Question #234904

A scientist assumes that there is a linear relationship between the amount of vitamin supplement supplied to cattle and subsequent yield of milk obtained. Eight cows of same kinds were selected at random and treated, weekly with water in which X grams of vitamin supplements was dissolved. The yield Y litter of milk was recorded. Cattle A B C D E F G H X 1.0 1.4 2.1 2.3 3.0 3.2 4.0 4.3 Y 3.9 4.4 4.5 5.0 5.1 5.3 5.5 5.4


Expert's answer

XY1.03.91.44.42.14.52.35.03.05.13.25.34.05.54.35.4\def\arraystretch{1.5} \begin{array}{c:c} X & Y \\ \hline 1.0 & 3.9 \\ \hdashline 1.4 & 4.4\\ \hdashline 2.1 & 4.5 \\ \hdashline 2.3 & 5.0\\ \hdashline 3.0 & 5.1 \\ \hdashline 3.2 & 5.3\\ \hdashline 4.0 & 5.5 \\ \hdashline 4.3 & 5.4\\ \end{array}

xˉ=∑ixin=21.38=2.6625\bar{x}=\dfrac{\sum_ix_i}{n}=\dfrac{21.3}{8}=2.6625

yˉ=∑iyin=39.18=4.8875\bar{y}=\dfrac{\sum_iy_i}{n}=\dfrac{39.1}{8}=4.8875

SSxx=∑ixi2−1n(∑ix1)2=66.39−18(21.3)2SS_{xx}=\sum_ix_i^2-\dfrac{1}{n}(\sum_ix_1)^2=66.39−\dfrac{1}{8}(21.3)^2

=9.67875=9.67875

SSyy=∑iyi2−1n(∑iy1)2=193.33−18(39.1)2SS_{yy}=\sum_iy_i^2-\dfrac{1}{n}(\sum_iy_1)^2=193.33-\dfrac{1}{8}(39.1)^2

=2.22875=2.22875




SSxy=∑ixiyi−1n(∑ix1)(∑iyi)SS_{xy}=\sum_ix_iy_i-\dfrac{1}{n}(\sum_ix_1)(\sum_iy_i)

=108.49−18(21.3)(39.1)=4.38625=108.49-\dfrac{1}{8}(21.3)(39.1)=4.38625

slope=m=SSxySSxx=4.386259.67875=0.4532slope=m=\dfrac{SS_{xy}}{SS_{xx}}=\dfrac{4.38625}{9.67875}=0.4532

b=yˉ−m⋅xˉ=4.8875−0.4532(2.6625)b=\bar{y}-m\cdot\bar{x}=4.8875-0.4532(2.6625)

=3.6809=3.6809

The equation of the trend line is


Y=0.4532X+3.6809Y=0.4532X+3.6809





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