿﺔ وﻧ ﺴﺠﻞ ﻋ ﺪد اﻻﻓ ﺮاد اﻟﻤﯿﺘ ﺔ ﻟﻜ ﻞ
ﻣﺠﻤﻮﻋﺔ اﺧﺬت اﻟﺠﺮﻋﺔ وﻧﺮﯾﺪ ان ﻧﻘﺪر اﻟﺠﺮﻋﺔ اﻟﺘﻲ ﺗﺆدي اﻟﻰ ﻗﺘﻞ ﻧﺴﺒﺔ ﻣﻌﯿﻨﺔ
ﻣﻦ اﻓﺮاد اﻟﻤﺠﻤﻮﻋﺔ ﻓﺎﻧﻨﺎ ﻧﻄﺒ ﻖ ﻣﻌﺎدﻟ ﺔ اﻻﻧﺤ ﺪار ﻟﻐ ﺮض اﻟﺘﻨﺒ ﻮء ﺑﺘﻠ ﻚ اﻟﻘﯿﻤ ﺔ ،
وﻟﺘﻮﺿﯿﺢ اﻟﻔﻜﺮة اﻛﺜﺮ ﺳﻨﺤﺎول ﺣﻞ اﻟﻤﺜﺎل اﻵﺗﻲ:
ﻣﺜ ﺎل )2( ﺟ ﺪ ﻗﯿﻤ ﺔ 05 LDﻟﻠﻤ ﺎدة xاذا ﻋﻠﻤ ﺖ ﺑﺄﻧﮭ ﺎ اﺿ ﯿﻔﺖ ﺑﺠ ﺮع ﻣﺨﺘﻠﻔ ﺔ
وﺳﺠﻠﺖ اﻋﺪاد اﻟﺤﯿﻮاﻧﺎت اﻟﮭﺎﻟﻜﺔ ازاء ﻛﻞ ﺟﺮﻋﺔ؟
Total No
03
03
03
03
03
03
03
03
03
03
No.Dead
0
2
3
5
8
21
02
62
72
03
)dose(x
481.0
942.0
592.0
633.0
763.0
105.0
495.0
776.0
047.0
208.0
21
13. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
أ- اﻟﺤﻞ اﻟﯿﺪوي
ﻧﺤﻮل اﻟﺠﺮع اﻟﻰ ﻟﻮﻏ ﺎرﯾﺘﯿﻢ اﻻﺳ ﺎس 01 او اﻟﻠﻮﻏ ﺎرﯾﺘﯿﻢ اﻟﻄﺒﯿﻌ ﻲ وذﻟ ﻚ ﻟﺠﻌ ﻞ
اﻟﻌﻼﻗ ﺔ اﻛﺜ ﺮ اﺳ ﺘﻘﺎﻣﺔ وﻧ ﺴﺘﺨﺮج ﻧ ﺴﺒﺔ اﻟﺤﯿﻮاﻧ ﺎت اﻟﮭﺎﻟﻜ ﺔ اﻟ ﻰ ﻋ ﺪد اﻟﺤﯿﻮاﻧ ﺎت
اﻟﻜﻠﻲ:
Y
0
60.0
01.0
61.0
62.0
04.0
66.0
68.0
09.0
1
X
– 96.1
– 93.1
– 22.1
– 90.1
–1
– 96.0
– 25.0
– 93.0
– 03.0
– 22.0
ﻧﻼﺣ ﻆ ﻣ ﻦ اﻟﺒﯿﺎﻧ ﺎت ﻋ ﺪم وﺟ ﻮد ﻧ ﺴﺒﺔ ھﻼﻛ ﺎت 05.0 )وﺣﺘ ﻰ ان وﺟ ﺪت ھ ﺬه
اﻟﻨ ﺴﺒﺔ ﻓﻠ ﯿﺲ ﺻ ﺤﯿﺤﺎ اﻋﺘﺒﺎرھ ﺎ اﻟﻘﯿﻤ ﺔ اﻟﻤﺘﻮﻗﻌ ﺔ واﻧﻤ ﺎ ﯾﺠ ﺐ اﺳ ﺘﻌﻤﺎل ﻣﻌﺎدﻟ ﺔ
اﻻﻧﺤﺪارﻟﺘﻘﺪﯾﺮھﺎ( وﻻﻧﺠﺎز ذﻟﻚ ﻓﺎﻧﻨﺎ ﻧﻘﻮم ﺑﺤﺴﺎب ﻣﻜﻮﻧﺎت ﻣﻌﺎدﻟﺔ اﻻﻧﺤﺪار:
4.4 = ∑yوﺗﻤﺜﻞ ﻣﺠﻤﻮع ﻗﯿﻢ y
8.51– = ∑xوﺗﻤﺜﻞ ﻣﺠﻤﻮع ﻗﯿﻢ x
2.084 = ∑ xyﺗﻤﺜﻞ ﻣﺠﻤﻮع ﺣﺎﺻﻞ ﺿﺮب ﻛﻞ ﻗﯿﻤﺔ ﻣﻦ ﻗﯿﻢ xﻣﻊ ﻗﯿﻢ y
37.44 – = ∑x ∑yوﺗﻤﺜﻞ ﺣﺎﺻﻞ ﺿﺮب ﻣﺠﻤﻮع ﻗﯿﻢ xوﻣﺠﻤﻮع ﻗﯿﻢ y
² 9.50 = ∑xوﺗﻤﺜﻞ ﻣﺠﻤﻮع ﻣﺮﺑﻌﺎت ﻗﯿﻢ x
²) 72.42 = (∑ xوﺗﻤﺜﻞ ﻣﺮﺑﻊ ﻣﺠﻤﻮع ﻗﯿﻢ x
437.0 = 66.1 = )01/24.73 –( – 480.2 – = b
01/24.27 – 05.9
852.2
a = − b
) = اﻟﻤﻌﺪل( ﻣﺠﻤﻮع ﻗﯿﻢ yﻋﻠﻰ ﻋﺪدھﺎ = 4.4÷01 = 44.0
) = اﻟﻤﻌﺪل( ﻣﺠﻤﻮع ﻗﯿﻢ xﻋﻠﻰ ﻋﺪدھﺎ = – 15.8 ÷ 01= – 158.0
60.1 = )158.0 – × 437.0( – 44.0 = a
31
14. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
ﻣﻌﺎدﻟﺔ اﻟﺘﻨﺒﻮء
Ŷ =1.065 + 0.73X
وھﺬه ﺗﻤﺜﻞ ﻣﻌﺎدﻟﺔ اﻻﻧﺤﺪار ﻟﻠﻤﺘﻐﯿﺮﯾﻦ ﻟﻮﻏﺎرﯾﺘﯿﻢ اﻟﺠﺮﻋﺔ وﻧﺴﺒﺔ اﻟﮭﻼﻛﺎت اذ ان
اﻟﺤﺮف xﯾﻤﺜﻞ اي ﻗﯿﻤﺔ ﻣﻦ ﻗﯿﻢ اﻟﻤﺘﻐﯿﺮ اﻟﻤﺴﺘﻘﻞ واﻟﺬي ﯾﻤﻜ ﻦ ﺗﻌﻮﯾ ﻀﮫ ﻟﻐ ﺮض
ﺗﻘ ﺪﯾﺮ اﻻﺳ ﺘﺠﺎﺑﺔ ) ، (Ŷﻓﻤ ﺜﻼ ﻣ ﺎھﻲ اﻟﻨ ﺴﺒﺔ اﻟﻤﺘﻮﻗﻌ ﺔ ﻟﻼﺳ ﺘﺠﺎﺑﺔ اذا ﻛ ﺎن
ﻟﻮﻏﺎرﯾﺘﯿﻢ اﻟﺠﺮﻋﺔ ﯾﺴﺎوي – 08.0 :
%84 =84.0 = 485.0 – 560.1 =08.0 – Ŷ = 1.065 + 0.73 x
ﻟﻘ ﺪ اﺳ ﺘﻌﻤﻠﻨﺎ اﻟﻤﻌﺎدﻟ ﺔ اﻟﻤ ﺬﻛﻮرة ﻟﻐ ﺮض ﺗﻘ ﺪﯾﺮ ﻗﯿﻤ ﺔ Ŷﻋﻨ ﺪﻣﺎ ﻛﺎﻧ ﺖ ﻗﯿﻤ ﺔ x
ﻣﻌﻠﻮﻣﺔ ، ﻛﻤﺎ ﯾﻤﻜﻦ اﺳﺘﻌﻤﺎل ﻧﻔﺲ اﻟﻤﻌﺎدﻟﺔ ﻟﺘﻘﺪﯾﺮ ﻗﯿﻤﺔ xﺑﺪﻻﻟﺔ .Ŷ
وﻧﻈﺮا ﻟﻜﻮن ﻣﻄﻠ ﻮب اﻟ ﺴﺆال ھ ﻮ ﺗﻘ ﺪﯾﺮ ﻗﯿﻤ ﺔ 05) LDاي ﻗﯿﻤ ﺔ ( xوھ ﻲ ﻗﯿﻤ ﺔ
ﻣﺠﮭﻮﻟﺔ ﻟﺬا ﻧﻌﻮض ﻋﻦ Ŷﺑﺎﻟﻘﯿﻤﺔ 5.0 وھﻲ ﻗﯿﻤ ﺔ ﻣﻌﻠﻮﻣ ﺔ واﻟﺘ ﻲ ﺗﻤﺜ ﻞ ﻧ ﺼﻒ
اﻟﺤﯿﻮاﻧﺎت اﻟﻤﯿﺘﺔ ﻻﯾﺠﺎد اﻟﺠﺮﻋﺔ اﻟﻤﻨﺎﻇﺮ ﻟﮭﺎ.
)0.5 – 1.065 = 0.734(x
05x = – 0.565/0.734 = – 0.769 = LD
71.0 = )967.0 –( 01Antilog
وﯾﻤﻜ ﻦ ﺗﻘ ﺪﯾﺮ 09 LDﺑ ﺎﻟﺘﻌﻮﯾﺾ ﻋ ﻦ Ŷﺑﺎﻟﻘﯿﻤ ﺔ 09.0 ﻟﻐ ﺮض ﺗﻘ ﺪﯾﺮ xاﻟﺘ ﻲ
ﺗﻤﺜ ﻞ ﻗﯿﻤ ﺔ 09 ، LDاوﯾﻤﻜ ﻦ ﺗﻘ ﺪﯾﺮ 01 LDﺑ ﺎﻟﺘﻌﻮﯾﺾ ﻋ ﻦ Ŷﺑﺎﻟﻘﯿﻤ ﺔ 01.0
وھﻜﺬا ﺑﺎﻟﻨﺴﺒﺔ ﻷي ﻗﯿﻤﺔ ﻧﺮﻏﺐ ﻓﻲ ﺗﻘﺪﯾﺮھﺎ.
ﻣﻼﺣﻈﺔ : ان اﻟﺤﻞ اﻟﯿﺪوي ﻓ ﻲ ﺑﻌ ﺾ اﻟﺘﺠ ﺎرب ﻗ ﺪ ﯾﻜ ﻮن ﺻ ﻌﺐ اﻟﺘﻨﻔﯿ ﺬ وھﻨ ﺎك
اﺣﺘﻤﺎل ﻛﺒﯿﺮ ﻟﻠﻮﻗﻮع ﻓﻲ ﺧﻄﺄ ﻋﻨﺪ اﺟ ﺮاء اﻟﻌﻤﻠﯿ ﺎت اﻟﺤ ﺴﺎﺑﯿﺔ ﻻﺳ ﯿﻤﺎ ﻋﻨ ﺪ زﯾ ﺎدة
ﻋ ﺪد ﻣ ﺴﺘﻮﯾﺎت اﻟﺠ ﺮع ﻟ ﺬا ﻓﻘ ﺪ ﺣﺎوﻟﻨ ﺎ ان ﻧﻮﺿ ﺢ ﻃﺮﯾﻘ ﺔ اﻟﺘﻘ ﺪﯾﺮ ﺑﺎﺳ ﺘﻌﻤﺎل
اﻟﺤﺎﺳ ﻮب وﻻﻛﺜ ﺮ ﻣ ﻦ ﺑﺮﻧ ﺎﻣﺞ اﺣ ﺼﺎﺋﻲ ﻟﺘﻌﻤ ﯿﻢ اﻟﻔﺎﺋ ﺪة ﻻﻛﺒ ﺮ ﻋ ﺪد ﻣﻤﻜ ﻦ ﻣ ﻦ
اﻟﺒ ﺎﺣﺜﯿﻦ واﻟﻄ ﻼب وﻟ ﻀﻤﺎن اﻟﺤ ﺼﻮل ﻋﻠ ﻰ ﺗﻘ ﺪﯾﺮات ﺻ ﺤﯿﺤﺔ ودﻗﯿﻘ ﺔ وﺑﻮﻗ ﺖ
اﻗﺼﺮ.
41
15. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
SAS ب- اﻟﺤﻞ ﺑﺎﺳﺘﻌﻤﺎل ﺑﺮﻧﺎﻣﺞ
ﻟﻐﺮض ﺗﻄﺒﯿﻖ ﻣﻌﺎدﻟﺔ اﻻﻧﺤﺪار اﻟﺨﻄﻲ اﻟﺒ ﺴﯿﻂ واﻟﺤ ﺼﻮل ﻋﻠ ﻰ ﻣﻌﺎدﻟ ﺔ اﻟﺘﻨﺒ ﻮء
.ﻓﻼﺑﺪ ﻣﻦ ﻛﺘﺎﺑﺔ اﻟﺒﯿﺎﻧﺎت ﺑﺎﻟﺼﯿﻐﺔ اﻟﻤﻮﺿﺤﺔ ادﻧﺎه ﺛﻢ اﻋﻄﺎء اﯾﻌﺎز اﻟﺘﻨﻔﯿﺬ
:(2 )ﻣﺜﺎل
data s;
input x y;
cards;
-1.69
0
-1.39
0.06
-1.22
0.10
-1.09
0.16
-1
0.26
-0.69
0.40
-0.52
0.66
-0.39
0.86
-0.30
0.90
-0.22
1
proc reg;
model y=x;
run;
The SAS System
Model: MODEL1
Dependent Variable: Y
Source
DF
Model
Error
C Total
1
8
9
Analysis of Variance
Sum of
Mean
Squares
Square
1.21946
0.09654
1.31600
Root MSE
Dep Mean
C.V.
