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ISBN_978-4-7806-0852-6_Edition1_Printing4_p175_20_6.ipynb
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"name": " ISBN_978-4-7806-0852-6_Edition1_Printing4_p175_20_6.ipynb", | |
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"authorship_tag": "ABX9TyNZ6lCjt2kaMR0vO3OntYVL", | |
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"<a href=\"https://colab.research.google.com/gist/axjack/c9e0274acab5760051eade6a0cabf4cc/-isbn_978-4-7806-0852-6_edition1_printing4_p175_20_6.ipynb\" target=\"_parent\"><img src=\"https://colab.research.google.com/assets/colab-badge.svg\" alt=\"Open In Colab\"/></a>" | |
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{ | |
"cell_type": "markdown", | |
"metadata": { | |
"id": "BNI_te9MmYU-" | |
}, | |
"source": [ | |
"# 書籍情報\n", | |
" - 『統計検定準1級対応 統計学実践ワークブック』 日本統計学会編,学術図書出版社\n", | |
" - 第1版第4刷\n", | |
" - p.175の式(20.6)\n", | |
"\n", | |
"# 問い合わせ箇所\n", | |
"式(20.6)\n", | |
"\n", | |
"$$\n", | |
"S_{[k]} = \\sum_{i=1}^N \\sum_{j=1}^2 (\\bar{y}_{[k]j} - \\bar{y} )^2 = \\frac{N}{4}(\\bar{y}_{[k]1} - \\bar{y}_{[k]2})^2\n", | |
"$$\n", | |
"\n", | |
"の真ん中の式から一番右の式は導出できない。\n", | |
"\n", | |
"# どうあるべきか\n", | |
"\n", | |
"$$\n", | |
"S_{[k]} = \\sum_{i=1}^{N/2} \\sum_{j=1}^2 (\\bar{y}_{[k]j} - \\bar{y} )^2 = \\frac{N}{4}(\\bar{y}_{[k]1} - \\bar{y}_{[k]2})^2\n", | |
"$$\n", | |
"\n", | |
"もしくは、\n", | |
"\n", | |
"$$\n", | |
"S_{[k]} = \\frac{1}{2}\\sum_{i=1}^{N} \\sum_{j=1}^2 (\\bar{y}_{[k]j} - \\bar{y} )^2 = \\frac{N}{4}(\\bar{y}_{[k]1} - \\bar{y}_{[k]2})^2\n", | |
"$$\n", | |
"\n", | |
"ではないかと考えます。\n", | |
"\n", | |
"# 問い合わせに至った理由\n", | |
"\n", | |
"応答 $ y_i $ の総平均を、\n", | |
"$$\\bar{y} = \\frac{1}{N}\\sum_{i=1}^N y_i \\tag{0}$$\n", | |
"\n", | |
"とし、以下が成り立つとする。\n", | |
"\n", | |
"$$\n", | |
" \\frac{N}{2} \\bar{y}_{[k]1} + \\frac{N}{2} \\bar{y}_{[k]2} = \\sum_{i=1}^N y_i \\tag{1}\n", | |
"$$\n", | |
"\n", | |
"このとき、\n", | |
"$$\n", | |
" \\frac{1}{2} \\Biggl( \\bar{y}_{[k]1} + \\bar{y}_{[k]2} \\Biggr) = \\frac{1}{N}\\sum_{i=1}^N y_i = \\bar{y} \\tag{2}\n", | |
"$$\n", | |
"\n", | |
"である。すると、p.175 (20.6)は、\n", | |
"\n", | |
"$$\n", | |
"\\begin{aligned}\n", | |
"S_{[k]} &= \\sum_{i=1}^N \\sum_{j=1}^2 (\\bar{y}_{[k]j} - \\bar{y} )^2 \\\\\n", | |
" &= \\sum_{i=1}^N \\Biggl[\n", | |
" (\\bar{y}_{[k]1} - \\bar{y} )^2 \n", | |
" +\n", | |
" (\\bar{y}_{[k]2} - \\bar{y} )^2 \n", | |
" \\Biggr]\\\\\n", | |
" &= \\sum_{i=1}^N \\Biggl[\n", | |
" (\\bar{y}_{[k]1} - \\frac{1}{2} \\Bigl( \\bar{y}_{[k]1} + \\bar{y}_{[k]2} \\Bigr) )^2 \n", | |
" +\n", | |
" (\\bar{y}_{[k]2} - \\frac{1}{2} \\Bigl( \\bar{y}_{[k]1} + \\bar{y}_{[k]2} \\Bigr) )^2 \n", | |
" \\Biggr]\\\\\n", | |
" &= \\sum_{i=1}^N \\Biggl[\n", | |
" (\\frac{1}{2} \\Bigl( \\bar{y}_{[k]1} - \\bar{y}_{[k]2} \\Bigr) )^2 \n", | |
" +\n", | |
" (-\\frac{1}{2} \\Bigl( \\bar{y}_{[k]1} - \\bar{y}_{[k]2} \\Bigr) )^2 \n", | |
" \\Biggr]\\\\\n", | |
" &= \\sum_{i=1}^N \\Biggl[\n", | |
" \\frac{2}{2^2} \\Bigl( \\bar{y}_{[k]1} - \\bar{y}_{[k]2} \\Bigr)^2 \\Biggr]\\\\\n", | |
" &= \n", | |
" \\frac{N}{2} \\Bigl( \\bar{y}_{[k]1} - \\bar{y}_{[k]2} \\Bigr)^2 \\\\\n", | |
"\\end{aligned}\n", | |
"$$\n", | |
"\n", | |
"となって、式(20.6)のようにイコールでは結べない、と考えました。\n", | |
"\n", | |
"以上。\n" | |
] | |
} | |
] | |
} |
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問い合わせ結果