Variable
DF
INTERCEP
X
1
1
1.21946
0.01207
0.10985
0.44000
24.96649
F Value
101.052
R-square
Adj R-sq
Parameter Estimates
Parameter
Standard
T for H0:
Estimate
Error
Parameter=0
1.065157
0.734614
0.07123385
0.07307785
15
14.953
10.052
Prob>F
0.0001
0.9266
0.9175
Prob > |T|
0.0001
0.0001
16. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
وﻣﻦ اﻟﻨﺘﺎﺋﺞ اﻋﻼه ﯾﺘﻀﺢ ﻟﻨﺎ ﺑﺄن ﻣﻌﺎدﻟﺔ اﻻﻧﺤﺪار اﻟﺨﻄﻲ اﻟﺒﺴﯿﻂ ھﻲ :
Ŷ = 1.065 + 0.73X
ﺛﻢ ﻧﻘﻮم ﺑﺘﻄﺒﯿﻖ اﻟﺨﻄﻮات اﻟﻤﺬﻛﻮرة ﻓﻲ اﻟﺤﻞ اﻟﯿﺪوي ﻻﺳﺘﺨﺮاج ﻗﯿﻤﺔ 05.LD
ﺠ- اﻟﺤﻞ ﺑﺎﺳﺘﻌﻤﺎل ﺑﺮﻧﺎﻣﺞ Minitab
ﻧﺪرج اﻟﺒﯿﺎﻧﺎت )ﻣﺜﺎل 2( ﻓﻲ اﻟﺼﻔﺤﺔ اﻟﺮﺋﯿﺴﺔ ﻟﻠﺒﺮﻧﺎﻣﺞ وﻛﻤﺎ ﻣﻮﺿﺢ ادﻧﺎه:
Y
X
0
60.0
01.0
61.0
62.0
04.0
66.0
68.0
09.0
1
96.1-
93.1-
22.1-
90.1-
1-
96.0-
25.0-
93.0-
03.0-
22.0-
ﺛ ﻢ ﻧﻨﻘ ﺮ ﻋﻠ ﻰ اﯾﻘﻮﻧ ﺔ statﻓ ﻲ ﺷ ﺮﯾﻂ اﻟﻤﮭ ﺎم ﻓﻨﺤ ﺼﻞ ﻋﻠ ﻰ اﺧﺘﯿ ﺎرات ﻧﺆﺷ ﺮ
اﻻﺧﺘﯿ ﺎر regressionﻓﺘﻈﮭ ﺮ اﺧﺘﯿ ﺎرات اﺧ ﺮى ﻧﺆﺷ ﺮ ﻣﻨﮭ ﺎ اﻻﺧﺘﯿ ﺎر
Fitted line plotﻓﯿﻈﮭ ﺮ ﺻ ﻨﺪوق ﺣ ﻮار ﻓﯿ ﮫ ﻣ ﺴﺘﻄﯿﻼن ﻋﻠ ﻰ ﺟﮭ ﺔ اﻟﯿﻤ ﯿﻦ
اﻻول ﻣﺆﺷ ﺮ اﻣﺎﻣ ﮫ ) response (Yواﻟﺜ ﺎﻧﻲ ﻣﺆﺷ ﺮ اﻣﺎﻣ ﮫ )predictor (x
وﻋﻠ ﻰ ﺟﮭ ﺔ اﻟﯿ ﺴﺎر ﻧﻼﺣ ﻆ وﺟ ﻮد ﻣ ﺴﺘﻄﯿﻞ ﯾﺘ ﻀﻤﻦ اﻟﻤﺘﻐﯿ ﺮان xو yﺛ ﻢ ﻧﻨﻘ ﺮ
ﻋﻠ ﻰ xﻓﯿﻈﮭ ﺮ ﻣ ﺴﺘﻄﯿﻞ ﯾﺤ ﻮي ﻛﻠﻤ ﺔ selectﻧﻨﻘ ﺮ ﻋﻠﯿ ﮫ ﻓﯿﻨﺘﻘ ﻞ اﻟﻤﺘﻐﯿ ﺮ xاﻟ ﻰ
اﻟﺤﻘ ﻞ اﻟﻤﺆﺷ ﺮ ﻋﻠﯿ ﮫ responseﺛ ﻢ ﻧﻨﻘ ﺮ ﻋﻠ ﻰ اﻟﻤﺘﻐﯿ ﺮ اﻟﺜ ﺎﻧﻲ وﺑﻌ ﺪھﺎ ﻧﻨﻘ ﺮ
selectﻟﻨﻘ ﻞ اﻟﻤﺘﻐﯿ ﺮ اﻟ ﻰ اﻟﻤﺮﺑ ﻊ اﻟﻤﺆﺷ ﺮ اﻣﺎﻣ ﮫ predictorوﻧﺆﺷ ﺮﻧﻮع
اﻻﻧﺤ ﺪار Linearﻣ ﻦ ﺻ ﻨﺪوق اﻟﺤ ﻮارﺛﻢ ﻧﻨﻘ ﺮ okﻓﺘﻈﮭ ﺮ اﻟﻨﺘ ﺎﺋﺞ ﻋﻠ ﻰ ﺷ ﻜﻞ
رﺳﻢ ﺑﯿﺎﻧﻲ ﯾﺘﻀﻤﻦ ﻣﻌﺎدﻟﺔ اﻟﺘﻨﺒﻮء .
61
17. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
Regression Analysis: y versus x
The regression equation is
ﻣﻌﺎدﻟﺔ اﻟﺘﻨﺒﻮء )ﻣﻌﺎدﻟﺔ اﻻﻧﺤﺪار( ←
% 7.19 = )R-Sq(adj
y = 1.06516 + 0.734614 x
358901.0 = S
% 7.29 = R-Sq
Analysis of Variance
P
000.0
F
250.101
SS
64912.1
45690.0
00613.1
MS
64912.1
70210.0
Source
Regression
Error
Total
DF
1
8
9
Fitted Line Plot: y versus x
Fitted Line Plot
x = 1.065 + 0.7346 ldose
358901.0
%7.29
%7.19
S
R-Sq
)R-Sq(adj
00.1
57.0
05.0
x
52.0
00.0
0.0
2.0-
4.0-
6.0-
8.0- 0.1-
ldose
2.1-
4.1-
6.1-
8.1-
ﺷﻜﻞ 6: ﻣﻌﺎدﻟﺔ اﻟﺨﻂ اﻟﻤﺴﺘﻘﯿﻢ ﻟﻮﺻﻒ اﻻﺳﺘﺠﺎﺑﺔ
ﯾﻤﻜﻦ ﺗﻘﺪﯾﺮ ﻗﯿﻤﺔ 05 LDاﻋﺘﻤﺎدا ﻋﻠﻰ اﻟﺮﺳﻢ اﻋﻼه وذﻟﻚ ﺑﻌﻤ ﻞ ﺧ ﻂ ﻣ ﺴﺘﻘﯿﻢ ﻣ ﻦ
اﻟﻤﺤﻮر اﻟﺼﺎدي وﻋﻨﺪ اﻟﻘﯿﻤﺔ 5.0 ﺑﺼﻮرة ﻣﻮازﯾﺔ ﻟﻠﻤﺤﻮر اﻟﺴﯿﻨﻲ ﻓﯿﻘﻄ ﻊ اﻟﺨ ﻂ
اﻟﻤﺴﺘﻘﯿﻢ وﻣﻦ ﻧﻘﻄﺔ اﻟﺘﻘﺎﻃﻊ ﻧﺮﺳﻢ ﻣﺴﺘﻘﯿﻢ ﻋﻤﻮدﯾ ﺎ ﻋﻠ ﻰ اﻟﻤﺤ ﻮر اﻟ ﺴﯿﻨﻲ وﺗﻜ ﻮن
ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ ذﻟﻚ اﻟﻌﻤﻮد ﻣﻤﺜﻠﺔ ﻟﻘﯿﻤﺔ 05.LD
71
18. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
د- اﻟﺤﻞ ﺑﺎﺳﺘﻌﻤﺎل ﺑﺮﻧﺎﻣﺞ SPSS
ﺗﺪرج اﻟﺒﯿﺎﻧﺎت )ﻣﺜﺎل 2( اﻟﺼﻔﺤﺔ اﻟﺮﺋﯿ ﺴﯿﺔ ﻟﻠﺒﺮﻧ ﺎﻣﺞ ﺑﻌ ﺪ ﺗﻌﺮﯾ ﻒ اﻟﻤﺘﻐﯿ ﺮات ﺛ ﻢ
ﻧﻨﻘﺮ ﻋﻠ ﻰ اﻻﯾﻘﻮﻧ ﺔ analyzeﻓﺘﻈﮭ ﺮ ﻋ ﺪة اﺧﺘﯿ ﺎرات ﻧﺨﺘ ﺎر ﻣﻨﮭ ﺎ regression
ﻓﺘﻈﮭ ﺮ اﺧﺘﯿ ﺎرات اﺧ ﺮى ﻧﺨﺘ ﺎر ﻣﻨﮭ ﺎ curve estimationﻓﯿﻈﮭ ﺮ ﺻ ﻨﺪوق
ﺣﻮار ﻣﺆﺷﺮ ﻓﯿﮫ ﻋﻠﻰ اﺣﺪ اﻟﺤﻘﻮل dependentﻧﻀﻊ ﻓﯿﮫ Yوﻧﻀﻊ ﻓ ﻲ اﻟﺤﻘ ﻞ
اﻟﻤﺆﺷ ﺮ ﻓﯿ ﮫ independentاﻟﻤﺘﻐﯿ ﺮ Xوﻓ ﻲ اﻟﺤﻘ ﻮل اﻟﺘ ﻲ ﺗﺘ ﻀﻤﻦ اﻟﻜﻠﻤ ﺔ
Modelsﻧﺨﺘ ﺎر Linearﻟﻐ ﺮض اﻟﺤ ﺼﻮل ﻋﻠ ﻰ اﻟﺮﺳ ﻢ اﻟﺒﯿ ﺎﻧﻲ ﺛ ﻢ ﻧﻨﻘ ﺮ ﻋﻠ ﻰ
ﻛﻠﻤ ﺔ okوﺑﻌ ﺪ اﻟﺘﻨﻔﯿ ﺬ ﻧﺘﺒ ﻊ ﻧﻔ ﺲ اﻟﺨﻄ ﻮات ﺛ ﻢ ﻧﺨﺘ ﺎر regressionﻟﻠﺤ ﺼﻮل
ﻋﻠﻰ ﻣﻌﺎدﻟﺔ اﻟﺘﻨﺒﻮء.
Unstandardized
Coefficients
Model
x
1
)(Constant
B
560.1
Std.Error
170.
537.
370.
Standardized
Coefficients
.Sig
t
Beta
359.41
369.
000.
250.01
000.
a Dependent Variable: y
Y
Observed
00.1
Linear
08.0
06.0
04.0
02.0
00.0
05.0-
00.1-
05.1-
X
ﺷﻜﻞ 6: ﻣﻌﺎدﻟﺔ اﻟﺨﻂ اﻟﻤﺴﺘﻘﯿﻢ ﻟﻮﺻﻒ اﻻﺳﺘﺠﺎﺑﺔ
81
19. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
ھ- اﻟﺤﻞ ﺑﺎﺳﺘﻌﻤﺎل ﺑﺮﻧﺎﻣﺞ STATISTICA
ﻟﻐﺮض ادراج اﻟﺒﯿﺎﻧﺎت ﻻﺑﺪ ﻣﻦ ﺗﺤﺪﯾﺪ اﺳﻤﺎء اﻟﻤﺘﻐﯿﺮات وﯾﻤﻜﻦ اﺟ ﺮاء ذﻟ ﻚ ﻣ ﻦ
ﺧﻼل اﻟﻨﻘﺮ ﻋﻠ ﻰ اول ﺧﻠﯿ ﺔ ﻓ ﻲ اﻟﻌﻤ ﻮد اﻻول ﻓﯿﻈﮭ ﺮ ﻣﺮﺑ ﻊ ﺣ ﻮار ﻧﻘ ﻮم ﺑﺘ ﺪوﯾﻦ
اﺳﻢ اﻟﻤﺘﻐﯿﺮ ﻓﻲ اﻟﺤﻘﻞ اﻟﻤﺆﺷﺮ اﻣﺎﻣﮫ nameﺛﻢ ﻧﻀﻐﻂ okوﻧﻘﻮم ﺑ ﻨﻔﺲ اﻟﻌﻤﻠﯿ ﺔ
ﻋﻠﻰ اﻟﻌﻤﻮد اﻟﺜﺎﻧﻲ ﺛﻢ ﻧﻨﻘﺮ ﻋﻠﻰ اﻻﺧﺘﯿﺎر Graphsﻓﻲ ﺷﺮﯾﻂ اﻟﻤﮭﺎم ← stats
Line plots (variable) ←2D Graphsﻓﯿﻈﮭﺮ ﻣﺮﺑﻊ ﺣ ﻮار ، ﻧﺆﺷ ﺮ ﻋﻠ ﻰ
Linearﻓ ﻲ اﻻﺧﺘﯿ ﺎر Fitوﻧﺆﺷ ﺮ ﻋﻠ ﻰ trace
ﻓ ﻲ اﻻﺧﺘﯿ ﺎر
xy
Graph typeﺛ ﻢ ﻧﻨﻘ ﺮ ﻋﻠ ﻰ زر Variablesﻓﯿﻈﮭ ﺮ ﻣﺮﺑ ﻊ ﺣ ﻮار ﻓﯿ ﮫ اﺳ ﻤﺎء
اﻟﻤﺘﻐﯿ ﺮان وﻣ ﺮﺑﻌﯿﻦ ﯾﺘ ﻀﻤﻦ اﺣ ﺪھﻤﺎ xوﻧ ﻀﻊ ﻓﯿ ﮫ اﻟﻌﺎﻣ ﻞ اﻟﻤ ﺴﺘﻘﻞ logdose
واﻻﺧﺮ yوﻧﻀﻊ ﻓﯿﮫ ratioﺛ ﻢ okﻓﯿﺨﺘﻔ ﻲ اﻟﻤﺮﺑ ﻊ ﺛ ﻢ okﻓﺘﻈﮭ ﺮ اﻟﻨﺘ ﺎﺋﺞ ﻋﻠ ﻰ
ﺷﻜﻞ رﺳﻢ ﺑﯿﺎﻧﻲ ﻣﻊ ﻣﻌﺎدﻟﺔ اﻻﻧﺤﺪار.
)Line Plot (YOUDEN.STA 7v*12c
y=1.065+0.735*x+eps
2.1
0.1
8.0
4.0
2.0
0.0
0.0
2.0-
4.0-
6.0-
8.0-
0.1-
2.1-
4.1-
LOGDOSE
ﺷﻜﻞ 7: ﻣﻌﺎدﻟﺔ اﻟﺨﻂ اﻟﻤﺴﺘﻘﯿﻢ ﻟﻮﺻﻒ اﻻﺳﺘﺠﺎﺑﺔ
91
6.1-
2.0-
8.1-
RATIO
6.0
20. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
STATGRAPHICS Plusٍ و- اﻟﺤﻞ ﺑﺎﺳﺘﻌﻤﺎل ﺑﺮﻧﺎﻣﺞ
ﻧﺤﺪد اﺳﻢ ﻛﻞ ﻣﺘﻐﯿﺮ وذﻟﻚ ﺑﺘﺎﺷﯿﺮ اﻟﻌﻤﻮد اﻻول وﻋﻤﻞ ﻛﻠﻚ اﯾﻤﻦ ﻓﯿﻈﮭ ﺮ ﺷ ﺮﯾﻂ
ﻓﯿﻈﮭ ﺮ ﻣﺮﺑ ﻊ ﺣ ﻮار ﻧﺜﺒ ﺖModify column اﺧﺘﯿ ﺎرات ﻧﺆﺷ ﺮ ﻋﻠ ﻰ اﻻﺧﺘﯿ ﺎر
. ﺛ ﻢok ﺛ ﻢnumeric ﻋﻠ ﻰ اﻻﺧﺘﯿ ﺎرtype ﻓﯿ ﮫ اﺳ ﻢ اﻟﻤﺘﻐﯿ ﺮ وﻧﺆﺷ ﺮ ﻓ ﻲ ﺣﻘ ﻞ
ﻓﯿﻈﮭ ﺮx – y plots
← scatterplot ← plot ﻧﺨﺘ ﺎر ﻣ ﻦ ﺷ ﺮﯾﻂ اﻟﻘ ﻮاﺋﻢ
x ﻓ ﻲ اﻟﺤﻘ ﻞlogdose وﻧﻀﻊy ﻓﻲ اﻟﺤﻘﻞratio ﻣﺮﺑﻊ ﯾﺘﻀﻤﻦ اﻟﻌﺎﻣﻠﯿﻦ ﻧﻀﻊ
. ﻓﺘﻈﮭﺮ اﻟﻨﺘﺎﺋﺞ ﻣﻊ اﻟﺮﺳﻢok ﺛﻢ ﻧﻨﻘﺮ ﻋﻠﻰ
Multiple Regression Analysis
----------------------------------------------------------------------------Dependent variable: ratio
----------------------------------------------------------------------------Standard
T
Parameter
Estimate
Error
Statistic
P-Value
---------------------------------------------------------- -----------------------CONSTANT
1.06516
0.0712339
14.953
0.0000
logdose
0.734614 0.0730779
10.0525
0.0000
---------------------------------------------------------- ------------------------Analysis of Variance
-----------------------------------------------------------------------------------Source
Sum of Squares Df Mean Square F-Ratio P-Value
----------------------------------------------------------- ------------------------Model
1.21946
1 1.21946
101.05
0.0000
Residual
0.0965407
8 0.0120676
-----------------------------------------------------------------------------------Total (Corr.)
1.316
9
R-squared = 92.6641 percent
R-squared (adjusted for d.f.) = 91.7471 percent
Standard Error of Est. = 0.109853
Mean absolute error = 0.0850655
Durbin-Watson statistic = 0.772648
= 560.1 وﺗﻤﺜ ﻞ اﻟﻘ ﯿﻢ اﻟﺘ ﻲb = 37.0 وﻗﯿﻤ ﺔa وﻧﻼﺣ ﻆ ﻣ ﻦ اﻟﻨﺘ ﺎﺋﺞ ان ﻗﯿﻤ ﺔ
y = 1.065 + 0.73 x
ﺗﺤﺘﮭﺎ ﺧﻂ ، وﺑﺬﻟﻚ ﻓﺎن ﻣﻌﺎدﻟﺔ اﻟﺘﻨﺒﻮء ﺳﺘﻜﻮن
20
21. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
Component+Residual Plot for ratio
7.0
3.0
1.0
1.0-
3.0-
component effect
5.0
5.0-
2.0-
5.0-
8.0-
1.1-
4.1-
7.1-
logdose
ﺷﻜﻞ 8: ﻣﻌﺎدﻟﺔ اﻟﺨﻂ اﻟﻤﺴﺘﻘﯿﻢ ﻟﻮﺻﻒ اﻻﺳﺘﺠﺎﺑﺔ
2-2-2 ﻃﺮﯾﻘﺔ Probit
ﺗﻌﺪ ﻣﻦ اھﻢ ﻃﺮق اﻟﺘﻘﺪﯾﺮ واﻛﺜﺮھﺎ ﺷﯿﻮﻋﺎ وھﻲ ﺗﻤﺜﻞ اﺣ ﺪ اﻧ ﻮاع اﻻﻧﺤ ﺪار اﻟ ﺬي
ﯾ ﺴﺘﻌﻤﻞ ﻟﺘﺤﻠﯿ ﻞ ﻣﺘﻐﯿ ﺮات اﻻﺳ ﺘﺠﺎﺑﺔ اﻟﺜﻨﺎﺋﯿ ﺔ ) ( Binomialوﻓﯿ ﮫ ﯾ ﺘﻢ ﺗﺤﻮﯾ ﻞ
ﺑﯿﺎﻧ ﺎت اﻻﺳ ﺘﺠﺎﺑﺔ ﻓ ﻲ ﻣﻨﺤﻨ ﻰ اﻻﺳ ﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋ ﺔ اﻟ ﻰ اﻟﻮﺣ ﺪات اﻻﺣﺘﻤﺎﻟﯿ ﺔ
ﻟﻠﺤﺼﻮل ﻋﻠﻰ ﻋﻼﻗﺔ ﺧﻄﯿﺔ ﺗﻘﺮﯾﺒﺎ ﻟﻜﻲ ﯾﺘﻢ ﺗﺤﻠﯿﻠﮭﺎ ﺑﺎﻻﻧﺤﺪار اﻟﺨﻄ ﻲ ﺑﺎﺳ ﺘﻌﻤﺎل
ﻃﺮﯾﻘﺔ اﻟﻤﺮﺑﻌﺎت اﻟﺼﻐﺮى ) (Least Squaresاو ﻃﺮﯾﻘﺔ ﺗﻌﻈﯿﻢ اﻻﺣﺘﻤ ﺎﻻت
) ( Maximum Likelihoodوﯾﻤﻜ ﻦ ﺗﻄﺒﯿ ﻖ ھ ﺬه اﻟﻄﺮﯾﻘ ﺔ ﺑﺘﺤﻮﯾ ﻞ ﻧ ﺴﺒﺔ
اﻟﮭﻼﻛﺎت اﻟﻰ ﻗﯿﻢ اﺣﺘﻤﺎﻟﯿﺔ اﻋﺘﻤﺎدا ﻋﻠﻰ ﺟﺪول ﻗﯿﻢ اﻟﻮﺣ ﺪات اﻻﺣﺘﻤﺎﻟﯿ ﺔ وﻣ ﻦ ﺛ ﻢ
ﺗﻄﺒﯿ ﻖ ﻣﻌﺎدﻟ ﺔ اﻻﻧﺤ ﺪار اﻟﺨﻄ ﻲ اﻟﺒ ﺴﯿﻂ ، ﻛﻤ ﺎ ﯾﻤﻜ ﻦ ﺗﻄﺒﯿ ﻖ ﻃﺮﯾﻘ ﺔ probit
ﺑﺎﺳﺘﻌﻤﺎل اﻟﺒﺮاﻣﺞ اﻻﺣﺼﺎﺋﯿﺔ دون اﺟﺮاء ﺗﺤﻮﯾﻞ ﻟﻨﺴﺒﺔ اﻟﮭﻼﻛ ﺎت اﻟ ﻰ اﻟﻮﺣ ﺪات
12
22. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
اﻻﺣﺘﻤﺎﻟﯿﺔ اذ ان ﻋﻤﻠﯿﺔ اﻟﺘﺤﻮﯾﻞ ﺗﺠﺮى ﺗﻠﻘﺎﺋﯿﺎ ﻓ ﻲ ﺗﻠ ﻚ اﻟﺒ ﺮاﻣﺞ. وﺳ ﻨﺤﺎول ﺑﺪاﯾ ﺔ
ﺗﻮﺿﯿﺢ ﻃﺮﯾﻘﺔ ﺗﺤﻮﯾﻞ اﻟﻨﺴﺐ اﻟﻰ اﻟﻮﺣﺪات اﻻﺣﺘﻤﺎﻟﯿﺔ.
2-2-2-1- ﺗﻘﺪﯾﺮ اﻟﻮﺣﺪات اﻻﺣﺘﻤﺎﻟﯿﺔ Estimation of probit units
ان ﺗﺤﻮﯾﻞ ﻧﺴﺒﺔ اﻟﺤﯿﻮاﻧﺎت اﻟﻤﯿﺘ ﺔ اﻋﺘﻤ ﺎدا ﻋﻠ ﻰ اﻟﺠ ﺪول )1( اﻟﺨ ﺎص ﺑﺎﻟﻮﺣ ﺪات
اﻻﺣﺘﻤﺎﻟﯿﺔ ﯾﺘﻢ ﻛﺎﻻﺗﻲ : ﻟﻮ ﻓﺮﺿﻨﺎ ان ﻧﺴﺒﺔ اﻟﺤﯿﻮاﻧﺎت اﻟﻤﯿﺘﺔ ﻓ ﻲ اﻟﻤ ﺴﺘﻮى اﻻول
ﻟﻠﺠﺮﻋ ﺔ ﻛ ﺎن 02% ﻓ ﺄن اﻟ ﺮﻗﻢ اﻟﻤﻨ ﺎﻇﺮ ﻟﮭ ﺎ ﻓ ﻲ اﻟﺠ ﺪول ھ ﻮ 61.4 اﻟ ﺬي ﯾﻤﺜ ﻞ
ﺗﻘﺎﻃﻊ اﻟﻨﺴﺒﺔ 02% ﻣﻊ اﻟﻘﯿﻤﺔ ﺻﻔﺮ ﻓﻲ اﻟﻌﻤﻮد اﻟﺜﺎﻧﻲ ﻟﻠﺠﺪول وﻟﻮ ﻛﺎﻧ ﺖ اﻟﻨ ﺴﺒﺔ
33% ﻓ ﺎن اﻟ ﺮﻗﻢ اﻟﻤﻨ ﺎﻇﺮ ﺳ ﯿﻜﻮن ﺗﻘ ﺎﻃﻊ 03% ﻣ ﻊ اﻟ ﺮﻗﻢ 3 ﻓ ﻲ اﻟ ﺼﻒ اﻻول
ﻟﻠﺠﺪول واﻟﺘﻲ ﺗﻤﺜﻞ 65.4 وھﻜﺬا.
ﺟﺪول )1( ﻗﯿﻢ اﻟﻮﺣﺪات اﻻﺣﺘﻤﺎﻟﯿﺔ
9
8
7
6
5
4
3
2
1
0
%
66.3
95.3
25.3
54.3
63.3
52.3
21.3
59.2
76.2
-
0
21.4
80.4
50.4
10.4
69.3
29.3
78.3
28.3
77.3
↓27.3
01
54.4
24.4
93.4
63.4
33.4
92.4
62.4
32.4
91.4
61.4
→02
27.4
96.4
76.4
46.4
16.4
95.4
65.4
35.4
05.4
84.4
03
79.4
59.4
29.4
09.4
78.4
58.4
28.4
08.4
77.4
57.4
04
32.5
02.5
81.5
51.5
31.5
01.5
80.5
50.5
30.5
00.5
05
05.5
74.5
44.5
14.5
93.5
63.5
33.5
13.5
82.5
52.5
06
18.5
77.5
47.5
17.5
76.5
46.5
16.5
85.5
55.5
25.5
07
32.6
81.6
31.6
80.6
40.6
99.5
59.5
29.5
88.5
48.5
08
33.7
50.7
88.6
57.6
46.6
55.6
84.6
14.6
43.6
82.6
09
اﻟﻤﺼﺪر: :)2591 (Finney
ﺑﺎﺗﺒﺎع ﻧﻔﺲ اﻟﻄﺮﯾﻘﺔ ﺳﻨﺠﺪ ان اﻟﻘﯿﻢ اﻻﺣﺘﻤﺎﻟﯿﺔ ﻟﻤﺠﻤﻮﻋﺔ ﻣ ﻦ اﻟﻨ ﺴﺐ اﻻﻓﺘﺮاﺿ ﯿﺔ
اﻋﺘﻤﺎدا ﻋﻠﻰ اﻟﺠﺪول اﻋﻼه ﺳﺘﻜﻮن ﻛﻤﺎ ﯾﻠﻲ:
22
23. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
اﻟﻘﯿﻤﺔ اﻻﺣﺘﻤﺎﻟﯿﺔ
ﻧﺴﺒﺔ اﻟﺤﯿﻮاﻧﺎت اﻟﻤﯿﺘﺔ
73%
76.4
54%
78.4
97%
18.5
99%
33.7
ﺳﺘﻜﻮن ﻧﺴﺐ اﻟﺤﯿﻮاﻧﺎت اﻟﻤﯿﺘ ﺔ )ﻣﺜ ﺎل 2( ﺑﻌ ﺪ ﺗﺤﻮﯾﻠﮭ ﺎ اﻟ ﻰ اﻟﻮﺣ ﺪات اﻻﺣﺘﻤﺎﻟﯿ ﺔ
اﻋﺘﻤﺎدا ﻋﻠﻰ ﺟﺪول ﻗﯿﻢ اﻟﻮﺣﺪات اﻻﺣﺘﻤﺎﻟﯿﺔ ﻛﺎﻻﺗﻲ:
ﺟﺪول ﺑﯿﺎﻧﺎت اﻟﻤﺜﺎل 2 ﺑﻌﺪ ﺗﺤﻮﯾﻠﮭﺎ اﻟﻰ اﻟﻮﺣﺪات اﻻﺣﺘﻤﺎﻟﯿﺔ
اﻟﺠﺮﻋﺔ
ﻟﻮﻏﺎرﺗﯿﻢ اﺳﺎس
01
)(x
اﻟﻮﺣﺪات
اﻻﺣﺘﻤﺎﻟﯿﺔ
)(y
ﻋﺪد اﻟﻜﺎﺋﻨﺎت
اﻟﻤﯿﺘﺔ
اﻟﻨﺴﺒﺔ
20.0
40.0
60.0
80.0
01.0
02.0
03.0
04.0
05.0
06.0
– 96.1
– 93.1
– 22.1
– 90.1
–1
– 96.0
– 25.0
– 93.0
– 03.0
– 22.0
0
54.3
27.3
10.4
63.4
57.4
14.5
80.6
82.6
33.7
0
2
3
5
8
21
02
62
72
03
0
60.0
01.0
61.0
62.0
04.0
66.0
68.0
09.0
00.1
15.8 – = ∑x = ldose
93.54 = ∑y
32
24. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
أ- اﻟﺤﻞ اﻟﯿﺪوي ﺑﻄﺮﯾﻘﺔ Probit
ﻟﺤﻞ ﺑﯿﺎﻧﺎت اﻟﻤﺜﺎل 2 ﯾﻤﻜﻦ اﺳﺘﻌﻤﺎل اﻟﺤﺎﺳﺒﺔ اﻟﯿﺪوﯾﺔ ﻟﺤﺴﺎب ﻣﺎﯾﻠﻲ:
93.54 = 33.7 +……54.3 + 0 = ∑y
ﻣﻌﺪل = ﻣﺠﻤﻮع ﻗﯿﻢ ) yﻣﺠﻤﻮع اﻟﻮﺣﺪات اﻻﺣﺘﻤﺎﻟﯿﺔ( ﻣﻘﺴﻮﻣﺎ ﻋﻠﻰ ﻋﺪدھﺎ
935.4 =01÷ 93.54
15.8 – = )22.0 –( + ..… )93.1 –( + 96.1 – = ∑x = ldose
ﻣﻌﺪل = ﻣﺠﻤﻮع ﻗﯿﻢ ) xﻟﻮﻏﺎرﯾﺘﯿﻢ اﻟﺠﺮﻋﺔ( ﻣﻘﺴﻮﻣﺎ ﻋﻠﻰ ﻋﺪدھﺎ
158.0 – = 01 ÷ 15.8 –
7105.9 = ²)22.0 – ( +..…… + ²)93.1–( + ²)96.1–( = ²∑x
1024.27 = ²)15.8 –( = ²)(∑ x
)33.7 ∑ xy = (–1.69 x 0) + ( –1.39 x 3.45) + …..( – 0.22 x
3320.03 – =
01 = n
01/])93.54()15.8 –([ – 3320.03 – =b
]01/1024.27[ – 7105.9
4708.3=b
a = − b
]158.0 – × 4708.3 [ – 935.4 = a
977.7 = a
ﻣﻌﺎدﻟﺔ اﻟﺘﻨﺒﻮء…………… Ŷ = 7.779 + 3.80 x
5= 7.779 + 3.80x
37.0 – = 08.3/977.2 – =x
681.0 = )37.0 –( 01 Anti log
681.0 = 05LD
42
25. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
SAS ب- اﻟﺤﻞ ﺑﺎﺳﺘﻌﻤﺎل ﺑﺮﻧﺎﻣﺞ
ﻧﺤ ﻮل ﻧ ﺴﺐ اﻟﺤﯿﻮاﻧ ﺎت اﻟﻤﯿﺘ ﺔ )ﻣﺜ ﺎل 2( اﻟ ﻰ اﻟﻮﺣ ﺪات اﻻﺣﺘﻤﺎﻟﯿ ﺔ ﺗ ﻢ ﻧﻄﺒ ﻖ
: ﻻﺳﺘﺨﺮاج ﻣﻌﺎدﻟﺔ اﻟﺘﻨﺒﻮء ) ﻣﻌﺎدﻟﺔ اﻻﻧﺤﺪاراﻟﺨﻄﻲ( وﻛﺎﻻﺗﻲSAS ﺑﺮﻧﺎﻣﺞ
data s;
input ldose y;
N=30;
cards;
-1.69 0
-1.39 3.45
-1.22 3.72
-1.09 4.01
-1
4.36
-0.69 4.75
-0.52 5.41
-0.39 6.08
-0.30 6.28
-0.22 7.33
proc reg;
model y=ldose;
run;
The SAS System
Model: MODEL1
Dependent Variable: Y
Analysis of Variance
Source
Model
Error
C Total
DF
1
8
9
Root MSE
Dep Mean
Sum of
Squares
32.75748
4.01221
36.76969
0.70818
4.53900
C.V.
Mean
Square
32.75748
0.50153
F Value
65.316
R-square
Adj R-sq
15.60222
Prob>F
0.0001
0.8909
0.8772
Parameter Estimates
Variable
DF
Parameter
Estimate
Standard
Error
T for H0:
Parameter=0
Prob > |T|
INTERCEP
LDOSE
1
1
7.779115
3.807420
0.45922212
0.47110980
16.940
8.082
0.0001
0.0001
25
26. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
وﺑﺬﻟﻚ ﻓﺎن ﻣﻌﺎدﻟﺔ اﻟﺘﻨﺒﻮء ھﻲ:
Ŷ = 7.779 + 3.80 x
ﺛﻢ ﻧﺘﺒﻊ ﻧﻔﺲ اﻟﺨﻄﻮات ﻓﻲ اﻟﺤﻞ اﻟﯿﺪوي ﻟﻐﺮض اﻟﺘﻮﺻﻞ اﻟﻰ ﻗﯿﻤﺔ
05.LD
ﺠ- اﻟﺤﻞ ﺑﺎﺳﺘﻌﻤﺎل اﻟﺒﺮﻧﺎﻣﺞ Minitab
ﻧﺪرج ﺑﯿﺎﻧﺎت اﻟﻤﺜﺎل 2 اﻟﺨﺎﺻﺔ ﺑﻘﯿﻢ اﻟﻠﻮﻏ ﺎرﯾﺘﯿﻢ واﻟﻮﺣ ﺪات اﻻﺣﺘﻤﺎﻟﯿ ﺔ ، ﺛ ﻢ ﻧﻨﻔ ﺬ
ﺑﻨﻔﺲ اﻻﺳﻠﻮب اﻟﺴﺎﺑﻖ اﻟﺨﺎص ﺑﺘﻘﺪﯾﺮ ﻣﻌﺎدﻟﺔ اﻻﻧﺤﺪار.
Logdose probit
96.1-
0
93.1-
54.3
22.1-
27.3
90.1-
10.4
1-
63.4
96.0-
57.4
25.0-
14.5
93.0-
80.6
03.0-
82.6
22.0-
33.7
The regression equation is
probit = 7.78 + 3.81 Logdose
P
T
SE Coef
Coef
000.0
49.61
2954.0
1977.7
Constant
000.0
80.8
1174.0
4708.3
Logdose
%7.78 = )R-Sq(adj
%1.98 = R-Sq
62
Predictor
2807.0 = S
27. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
Fitted Line Plot
x = 7.779 + 3.807 ldose
581807.0
%1.98
%7.78
8
S
R-Sq
)R-Sq(adj
7
6
5
x
4
3
2
1
0
0.0
2.0-
4.0-
6.0-
8.0- 0.1-
ldose
2.1-
4.1-
6.1-
8.1-
ﺷﻜﻞ 9: ﻣﻌﺎدﻟﺔ اﻟﺨﻂ اﻟﻤﺴﺘﻘﯿﻢ ﻟﻮﺻﻒ اﻻﺳﺘﺠﺎﺑﺔ
د- اﻟﺤﻞ ﺑﺎﺳﺘﻌﻤﺎل اﻟﺒﺮﻧﺎﻣﺞ SPSS
ﻧﺪرج ﺑﯿﺎﻧﺎت اﻟﻤﺜﺎل 2 اﻟﺨﺎﺻﺔ ﺑﻘﯿﻢ اﻟﻠﻮﻏ ﺎرﯾﺘﯿﻢ واﻟﻮﺣ ﺪات اﻻﺣﺘﻤﺎﻟﯿ ﺔ ، ﺛ ﻢ ﻧﻨﻔ ﺬ
ﺑﻨﻔﺲ اﻻﺳﻠﻮب اﻟﺴﺎﺑﻖ اﻟﺨﺎص ﺑﺘﻘﺪﯾﺮ ﻣﻌﺎدﻟﺔ اﻻﻧﺤﺪار.
)Coefficients(a
Unstandardized
Coefficients
.Std
Error
B
Model
1
Standardized
Coefficients
t
Beta
)(Constant
977.7
954.
logdose
708.3
280.8
449.
174.
a Dependent Variable: probit
72
.Sig
049.61
000.
000.
28. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
probit
Observed
00.8
Linear
00.6
00.4
00.2
00.0
05.0-
05.1-
00.1-
logdose
ﺷﻜﻞ 01: ﻣﻌﺎدﻟﺔ اﻟﺨﻂ اﻟﻤﺴﺘﻘﯿﻢ ﻟﻮﺻﻒ اﻻﺳﺘﺠﺎﺑﺔ
ھ- اﻟﺤﻞ ﺑﺎﺳﺘﻌﻤﺎل ﺑﺮﻧﺎﻣﺞ STATISTICA
ﻋﻨﺪ اﻟﺘﻨﻔﯿﺬ ﺑﺎﺳﺘﻌﻤﺎل ھ ﺬا اﻟﺒﺮﻧ ﺎﻣﺞ ﺳ ﺘﻈﮭﺮ اﻟﻨﺘ ﺎﺋﺞ )ﻣﻌﺎدﻟ ﺔ اﻟﺘﻨﺒ ﻮء( ﻣ ﻊ اﻟﺮﺳ ﻢ
ﻛﺎﻵﺗﻲ :
)Line Plot (YOUDEN.STA 7v*12c
y=7.779+3.807*x+eps
8
7
6
5
3
2
1
0
0.0
2.0-
4.0-
6.0-
8.0-
0.1-
2.1-
4.1-
6.1-
1-
8.1-
LOGDOSE
ﺷﻜﻞ 11: ﻣﻌﺎدﻟﺔ اﻟﺨﻂ اﻟﻤﺴﺘﻘﯿﻢ ﻟﻮﺻﻒ اﻻﺳﺘﺠﺎﺑﺔ
82
PROBIT
4
29. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
STATGRAPHICS Plusٍ و- اﻟﺤﻞ ﺑﺎﺳﺘﻌﻤﺎل ﺑﺮﻧﺎﻣﺞ
:ﻋﻨﺪ اﺳﺘﻌﻤﺎل ھﺬا اﻟﺒﺮﻧﺎﻣﺞ ﻓﺎن اﻟﻨﺘﺎﺋﺞ ﺑﻌﺪ اﻟﺘﻨﻔﯿﺬ ﺳﺘﻜﻮن ﻛﻤﺎ ﻣﻮﺿﺢ ادﻧﺎه
Multiple Regression Analysis
----------------------------------------------------------------------------Dependent variable: probit
----------------------------------------------------------------------------Standard
T
Parameter
Estimate
Error
Statistic
P-Value
----------------------------------------------------------------------------CONSTANT
7.77911
0.459222 16.9398
0.0000
logdose
3.80742
0.47111
8.08181
0.0000
----------------------------------------------------------------------------Analysis of Variance
----------------------------------------------------------------------------Source
Sum of Squares Df Mean Square F-Ratio P-Value
----------------------------------------------------------------------------Model
32.7575
1 32.7575 65.32
0.0000
Residual
4.01221
8 0.501526
----------------------------------------------------------------------------Total (Corr.)
36.7697 9
R-squared = 89.0883 percent
R-squared (adjusted for d.f.) = 87.7243 percent
Standard Error of Est. = 0.708185
Mean absolute error = 0.541387
Durbin-Watson statistic = 1.68027
Component+Residual Plot for probit
component effect
3.4
1.4
-0.6
-2.6
-4.6
-1.7
-1.4
-1.1
-0.8
-0.5
-0.2
logdose
ﺷﻜﻞ 21: ﻣﻌﺎدﻟﺔ اﻟﺨﻂ اﻟﻤﺴﺘﻘﯿﻢ ﻟﻮﺻﻒ اﻻﺳﺘﺠﺎﺑﺔ
29
30. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
ز- اﻟﺘﺤﻮﯾﻞ اﻟﺘﻠﻘﺎﺋﻲ ﺑﺎﺳﺘﻌﻤﺎل ﺑﺮﻧﺎﻣﺞ SAS
ھﻨﺎﻟ ﻚ ﻋ ﺪة اواﻣ ﺮﯾﻤﻜﻦ ﺗﻄﺒﯿﻘﮭ ﺎ ﻗ ﻲ ﺑﺮﻧ ﺎﻣﺞ SASﻟﻐ ﺮض ﺗﻘ ﺪﯾﺮ ﻗﯿﻤ ﺔ 05LD
)ﻣﺜﺎل 2( دون اﻟﺤﺎﺟﺔ اﻟﻰ ﺗﻘ ﺪﯾﺮ اﻟﻮﺣ ﺪات اﻻﺣﺘﻤﺎﻟﯿ ﺔ اذ ﯾﺠ ﺮي ﺗﺤﻮﯾﻠﮭ ﺎ ﺗﻠﻘﺎﺋﯿ ﺎ
ﻣﻦ ﻗﺒﻞ اﻟﺒﺮﻧﺎﻣﺞ وﻟﻜﻦ ﯾﺠﺐ ﻣﺮاﻋﺎت ﻛﺘﺎﺑﺔ اﻟﺤﯿﻮاﻧﺎت اﻟﮭﺎﻟﻜﺔ ﻟﻜﻞ ﺟﺮﻋﺔ وﻋﺪد
اﻟﺤﯿﻮاﻧ ﺎت ﻟﻜ ﻞ ﺟﺮﻋ ﺔ ﺑ ﺪﻻ ﻋ ﻦ ﻧ ﺴﺒﺔ اﻟﺤﯿﻮاﻧ ﺎت اﻟﮭﺎﻟﻜ ﺔ ﻟﻜ ﻲ ﯾ ﺘﻢ اﻟﺘﻨﻔﯿ ﺬ وﻓ ﻲ
ﻣﺜﺎﻟﻨﺎ ھﺬا اﻓﺘﺮﺿﻨﺎ اﻧﻨﺎ اﺳﺘﻌﻤﻠﻨﺎ 03 ﺣﯿﻮان ﻟﻜﻞ ﺟﺮﻋﺔ.
1- اﻵﻣﺮ Proc probit
;data s
;input ldose x N
;cards
96.1-
03 0
93.1-
03 2
22.1-
03 3
90.1-
03 5
1-
03 8
96.0-
03 21
25.0-
03 02
93.0-
03 62
03.0-
03 72
22.0-
03 03
;proc probit data=s lackfit inversecl
;model x/N = lDose
;run
The SAS System
Probit Procedure
003 = Number of Trials
Data Set = WORK.S
Dependent Variable=X
Dependent Variable=N
01=Number of Observations
331 = Number of Events
9441085.311- Log Likelihood for NORMAL
The SAS System
Probit Procedure
Pr>Chi Label/Value
0.0001 Intercept
1000.0
DF
Estimate Std Err ChiSquare
587512.0 26077080.2 1
3389.29
2723.611 973062.0 67413808.2 1
03
Variable
INTERCPT
LDOSE
31. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
Probit Model in Terms of Tolerance Distribution
MU
SIGMA
-0.74093
0.356085
Estimated Covariance Matrix for Tolerance Parameters
MU
SIGMA
0.001121
0.000015211
0.000015211
0.001090
The SAS System
Probit Procedure
Probit Analysis on LDOSE
MU
SIGMA
Probability
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.85
0.90
0.91
0.92
0.93
0.94
0.95
0.96
0.97
0.98
0.99
LDOSE 95 Percent
-1.56931
-1.47224
-1.41066
-1.36433
-1.32664
-1.29456
-1.26644
-1.24126
-1.21836
-1.19727
-1.10999
-1.04062
-0.98111
-0.92766
-0.87814
-0.83115
-0.78568
-0.74093
-0.69619
-0.65072
-0.60373
-0.55420
-0.50076
-0.44124
-0.37187
-0.28459
-0.26351
-0.24061
-0.21542
-0.18730
-0.15522
-0.11754
-0.07121
-0.00962
0.08745
Fiducial
Lower
-1.76596
-1.64899
-1.57500
-1.51947
-1.47440
-1.43613
-1.40264
-1.37272
-1.34557
-1.32063
-1.21801
-1.13742
-1.06919
-1.00885
-0.95387
-0.90270
-0.85420
-0.80751
-0.76188
-0.71657
-0.67076
-0.62350
-0.57348
-0.51877
-0.45603
-0.37825
-0.35961
-0.33942
-0.31728
-0.29262
-0.26458
-0.23172
-0.19146
-0.13810
-0.05433
31
Limits
Upper
-1.42891
-1.34510
-1.29172
-1.25142
-1.21854
-1.19048
-1.16579
-1.14363
-1.12341
-1.10475
-1.02684
-0.96395
-0.90909
-0.85889
-0.81144
-0.76542
-0.71988
-0.67401
-0.62709
-0.57837
-0.52698
-0.47181
-0.41128
-0.34290
-0.26217
-0.15943
-0.13446
-0.10728
-0.07734
-0.04382
-0.00552
0.03957
0.09512
0.16915
0.28615
32. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
واﻟﻤﻌﺒ ﺮ ﻋﻨﮭﻤ ﺎProbit اﻟﻤﻘ ﺪرة ﺑﻄﺮﯾﻘ ﺔLD50 ﻧﻼﺣ ﻆ ﻣ ﻦ اﻟﻨﺘ ﺎﺋﺞ ان ﻗﯿﻤ ﺔ
ﻟﮭﺬه اﻟﻘﯿﻤ ﺔ ﺗ ﺴﺎويantilog ﺑﺎﻟﻠﻮﻏﺎرﯾﺘﯿﻢ ﻟﻸﺳﺎس 01 ھﻲ – 47.0 ، وان ﻗﯿﻤﺔ
81.0 ، ﻛﻤﺎ ﯾﻤﻜﻦ اﻟﺤﺼﻮل ﻋﻠﻰ ﺣ ﺪود اﻟﺜﻘ ﺔ ﺑﻤ ﺴﺘﻮى 59% ﻟﻠﻘﯿﻤ ﺔ اﻟﻤﻘ ﺪرة اذ
. ﻛﻤ ﺎ ﯾﻤﻜ ﻦ ﺗﻨﻔﯿ ﺬ اﻟﺒﺮﻧ ﺎﻣﺞ ﺑﺤﯿ ﺚ ﺗﻈﮭ ﺮLD50 ﺗﻤﺜﻠﮭﺎ اﻟﻘﯿﻤﺘﺎن اﻟﻤﺤﺎذﯾﺘﺎن ﻟﻘﯿﻤﺔ
ﻓ ﻲ أن واﺣ ﺪ ﺑﺎﺳ ﺘﻌﻤﺎلantilog ﻓ ﻲ اﻟﻨﺘ ﺎﺋﺞ ﺑﺪﻻﻟ ﺔ اﻟﻠﻮﻏ ﺎرﯾﺘﯿﻢ وLD50 ﻗﯿﻤ ﺔ
: inversecl اﻷﻣﺮ
data conc;
input dose r n;
cards;
0.02
0 30
0.04
2 30
0.06
3 30
0.08
5 30
0.10
8 30
0.20
12 30
0.30
20 30
0.40
26 30
0.50
27 30
0.60
30 30
proc probit data=conc lackfit log10 inversecl;
model r/N = Dose;
run;
The SAS System
Probit Procedure
Data Set =WORK.CONC
Dependent Variable=R
Dependent Variable=N
Number of Observations= 10
Number of Events = 133
Number of Trials = 300
Log Likelihood for NORMAL -113.475854
Goodness-of-Fit Tests
Statistic
-----------------Pearson Chi-Square
L.R.
Chi-Square
Response Levels:
Value
-------7.3241
9.2853
2
DF
-8
8
Prob>Chi-Sq
----------0.5021
0.3188
Number of Covariate Values:
32
10
33. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
NOTE: Since the chi-square is small (p > 0.1000), fiducial limits
will be calculated using a t value of
1.96.
The SAS System
Probit Procedure
Variable
DF
INTERCPT
1
Log10(DOS) 1
Estimate
Std Err ChiSquare
2.094052
0.217166 92.98049
2.80898256 0.260777 116.0272
Pr>Chi Label/Value
0.0001 Intercept
0.0001
Probit Model in Terms of Tolerance Distribution
MU
SIGMA
-0.74548
0.356001
Estimated Covariance Matrix for Tolerance Parameters
MU
MU
SIGMA
Probability
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.10
0.15
0.20
0.25
0.30
0.35
0.40
0.45
0.50
0.55
0.60
0.65
0.70
0.75
0.80
0.85
0.90
SIGMA
0.001122
0.000015541
0.000015541
0.001092
Probit Procedure
Probit Analysis on DOSE
Log10(DOSE)
DOSE
95 Percent Fiducial Limits
Lower
Upper
-1.57367
0.02669
0.01696
0.03688
-1.47662
0.03337
0.02220
0.04473
-1.41505
0.03845
0.02633
0.05058
-1.36873
0.04278
0.02992
0.05549
-1.33105
0.04666
0.03319
0.05985
-1.29898
0.05024
0.03625
0.06385
-1.27087
0.05360
0.03916
0.06758
-1.24569
0.05679
0.04195
0.07112
-1.22279
0.05987
0.04466
0.07451
-1.20172
0.06285
0.04730
0.07778
-1.11446
0.07683
0.05990
0.09305
-1.04510
0.09014
0.07212
0.10754
-0.98560
0.10337
0.08438
0.12202
-0.93217
0.11690
0.09696
0.13697
-0.88266
0.13102
0.11005
0.15278
-0.83568
0.14599
0.12381
0.16985
-0.79022
0.16210
0.13843
0.18863
-0.74548
0.17969
0.15414
0.20963
-0.70075
0.19918
0.17121
0.23355
-0.65529
0.22116
0.19004
0.26128
-0.60831
0.24643
0.21117
0.29410
-0.55880
0.27619
0.23544
0.33394
-0.50537
0.31235
0.26416
0.38387
-0.44587
0.35821
0.29961
0.44934
-0.37651
0.42023
0.34616
0.54114
-0.28925
0.51375
0.41403
0.68559
33
34. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
61627.0
60377.0
42828.0
96498.0
81779.0
90480.1
40232.1
20164.1
77219.1
71234.0
37254.0
04674.0
22405.0
38735.0
80085.0
14636.0
65917.0
75278.0
92935.0
94865.0
24206.0
17246.0
79196.0
86457.0
16938.0
05769.0
67902.1
71862.0-
82542.0-
01022.0-
89191.0-
19951.0-
42221.0-
29570.0-
53410.0-
07280.0
19.0
29.0
39.0
49.0
59.0
69.0
79.0
89.0
99.0
2- اﻵﻣﺮ Proc Logistic
ﻋﻨﺪ اﻟﺘﻨﻔﯿﺬ ﺳﻨﺤﺼﻞ ﻏﻠﻰ ﻧﻔﺲ ﻧﺘﺎﺋﺞ اﻵﻣﺮ اﻟﺴﺎﺑﻖ.
;data f
;input logdose x n
;cards
96.1-
03 0
93.1-
03 2
22.1-
03 3
90.1-
03 5
1-
03 8
96.0-
03 21
25.0-
03 02
93.0-
03 62
03.0-
03 72
22.0-
03 03
; proc logistic data=f
;model x/N = logdose/link=probit
;run
3- اﻵﻣﺮ Proc genmod
ﻋﻨﺪ اﻟﺘﻔﯿﺬ ﺳﻨﺤﺼﻞ ﻏﻠﻰ ﻧﻔﺲ ﻧﺘﺎﺋﺞ اﻵﻣﺰ اﻟﺴﺎﺑﻖ.
;data s
;input ldose x N
;cards
96.1-
03 0
93.1-
03 2
22.1-
03 3
90.1-
03 5
43
35. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
1-
03 8
96.0-
03 21
25.0-
03 02
93.0-
03 62
03.0-
03 72
22.0-
03 03
;proc genmod data=s
model x/N = lDose/d=binomial
;link=probit
;run
ﺤ- اﻟﺘﺤﻮﯾﻞ اﻟﺘﻠﻘﺎﺋﻲ ﺑﺎﺳﺘﻌﻤﺎل اﻟﺒﺮﻧﺎﻣﺞ Minitab
ﻧﺪرج اﻟﺒﯿﺎﻧﺎت )ﻣﺜﺎل 2( ﻓﻲ اﻟﺼﻔﺤﺔ اﻟﺮﺋﯿﺴﯿﺔ ﻛﺎﻵﺗﻲ:
n
03
03
03
03
03
03
03
03
03
03
x
0
2
3
5
8
21
02
62
72
03
Logdos
96.1-
93.1-
22.1-
90.1-
1-
96.0-
25.0-
93.0-
03.0-
22.0-
ﻧﻨﻘ ﺮ ﻋﻠ ﻰ Statﻓﯿﻈﮭ ﺮ ﺷ ﺮﯾﻂ ﻗ ﻮاﺋﻢ ﻧﺨﺘ ﺎر ﻣﻨ ﮫ Reliability/Survival
ﻓﯿﻈﮭ ﺮ ﺷ ﺮﯾﻂ ﻗ ﻮاﺋﻢ ﻧﺨﺘ ﺎر ﻣﻨ ﮫ Probit analysisﻓﯿﻈﮭ ﺮ ﺻ ﻨﺪوق ﺣ ﻮار
ﻧﺨﺘ ﺎر ﻣﻨ ﮫ Response in success/trail formatﺛ ﻢ ﻧﻨﻘ ﺮ ﻓ ﻲ اﻟﻔ ﺮاغ
اﻟﻤﺆﺷﺮ اﻣﺎﻣﮫ Number of successesﻓﺘﻈﮭﺮ اﻟﻤﺘﻐﯿﺮات ﻓﻲ اﻟﻔﺮاغ اﻻﯾ ﺴﺮ
، ﺑﻌﺪھﺎ ﻧﺨﺘﺎر اﻟﻌﻤﻮد اﻟﺬي ﯾﺘﻀﻤﻦ ﻋﺪد اﻟﮭﻼﻛﺎت ﻟﻜ ﻞ ﺟﺮﻋ ﺔ xﻓﯿﻤ ﺎ ﻧ ﻀﻊ n
ﻓﻲ اﻟﻔﺮاغ اﻟﻤﺆﺷﺮ اﻣﺎﻣﮫ Number of trailsاﻣ ﺎ اﻟﻤﺘﻐﯿ ﺮ ldoseﻓﻨ ﻀﻌﮫ ﻓ ﻲ
اﻟﺤﻘ ﻞ اﻟﻤﺆﺷ ﺮ اﻣﺎﻣ ﮫ ) stress(stimulusوﻧﺨﺘ ﺎر Normalﻓ ﻲ اﻟﻤﺮﺑ ﻊ
اﻟﻤﺆﺷﺮ Assumd disributionﺛﻢ ﻧﻨﻘﺮ .ok
53
36. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
Probit Analysis: x, N versus ldose
Distribution: Normal
Response Information
Variable
x
N
Value
Success
Failure
Total
Estimation Method:
Count
133
167
300
Maximum Likelihood
Regression Table
Variable
Constant
ldose
Natural
Response
Coef
2.0808
2.8083
Standard
Error
0.2158
0.2604
Z
P
9.64 0.000
10.79 0.000
0.000
Log-Likelihood = -113.580
Goodness-of-Fit Tests
Method
Pearson
Deviance
Chi-Square
7.483
9.494
DF
8
8
P
0.486
0.302
Parameter Estimates
Parameter
Location
Scale
Estimate
-0.74093
0.35609
Standard
Error
0.03349
0.03302
95.0% Normal CI
Lower
Upper
-0.80657
-0.67530
0.29692
0.42705
Standard
Error
0.08336
0.07521
0.07014
0.06640
0.06341
0.06090
0.05874
0.05684
0.05514
0.05360
0.04322
0.03749
0.03441
0.03349
0.03463
0.03791
0.04381
0.05432
0.05587
0.05759
0.05950
95.0% Fiducial CI
Lower
Upper
-1.7660
-1.4289
-1.6490
-1.3451
-1.5750
-1.2917
-1.5195
-1.2514
-1.4744
-1.2185
-1.4361
-1.1905
-1.4026
-1.1658
-1.3727
-1.1436
-1.3456
-1.1234
-1.3206
-1.1048
-1.1374
-0.9640
-1.0088
-0.8589
-0.9027
-0.7654
-0.8075
-0.6740
-0.7166
-0.5784
-0.6235
-0.4718
-0.5188
-0.3429
-0.3782
-0.1594
-0.3596
-0.1345
-0.3394
-0.1073
-0.3173
-0.0773
Table of Percentiles
Percent
1
2
3
4
5
6
7
8
9
10
20
30
40
LD50 50
60
70
80
90
91
92
93
Percentile
-1.5693
-1.4722
-1.4107
-1.3643
-1.3266
-1.2946
-1.2664
-1.2413
-1.2184
-1.1973
-1.0406
-0.9277
-0.8311
-0.7409
-0.6507
-0.5542
-0.4412
-0.2846
-0.2635
-0.2406
-0.2154
36
37. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
6292.0-
6462.0-
7132.0-
5191.0-
1831.0-
3450.0-
8340.0-
5500.0-
5930.0
1590.0
1961.0
1682.0
76160.0
91460.0
02760.0
59070.0
40670.0
12480.0
49
59
69
79
89
99
3781.0-
2551.0-
5711.0-
2170.0-
6900.0-
4780.0
ﯾﻮﻓﺮھ ﺬا اﻟﺒﺮﻧ ﺎﻣﺞ ﻛﻤ ﺎ ﻗ ﻲ ﺑﺮﻧ ﺎﻣﺞ SASاﺧﺘﺒ ﺎران ﻟﺤ ﺴﻦ اﻟﻤﻄﺎﺑﻘ ﺔ
) Pearsonو (Devianceﻟﻐﺮض اﺧﺘﺒ ﺎر ﻣ ﺪى ﻗ ﺪرة ﻣﻌﺎدﻟ ﺔ اﻟﺨ ﻂ اﻟﻤ ﺴﺘﻘﯿﻢ
ﺑﺪﻻﻟ ﺔ اﻟﻮﺣ ﺪات اﻻﺣﺘﻤﺎﻟﯿ ﺔ ﻋﻠ ﻰ وﺻ ﻒ اﻟﺒﯿﺎﻧ ﺎت وﯾﺘ ﻀﺢ ﻣ ﻦ اﻟﻨﺘ ﺎﺋﺞ ان
اﻻﺣﺘﻤ ﺎل ﻓ ﻲ ﻛ ﻼ اﻻﺧﺘﺒ ﺎرﯾﻦ ) 84.0= Pو 03.0= (Pﻛ ﺎن ﻏﯿ ﺮ ﻣﻌﻨﻮﯾ ﺎ ﻣﻤ ﺎ
ﯾﻌﻨﻲ ان ﻣﻌﺎدﻟﺔ اﻟﺨﻂ اﻟﻤﺴﺘﻘﯿﻢ ﻣﻼﺋﻤﺔ ﻟﻮﺻﻒ اﻟﺒﯿﺎﻧﺎت.
ProbPlot for x
Probability Plot for r
Normal - 95% CI
Probit Data - ML Estimates
99
Table of Statistics
Mean
583227.0-
S tDev
148183.0
583227.0- Median
IQ R
690515.0
59
09
08
03
Percent
07
06
05
04
02
01
5
1
5.0
0.0
0.1-
5.0-
5.1-
0.2-
logdose
ﺷﻜﻞ 31: ﻣﻌﺎدﻟﺔ اﻟﺨﻂ اﻟﻤﺴﺘﻘﯿﻢ ﻟﻮﺻﻒ اﻻﺳﺘﺠﺎﺑﺔ ﻣﻊ ﺣﺪود اﻟﺜﻘﺔ
ان اﻟﺘﺤﻠﯿﻞ اﻻﺣﺼﺎﺋﻲ ﺑﮭﺬا اﻟﺒﺮﻧﺎﻣﺞ ﯾﻮﻓﺮ ﻟﻨﺎ ﺣﺪود اﻟﺜﻘﺔ ﺑﻤﺴﺘﻮى 59% اﺿﺎﻓﺔ
اﻟﻰ اﻟﺨﻄﺄ اﻟﻘﯿﺎﺳﻲ ﻟﻘﯿﻤﺔ 05 LDاﻟﻤﻘﺪرة.
73
38. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
ط- اﻟﺘﺤﻮﯾﻞ اﻟﺘﻠﻘﺎﺋﻲ ﺑﺎﺳﺘﻌﻤﺎل اﻟﺒﺮﻧﺎﻣﺞ SPSS
ﻧﺪرج اﻟﺒﯿﺎﻧﺎت )ﻣﺜﺎل 2( ﻛﻤﺎ ﻣﻮﺿﺢ ادﻧﺎه:
n
03
03
03
03
03
03
03
03
03
03
ﺛﻢ ﻧﺨﺘﺎر ﻃﺮﯾﻘﺔ Probitﻛﻤﺎ ﻣﻮﺿﺢ ﻓﻲ اﻟﺸﻜﻞ اﻵﺗﻲ :
ﺷﻜﻞ 41: اﻟﺼﻔﺤﺔ اﻟﺮﺋﯿﺴﯿﺔ ﻟﻠﺒﺮﻧﺎﻣﺞ SPSS
ﻓﯿﻈﮭﺮ ﻣﺮﺑﻊ ﺣﻮار اﺧﺮ ﻛﻤﺎ ﻣﻮﺿﺢ ﻓﻲ اﻟﺸﻜﻞ اﻵﺗﻲ:
83
x
0
2
3
5
8
21
02
62
72
03
Logdos
96.1-
93.1-
22.1-
90.1-
1-
96.0-
25.0-
93.0-
03.0-
22.0-
39. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
ﺷﻜﻞ 51: ﻣﺮﺑﻊ ﺣﻮار ﺧﺎص ﺑﺎﻻﯾﻌﺎز Probit
ﻧﻀﻊ اﻟﻤﺘﻐﯿﺮ xﻓﻲ اﻟﻤﺮﺑ ﻊ Response Frequencyواﻟﻤﺘﻐﯿ ﺮ nﻓ ﻲ اﻟﻤﺮﺑ ﻊ
Total Observedﻓﯿﻤ ﺎ ﻧ ﻀﻊ اﻟﻤﺘﻐﯿ ﺮ ldoseﻓ ﻲ اﻟﻤﺮﺑ ﻊ Covariates
وﻧﺆﺷ ﺮ ﻓ ﻲ اﻟﺤﻘ ﻞ Modelﻋﻠ ﻰ ﻛﻠﻤ ﺔ probitﺛ ﻢ ﻧﻨﻘ ﺮ Optionsوﻧﺆﺷ ﺮ
اﻟﺤﻘﻮل اﻟﻤﻮﺿﺤﺔ ﻓﻲ اﻟﺸﻜﻞ ادﻧﺎه:
ﺷﻜﻞ 61: ﻣﺮﺑﻊ ﺣﻮار ﺧﺎص ﺑﺎﻻﯾﻌﺎز Options
93
40. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
ﺛﻢ ﻧﻀﻐﻂ continueﻟﯿﻈﮭﺮ ﻣﺮﺑﻊ اﻟﺤﻮار اﻻول وﻧﻀﻐﻂ .ok
Probability
95% Confidence Limits for logdose
010.
Estimate
965.1-
Lower Bound
667.1-
Upper Bound
924.1-
020.
274.1-
946.1-
543.1-
030.
114.1-
575.1-
292.1-
040.
463.1-
915.1-
152.1-
050.
723.1-
474.1-
912.1-
060.
070.
592.1-
662.1-
634.1-
304.1-
091.1-
661.1-
080.
142.1-
373.1-
441.1-
090.
812.1-
643.1-
321.1-
001.
791.1-
123.1-
501.1-
051.
011.1-
812.1-
720.1-
002.
140.1-
731.1-
469.-
052.
189.-
960.1-
909.-
003.
053.
829.-
878.-
900.1-
459.-
958.-
118.-
004.
138.-
309.-
567.-
054.
687.-
458.-
027.-
005.
147.-
808.-
476.-
055.
696.-
267.-
726.-
006.
156.-
717.-
875.-
056.
406.-
176.-
725.-
007.
455.-
326.-
274.-
057.
008.
105.-
144.-
375.-
915.-
114.-
343.-
058.
273.-
654.-
262.-
009.
582.-
873.-
951.-
019.
462.-
063.-
431.-
029.
142.-
933.-
701.-
039.
512.-
713.-
770.-
049.
781.-
392.-
440.-
059.
069.
551.-
811.-
562.-
232.-
600.-
040.
079.
170.-
191.-
590.
089.
010.-
831.-
961.
099.
780.
450.-
682.
وﯾﻼﺣﻆ ان ﻗﯿﻤﺔ ﻟﻮﻏﺎرﯾﺘﯿﻢ 05 LDﯾﺴﺎوي – 47.0.
04
41. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
2-2-2-2- ﺗﻌﺪﯾﻞ اﻟﺒﯿﺎﻧﺎت
اوﻻ: اﻟﺘﻌﺪﯾﻞ ﻟﻼﺳﺘﺠﺎﺑﺔ اﻟﻄﺒﯿﻌﯿﺔ
ﻓ ﻲ ﺑﻌ ﺾ اﻟﺘﺠ ﺎرب ﯾﻤﻜ ﻦ ان ﺗﺤ ﺼﻞ اﺳ ﺘﺠﺎﺑﺔ )ﻣ ﻮت ﺣﯿﻮاﻧ ﺎت( ﻋﻨ ﺪ ﻣ ﺴﺘﻮى
اﻟﺠﺮﻋﺔ 0 وذﻟﻚ ﻗﺪ ﯾﻌ ﻮد اﻟ ﻰ اﺳ ﺒﺎب ﻃﺒﯿﻌﯿ ﺔ ، ﻟ ﺬا ﻓ ﺄن ﻣ ﻦ اﻟ ﻀﺮوري اﺟ ﺮاء
ﺗﻌﺪﯾﻞ ﻟﻠﺒﯿﺎﻧﺎت ﻋﻨﺪﻣﺎ ﺗﻜﻮن ھﻨﺎك ﻧﺴﺒﺔ ھﻼﻛﺎت ﻓﻲ ﻣﺠﻤﻮﻋﺔ اﻟﺴﯿﻄﺮة ﺗﺰﯾﺪ ﻋ ﻦ
5%. واﺣﺪى ﻃﺮق اﻟﺘﻌﺪﯾﻞ ھﻮ اﺳﺘﻌﻤﺎل ﻣﻌﺎدﻟﺔ )5291( ,. Abbott
001 Corrected = % Responded – % Responded in control x
100 – Responded in control
ﻣﺜ ﺎل : اذا ﻛﺎﻧ ﺖ ﻧ ﺴﺒﺔ اﻟﮭﻼﻛ ﺎت ﻓ ﻲ ﻣﺠﻤﻮﻋ ﺔ اﻟ ﺴﯿﻄﺮة 02% وﻓ ﻲ ﻣﺠﻤﻮﻋ ﺔ
اﻟﻤﻌﺎﻣﻠﺔ 06% ﻓﻌﻨ ﺪ ﺗﻄﺒﯿ ﻖ اﻟﻤﻌﺎدﻟ ﺔ ﺳﻨﺤ ﺼﻞ ﻋﻠ ﻰ اﻟﻘﯿﻤ ﺔ اﻟﻤﻌﺪﻟ ﺔ 05% ﺑ ﺪﻻ
ﻋﻦ 06 % وﻓﻖ اﻟﻤﻌﺎدﻟﺔ اﻵﺗﯿﺔ:
%05 = 08/04 =001 60% – 20% x
02 – 001
ﻣﺜ ﺎل ) 3(: اﺳ ﺘﻌﻤﻠﺖ 6 ﺟ ﺮع ﻣﺨﺘﻠﻔ ﺔ ﻣ ﻦ ﻣ ﺎدة ﻛﯿﻤﯿﺎوﯾ ﺔ وﺑﻤﻘﯿ ﺎس ﻣﻠﻐ ﻢ/ﻟﺘ ﺮ
وﺑﻮاﻗﻊ 02 ﻣﻜﺮر ﻟﻜﻞ ﺟﺮﻋﺔ وﺳﺠﻠﺖ اﻟﻨﺘﺎﺋﺞ ﺑﻌﺪ ﻣ ﺮور 42 ﺳ ﺎﻋﺔ ، اﻟﻤﻄﻠ ﻮب
ﺗﻘﺪﯾﺮ ﻗﯿﻤﺔ 05LD؟
اﻟﺠﺮﻋﺔ
ﻣﻠﻐﻢ/ﻟﺘﺮ
0
03.2
00.3
09.3
21.5
69.6
01Log
263.0
774.0
195.0
907.0
348.0
ﻋﺪد اﻟﮭﻼﻛﺎت
ﺧﻼل 42 ﺳﺎﻋﺔ
1
0
1
4
9
61
اﻟﻌﺪد
اﻟﻜﻠﻲ
02
02
02
02
02
02
14
ﻧﺴﺒﺔ
اﻟﮭﻼﻛﺎت
5
0
5
02
54
08
42. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
ﻧﻈﺮا ﻟﻮﺟﻮد ﻧﺴﺒﺔ ھﻼﻛﺎت ﻓﻲ ﻣﺠﻤﻮﻋ ﺔ اﻟ ﺴﯿﻄﺮة وﺗﻤﺜ ﻞ 5% ﻟ ﺬا ﯾﺠ ﺐ اﺟ ﺮاء
اﻟﺘﻌﺪﯾﻞ ﻟﻨﺴﺐ اﻟﺠﺮع اﻻﺧﺮى ﺛﻢ ﻧﻘﺪر ﻣﻌﺎدﻟﺔ اﻟﺘﻨﺒﻮء ﺑﺎﺳﺘﻌﻤﺎل اﻻﻧﺤﺪار اﻟﺨﻄﻲ
ﯾﺪوﯾﺎ او ﺑﺎﺳﺘﻌﻤﺎل اي ﺑﺮﻧﺎﻣﺞ اﺣﺼﺎﺋﻲ.
%0 = 001 0% – 5% x 100 = 5% 5% – 5% x
%5 – 001
%5 – 001
%51 = 001 20% – 5% x
%5 – 001
وھﻜﺬا ﺑﺎﻟﻨﺴﺒﺔ ﻟﺒﻘﯿﺔ اﻟﺠﺮع اذ ﺳﻨﺤﺼﻞ ﻋﻠﻰ 24% و 97% . ﺛ ﻢ ﻧﺤ ﻮل اﻟﺠ ﺮع
اﻟﺜﻼﺛ ﺔ اﻟ ﻰ ﻣﺎﯾﻘﺎﺑﻠﮭ ﺎ ﻣ ﻦ اﻟﻮﺣ ﺪات اﻻﺣﺘﻤﺎﻟﯿ ﺔ وھ ﻲ 69.3 ، 08.4 ، 18.5 .
وﺳﻨﺠﺪ ان ﻗﯿﻤﺔ 05 LDﻋﻨﺪ ﺗﻄﺒﯿﻖ ﻣﻌﺎدﻟﺔ اﻻﻧﺤﺪار ﺑﻌﺪ اﻟﺘﻌﺪﯾﻞ ﺳﺘ ﺴﺎوي 53.5
ﻓﯿﻤﺎ ان اﻟﻘﯿﻤﺔ ﺑﺪون ﺗﻌﺪﯾﻞ ﺳﺘﺴﺎوي 62.5.
أ- اﻟﺤﻞ ﺑﺎﺳﺘﻌﻤﺎل ﺑﺮﻧﺎﻣﺞ SASﺑﻌﺪ ﺗﻌﺪﯾﻞ وﺗﺤﻮﯾﻞ اﻟﺒﯿﺎﻧﺎت
;data c
;input dose pro
;)logdose=log10(dose
;cards
0
63.3
63.3 03.2
3
.
69.3 9.3
8.4 21.5
18.5 69.6
;proc reg
;model pro=logdose
;run
24
43. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
The SAS System
Model: MODEL1
Dependent Variable: PRO
Analysis of Variance
Source
Model
Error
C Total
Sum of
Squares
3.16617
0.22990
3.39607
DF
1
2
3
Root MSE
Dep Mean
C.V.
Variable DF
INTERCEP
1
LOGDOSE
1
Mean
Square
3.16617
0.11495
0.33905
4.48250
7.56375
R-square
Adj R-sq
F Value
27.544
Prob>F
0.0344
0.9323
0.8985
Parameter Estimates
Parameter
Standard
T for H0:
Estimate
Error
Parameter=0
1.330014
0.62414347
2.131
5.034570
0.95929633
5.248
Prob > |T|
0.1668
0.0344
LD50 = 1.33 + 5.034 x logdose
5 = 1.33 + 5.034 x logdose
Logdose = (5 – 1.33)/5.034 = 0.7289
Antilog(0.7289) = 5.35
SAS ب- اﻟﺘﻌﺪﯾﻞ اﻟﺘﻠﻘﺎﺋﻲ ﺑﺎﺳﺘﻌﻤﺎل ﺑﺮﻧﺎﻣﺞ
ﯾﻌﻨ ﻲ اﺟ ﺮاء ﺗﺤﻮﯾ ﻞ ﺗﻠﻘ ﺎﺋﻲ ﻟﻠﻨ ﺴﺐProc probit ان اﺳﺘﻌﻤﺎل اﻻﯾﻌ ﺎز
. اﻟﺬي ﯾﻌﻨﻲ اﻟﺘﻌﺪﯾﻞ ﻟﻼﺳﺘﺠﺎﺑﺔ اﻟﻄﺒﯿﻌﯿﺔOptc ﻓﯿﻤﺎ ﻧﺴﺘﻌﻤﻞ اﻻﯾﻌﺎز
data study;
input Dose number Respond;
datalines;
0.00
20
1
2.30
20
0
3.00
20
1
3.90
20
4
5.12
20
9
6.96
20
16
proc probit data=study log10 optc ;
model respond/number=dose;
run;
43
44. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
The SAS System
Probit Procedure
Data Set =WORK.STUDY
Dependent Variable=RESPOND
Dependent Variable=NUMBER
Number of Observations=
6
Number of Events = 31 Number of Trials =
Number of Events In Control Group = 1
Number of Trials In Control Group = 20
Log Likelihood for NORMAL
120
-42.5416874
The SAS System
Probit Procedure
Variable DF
Estimate
Std Err ChiSquare
INTERCPT
1 -5.3684678 1.11637
Log10(DOS) 1 7.36795736 1.549119
_C_
1 0.02502225 0.023681
23.12513
22.62167
Pr>Chi Label/Value
0.0001 Intercept
0.0001
Lower threshold
Probit Model in Terms of Tolerance Distribution
MU
SIGMA
0.728624
0.135723
5.35 ﻧﺴﺎوي 6827.0 اي ان اﻟﺠﺮﻋﺔ ﺗﺴﺎويLD50 ﯾﻼﺣﻆ ان ﻗﯿﻤﺔ ﻟﻮﻏﺎرﯾﺘﯿﻢ
ﺑ ﺪون ﺗﻌ ﺪﯾﻞ ﻓ ﺎن اﻟﻨﺘﯿﺠ ﺔ ﺳ ﺘﺨﺘﻠﻒ وﻛﻤ ﺎ ﻣﻮﺿ ﺢSAS اﻣﺎ ﻋﻨﺪ ﺗﻄﺒﯿﻖ ﺑﺮﻧﺎﻣﺞ
:ادﻧﺎه
data study;
input Dose number Respond;
datalines;
0.00
20
1
2.30
20
0
3.00
20
1
3.90
20
4
5.12
20
9
6.96
20
16
proc probit data=study log10;
model respond/number=dose;
run;
44
45. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
Probit Procedure
Data Set=WORK.STUDY
Dependent Variable=RESPOND
Dependent Variable=NUMBER
Number of Observations=5
Number of Events=30
Number of Trials =100
Log Likelihood for NORMAL -37.91081051
The SAS System
Probit Procedure
Variable
DF
Estimate
Std Err ChiSquare
INTERCPT
1 -5.0077336 0.884604
Log10(DOS) 1 6.93982983 1.279694
32.04682
29.40934
Pr>Chi Label/Value
0.0001 Intercept
0.0001
Probit Model in Terms of Tolerance Distribution
MU
0.721593
SIGMA
0.144096
5.26 ﻧﻼﺣﻆ ان ﻗﯿﻤﺔ ﻟﻮﻏﺎرﯾﺘﯿﻢ اﻟﺠﺮﻋﺔ 5127.0 اي ﻣﺎﯾﺴﺎوي
Minitab ﺠ- اﻟﺘﻌﺪﯾﻞ اﻟﺘﻠﻘﺎﺋﻲ ﺑﺎﺳﺘﻌﻤﺎل ﺑﺮﻧﺎﻣﺞ
ﻓﺘﻈﮭ ﺮ ﻋ ﺪةstat ﻧﻄﺒ ﻊ اﻟﺒﯿﺎﻧ ﺎت ﻓ ﻲ اﻟ ﺼﻔﺤﺔ اﻟﺮﺋﯿ ﺴﯿﺔ ﺛ ﻢ ﻧﻨﻘ ﺮ ﻋﻠ ﻰ اﯾﻘﻮﻧ ﺔ
ﻓﺘﻈﮭ ﺮ ﻗﺎﺋﻤ ﺔ اﺧﺘﯿ ﺎرات اﺧ ﺮىReliability/survival ﺧﯿ ﺎرات ﻧﺨﺘ ﺎر ﻣﻨﮭ ﺎ
ﻓﯿﻈﮭ ﺮ ﻣﺮﺑ ﻊ ﺣ ﻮار ﻓﯿ ﮫ ﻋ ﺪة ﻣﺮﺑﻌ ﺎت ، ﻧ ﻀﻊProbit analysis ﻧﺨﺘ ﺎر ﻣﻨﮭ ﺎ
ﻓﯿﻤ ﺎNumber of successes ﻋﺪد اﻻﻓ ﺮاد اﻟﻤﯿﺘ ﺔ ﻓ ﻲ اﻟﻤﺮﺑ ﻊ اﻟﻤﺆﺷ ﺮ اﻣﺎﻣ ﮫ
ﻛﻤ ﺎNumber of trails ﻧ ﻀﻊ ﻋ ﺪد اﻻﻓ ﺮاد اﻟﻜﻠ ﻲ ﻓ ﻲ اﻟﻤﺮﺑ ﻊ اﻟﻤﺆﺷ ﺮ اﻣﺎﻣ ﮫ
وﻧﺨﺘ ﺎر ﻛﻠﻤ ﺔStress(stimulus) ﻧ ﻀﻊ اﻟﺠﺮﻋ ﺔ ﻓ ﻲ اﻟﻤﺮﺑ ﻊ اﻟﻤﺆﺷ ﺮ اﻣﺎﻣ ﮫ
ﺛ ﻢ ﻧ ﻀﻐﻂ ﻋﻠ ﻰ اﻟ ﺰرAssumed distribution ﻟﻠﻤﺮﺑ ﻊ اﻟﻤﺆﺷ ﺮNormal
ﻓﯿﻈﮭ ﺮ ﻣﺮﺑ ﻊ ﺣ ﻮار ﻧﺆﺷ ﺮ ﻓﯿ ﮫ ﻋﻠ ﻰ اﻟﻤﺮﺑ ﻊ اﻟﻤﺆﺷ ﺮ اﻣﺎﻣ ﮫoptions
45
46. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
ﺛ ﻢ ﻧ ﻀﻊ اﻟ ﺮﻗﻢ 20.0 ﻓ ﻲ اﻟﻤﺮﺑ ﻊ اﻟﻤﺆﺷ ﺮ اﻣﺎﻣ ﮫNatural response rate
. وﻧﻀﻊ ﻧﻔﺲ اﻟﻘﯿﻤﺔ ﻓﻲ اﻟﺠﺮﻋﺔ 0 ﺛﻢ ﻧﻨﻔﺬset value
Estimation Method:
Maximum Likelihood
Regression Table
Variable
Constant
log
Natural
Response
Coef
-5.368
7.368
Standard
Error
1.116
1.549
0.02502
0.02368
Z
P
-4.81 0.000
4.76 0.000
Log-Likelihood = -42.542
ﯾﺴﺎوي 53.5 ﻓﯿﻤﺎ ﻧﺠﺪ ان اﻟﻘﯿﻤﺔAntilog اﻟﻤﻘﺪرة ﺗﺴﺎوي 827.0 وانLD50 ان ﻗﯿﻤﺔ
.5.26 دون ﺗﻌﺪﯾﻞ ﺗﺴﺎوي
Regression Table
Variable
Constant
log
Natural
Response
Coef
-5.0077
6.940
Standard
Error
0.8846
1.280
Z
P
-5.66 0.000
5.42 0.000
0.000
Log-Likelihood = -37.911
LD50= - a/b= 5.0077/6.94 = 0.721
Antilog (0.721)= 5.26
SPSS د- اﻟﺘﻌﺪﯾﻞ اﻟﺘﻠﻘﺎﺋﻲ ﺑﺎﺳﺘﻌﻤﺎل ﺑﺮﻧﺎﻣﺞ
Natural response rate ﻧﺘﺒﻊ ﻧﻔﺲ اﻟﺨﻄﻮات ﺑﺎﺳ ﺘﺜﻨﺎء ﺗﺄﺷ ﯿﺮ ﻣﺮﺑ ﻊ اﻟﺤ ﻮار
.وﻣﻦ ﺛﻢ اﻟﺘﻨﻔﯿﺬ
46
47. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
ﺷﻜﻞ 71: ﻣﺮﺑﻊ ﺣﻮار ﺧﺎص ﺑﺎﻻﯾﻌﺎز Natural response rate
Parameter Estimates
Param
)PROBIT(a
log
Intercept
Estimate
Lower
Bound
963.7
.Std
Error
Upper
Bound
945.1
Z
Lower
Bound
657.4
.Sig
Upper
Bound
000.
%59
Confidence
Interval
Upper
Lower
Bound Bound
233.4 504.01
584.6- 252.4-
000.
711.1 908.4-
963.5-
a PROBIT model: PROBIT(p) = Intercept + BX
ﺛﺎﻧﯿﺎ- اﻟﺘﻌﺪﯾﻞ ﻟﻨﺴﺒﺘﻲ اﻟﮭﻼك 0 و 001%
ﻋﻨ ﺪ اﺧﺘﯿ ﺎر ﻃﺮﯾﻘ ﺔ Probitﯾﻔ ﻀﻞ اﺟ ﺮاء ﺗﻌ ﺪﯾﻞ ﻟﻠﺒﯿﺎﻧ ﺎت ﻋﻨ ﺪ وﺟ ﻮد ﻧ ﺴﺒﺔ
ھﻼﻛﺎت 0 % او 001% وذﻟﻚ ﻟﻌﺪم وﺟﻮد ﻗﯿﻢ اﺣﺘﻤﺎﻟﯿﺔ ﻣﻨﺎﻇﺮة ﻟﮭﺎﺗﯿﻦ اﻟﻘﯿﻤﺘﯿﻦ
وﯾﻤﻜﻦ اﺳﺘﻌﻤﺎل اﻟﻤﻌﺎدﻟﺘﯿﻦ اﻟﺘﺎﻟﯿﺘﯿﻦ:
)0% death = 100 x (0.25/n
]100% death = 100 x [(100 – 0.25) /n
اذ ان : nﺗﻤﺜﻞ ﻋﺪد اﻟﺤﯿﻮاﻧﺎت ﻓﻲ ﻛﻞ ﻣﺠﻤﻮﻋﺔ.
74
48. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
ﻣﺜﺎل: ﻟﻮ ﺣﺎوﻟﻨﺎ ﺣﻞ ﺑﯿﺎﻧﺎت اﻟﻤﺜﺎل 2 ﺑﻌ ﺪ اﺟ ﺮاء اﻟﺘﻌ ﺪﯾﻞ ﻋﻠ ﻰ ﻧ ﺴﺒﺘﻲ اﻟﮭ ﻼك 0
و 001% واﻟﺘﻌ ﻮﯾﺾ ﻋﻨﮭﻤ ﺎ ﺑﻨ ﺴﺒﺔ 5.2% و5.79% ﻓ ﺎن اﻟﻘ ﯿﻢ اﻻﺣﺘﻤﺎﻟﯿ ﺔ
اﻟﻤﻨﺎﻇﺮة ﻟﻠﻨﺴﺒﺘﯿﻦ ﺳﺘﻜﻮن 30.3 و 69.6 ﺛﻢ ﻧﻄﺒﻖ اﻟﺤﻞ ﺑﻨﻔﺲ اﻻﺳ ﻠﻮب اﻟ ﺴﺎﺑﻖ
ﺳ ﻨﺠﺪ ان ﻗﯿﻤ ﺔ 05 LDﺳﺘ ﺴﺎوي ﻟﻮﻏ ﺎرﯾﺘﯿﻢ اﻻﺳ ﺎس 01 ) – 577.0 ( وان
Antilogﯾﺴﺎوي 761.0.
2-2-3- ﻃﺮﯾﻘﺔ Logistic
ﺗﻤﺜﻞ ھﺬه اﻟﻄﺮﯾﻘﺔ ﻧﻮع اﺧﺮ ﻣﻦ ﻃﺮق اﻻﻧﺤﺪاراذ ﯾﺠﺮي ﻓﯿﮭﺎ اﺳﺘﻌﻤﺎل ﻧﻮع اﺧﺮ
ﻣ ﻦ اﻟﺘﺤﻮﯾ ﻞ ﻟﻠﺒﯿﺎﻧ ﺎت اﻟﺜﻨﺎﺋﯿ ﺔ ) (Binomialﻣ ﺸﺎﺑﮫ ﺗﻘﺮﯾﺒ ﺎ ﻟﻄﺮﯾﻘ ﺔ ، Probit
وﻗﺪ اﻗﺘﺮح )2591( , Finneyاﺳﺘﻌﻤﺎل ھﺬه اﻟﻄﺮﯾﻘﺔ ﺑ ﺪﻻ ﻋ ﻦ اﺳ ﺘﻌﻤﺎل ﻃﺮﯾﻘ ﺔ
Probitﻋﻨ ﺪ ﻣ ﺎ ﯾﻜ ﻮن ﺗﻮزﯾ ﻊ اﻟﺒﯿﺎﻧ ﺎت ﺗﻮزﯾﻌ ﺎ ﻏﯿ ﺮ ﻃﺒﯿﻌﯿ ﺎ وﻓ ﻲ ھ ﺬه اﻟﻄﺮﯾﻘ ﺔ
ﯾﺠﺮي ﺗﺤﻮﯾﻞ ﻧﺴﺐ اﻟﺤﯿﻮاﻧﺎت ﺑﺎﻋﺘﻤﺎد داﻟﺔ اﺧﺮى :
)5+)2/)( Logit=(log(p/1-p
أ- اﻟﺤﻞ اﻟﯿﺪوي اﻻول
ﯾﻤﻜ ﻦ اﺟ ﺮاء ھ ﺬا اﻟﺤ ﻞ ﻟﺒﯿﺎﻧ ﺎت اﻟﻤﺜ ﺎل 2 ﻋ ﻦ ﻃﺮﯾ ﻖ ﺗﻘ ﺪﯾﺮ وﺣ ﺪات Logit
ﺑﺎﺳ ﺘﻌﻤﺎل ﻣﻌﺎدﻟ ﺔ اﻟﺘﺤﻮﯾ ﻞ اﻟﺘ ﻲ اﺷ ﺮﻧﺎ اﻟﯿﮭ ﺎ وﻟﻐ ﺮض ﺗﻮﺿ ﯿﺢ ﻋﻤﻠﯿ ﺔ اﻟﺘﺤﻮﯾ ﻞ
ﺳﻨﻘﻮم ﺑﺎﺟﺮاء اﻟﺨﻄﻮات ﺑﺼﻮرة ﻣﺘﺴﻠ ﺴﻠﺔ وﺣ ﺴﺐ ﻣﺎﻣﻮﺿ ﺢ ﻓ ﻲ اﻟﺠ ﺪول ادﻧ ﺎه
وھ ﺬه اﻟﻌﻤﻠﯿ ﺎت ﯾﻤﻜ ﻦ اﺟﺮاﺋﮭ ﺎ ﺑﺎﻟﺤﺎﺳ ﯿﺔ اﻻﻋﺘﯿﺎدﯾ ﺔ او اي ﻣ ﻦ اﻟﺒ ﺮاﻣﺞ
اﻻﺣﺼﺎﺋﯿﺔ.
logdose
96.1-
93.1-
22.1-
90.1-
00.1-
96.0-
25.0-
93.0-
03.0-
22.0-
5+)2/))(log(p/1 – p))/2 ((log(p/1 – p
*
90576.3
93109.3
88291.4
94294.4
72797.4
70543.5
20339.5
16890.6
*
*
19423.1-
16890.1-
21708.0-
15705.0-
37202.0-
70543.0
20339.0
16890.1
*
)log(p/1 – p
*
28946.2-
22791.2-
52416.1-
10510.1-
74504.0-
51096.0
50668.1
22791.2
*
84
)p/(1 – p
00000.0
66070.0
11111.0
40991.0
04263.0
76666.0
10499.1
96264.6
00000.9
*
1–p
000.1
439.0
009.0
438.0
437.0
006.0
433.0
431.0
001.0
000.0
P
000.0
660.0
001.0
661.0
662.0
004.0
666.0
668.0
009.0
000.1
49. وﺻﻒ ﻣﻨﺤﻨﯿﺎت اﻻﺳﺘﺠﺎﺑﺔ ﻟﻠﺠﺮﻋﺔ........................................... د- ﻓﺮاس اﻟﺴﺎﻣﺮاﺋﻲ
اﻟ ﺬي ﯾﮭﻤﻨ ﺎ ﻣ ﻦ اﻟﺠ ﺪول اﻋ ﻼه ھ ﻮ ﻗ ﯿﻢ اﻟﻌﻤ ﻮدان logdoseووﺣ ﺪات Logit
اﻟﺘ ﻲ ﺣ ﺼﻠﻨﺎ ﻋﻠﯿﮭ ﺎ ﻣ ﻦ ﺗﺤﻮﺑ ﻞ ﻧ ﺴﺒﺔ اﻟﺤﯿﻮاﻧ ﺎت اﻟﻤﯿﺘ ﺔ وﻓ ﻖ اﻟﺪاﻟ ﺔ
5+)2/) (( log(p/1 – pاذ ان اﻟﻌﻤ ﻮدان ﯾﻤ ﺜﻼن اﻟﺠﺮﻋ ﺔ واﻻﺳ ﺘﺠﺎﺑﺔ ﻋﻠ ﻰ
اﻟﺘﻮاﻟﻲ ﺛﻢ ﻧﻄﺒﻖ ﻗﯿﻤﮭﻤﺎ ﻋﻠ ﻰ ﻣﻌﺎدﻟ ﺔ اﻻﻧﺤ ﺪار اﻟﺨﻄ ﻲ اﻟﺒ ﺴﯿﻂ واﻟﺘ ﻲ ﺳ ﯿﻖ وان
ﺷﺮﺣﻨﺎ ﺧﻄﻮات اﻟﺤﻞ ﻓﯿﮭﺎ وﺳﻨﺤﺼﻞ ﻋﻠﻰ ﻣﻌﺎدﻟﺔ اﻟﺘﻮﻗﻊ اﻟﺘﺎﻟﯿﺔ :
Y = 6.627 + 2.209logdose
اذ ﻋﻮﺿﻨﺎ ﻋﻦ اﻟﻌﻤﻮد 5+)2/) (( log(p/1 – pﺑﺎﻟﺤﺮف .Y
وﺑﺬﻟﻚ ﻓﺄن ﻟﻮﻏﺎرﯾﺘﯿﻢ ﻗﯿﻤﺔ 05 LDھﻮ:
5= 6.627 + 2.209logdose
37.0 – = 902.2 /726.1 – = Logdose
ﻟﻘﺪ ﻋﻮﺿﻨﺎ ﻋﻦ ﻗﯿﻤﺔ Yﺑﺎﻟﺮﻗﻢ 5 وذﻟﻚ ﻻن ﻧﺴﺒﺔ اﻟﮭﻼﻛﺎت 05% اﻟﻤﺘﻮﻗﻌﺔ ﻋﻨﺪ
ﺗﻄﺒﯿﻘﮭﺎ ﻋﻠﻰ ﻣﻌﺎدﻟﺔ اﻟﺘﺤﻮﯾﻞ 5+)2/) (( log(p/1 – pﺗﺴﺎوي 5 .
)1(Log (0.50/(1 – 0.50)) = log
0 = )1(Log
5=5+2/0
ﯾﻤﻜ ﻦ اﺟ ﺮاء ﻣﻘﺎرﻧ ﺔ ﺑ ﯿﻦ ﻃﺮﯾﻘﺘ ﻲ Probitو Logisticﺑﻌ ﺪ اﺟ ﺮاء اﻟﺘﺤﻮﯾ ﻞ
وﻣ ﻦ ﺛ ﻢ اﻟﺘﻨﻔﯿ ﺬ ﻟﺘﺤﺪﯾ ﺪ اي اﻟﻄ ﺮﯾﻘﺘﯿﻦ اﻓ ﻀﻞ ﻓ ﻲ وﺻ ﻒ ﻣﻨﺤﻨ ﻰ اﻻﺳ ﺘﺠﺎﺑﺔ اذ
ﺳ ﺘﻈﮭﺮ ﻓ ﻲ اﻟﻨﺘ ﺎﺋﺞ ﺗﻘ ﺪﯾﺮات ² Rوھ ﻲ ﻛﻤ ﺎ ﯾﺒ ﺪو ﻛﺎﻧ ﺖ اﻋﻠ ﻰ )ﻣﺜ ﺎل 2( ﻋﻨ ﺪ
اﺳ ﺘﻌﻤﺎل ﻃﺮﯾﻘ ﺔ (%96) Logisticﻓﯿﻤ ﺎ ﺑﻠﻐ ﺖ 98% ﻓ ﻲ ﻃﺮﯾﻘ ﺔ .Probit
ﯾﻤﻜﻦ اﺳ ﺘﻌﻤﺎل اي ﻣ ﻦ اﻟﺒ ﺮاﻣﺞ اﻟﻤ ﺬﻛﻮرة ﻟﻠﺤ ﺼﻮل ﻋﻠ ﻰ ﻗﯿﻤ ﺔ ² Rوﻗ ﺪ اﺧﺘﺮﻧ ﺎ
ﺑﺮﻧﺎﻣﺞ Minitabﻟﻠﺘﻨﻔﯿﺬ.
Regression Analysis: f versus g
%4.69 = )R-Sq(adj
P
000.0
F
21.981
The regression equation is
f = 6.627 + 2.209 g
754271.0 = S
%9.69 = R-Sq
Analysis of Variance
MS
07426.5
47920.0
94
SS
07426.5
54871.0
51308.5
DF
1
6
7
Source
Regression
Error
Total
