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From these data, determine i. The mean, variance and standard deviation of the height of parents and of offspring. ii. The correlation and regression for a regression mean offspring height against mean parental height.

Mathematics

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Hapa kuna hatua za kutatua swali lako:

Data Zilizopewa:
Urefu wa Wazazi (X): 150,157,188,165,160,142,170,183,152,173150, 157, 188, 165, 160, 142, 170, 183, 152, 173
Urefu wa Watoto (Y): 152,163,193,163,152,157,183,175,163,180152, 163, 193, 163, 152, 157, 183, 175, 163, 180
Idadi ya familia, n=10n = 10.

Sehemu i: Maana, Tofauti (Variance) na Mkengeuko Wastani (Standard Deviation)

  1. Kokotoa Jumla na Maana kwa Urefu wa Wazazi (X).

    Jumla ya X:

    ∑X=150+157+188+165+160+142+170+183+152+173=1640\sum X = 150 + 157 + 188 + 165 + 160 + 142 + 170 + 183 + 152 + 173 = 1640

    Maana ya X:

    Xˉ=∑Xn=164010=164 cm\bar{X} = \frac{\sum X}{n} = \frac{1640}{10} = 164 \text{ cm}
  2. Kokotoa Tofauti (Variance) na Mkengeuko Wastani (Standard Deviation) kwa Urefu wa Wazazi (X).

    Jumla ya miraba ya mkengeuko kutoka maana:

    ∑(X−Xˉ)2=(150−164)2+(157−164)2+(188−164)2+(165−164)2+(160−164)2+(142−164)2+(170−164)2+(183−164)2+(152−164)2+(173−164)2\sum (X - \bar{X})^2 = (150-164)^2 + (157-164)^2 + (188-164)^2 + (165-164)^2 + (160-164)^2 + (142-164)^2 + (170-164)^2 + (183-164)^2 + (152-164)^2 + (173-164)^2 =(−14)2+(−7)2+(24)2+(1)2+(−4)2+(−22)2+(6)2+(19)2+(−12)2+(9)2= (-14)^2 + (-7)^2 + (24)^2 + (1)^2 + (-4)^2 + (-22)^2 + (6)^2 + (19)^2 + (-12)^2 + (9)^2 =196+49+576+1+16+484+36+361+144+81=1944= 196 + 49 + 576 + 1 + 16 + 484 + 36 + 361 + 144 + 81 = 1944

    Tofauti (Variance) ya X:

    sX2=∑(X−Xˉ)2n−1=194410−1=19449=216 cm2s_X^2 = \frac{\sum (X - \bar{X})^2}{n-1} = \frac{1944}{10-1} = \frac{1944}{9} = 216 \text{ cm}^2

    Mkengeuko Wastani (Standard Deviation) wa X:

    sX=sX2=216≈14.6969≈14.70 cms_X = \sqrt{s_X^2} = \sqrt{216} \approx 14.6969 \approx 14.70 \text{ cm}
  3. Kokotoa Jumla na Maana kwa Urefu wa Watoto (Y).

    Jumla ya Y:

    ∑Y=152+163+193+163+152+157+183+175+163+180=1681\sum Y = 152 + 163 + 193 + 163 + 152 + 157 + 183 + 175 + 163 + 180 = 1681

    Maana ya Y:

    Yˉ=∑Yn=168110=168.1 cm\bar{Y} = \frac{\sum Y}{n} = \frac{1681}{10} = 168.1 \text{ cm}
  4. Kokotoa Tofauti (Variance) na Mkengeuko Wastani (Standard Deviation) kwa Urefu wa Watoto (Y).

    Jumla ya miraba ya mkengeuko kutoka maana:

    ∑(Y−Yˉ)2=(152−168.1)2+(163−168.1)2+(193−168.1)2+(163−168.1)2+(152−168.1)2+(157−168.1)2+(183−168.1)2+(175−168.1)2+(163−168.1)2+(180−168.1)2\sum (Y - \bar{Y})^2 = (152-168.1)^2 + (163-168.1)^2 + (193-168.1)^2 + (163-168.1)^2 + (152-168.1)^2 + (157-168.1)^2 + (183-168.1)^2 + (175-168.1)^2 + (163-168.1)^2 + (180-168.1)^2 =(−16.1)2+(−5.1)2+(24.9)2+(−5.1)2+(−16.1)2+(−11.1)2+(14.9)2+(6.9)2+(−5.1)2+(11.9)2= (-16.1)^2 + (-5.1)^2 + (24.9)^2 + (-5.1)^2 + (-16.1)^2 + (-11.1)^2 + (14.9)^2 + (6.9)^2 + (-5.1)^2 + (11.9)^2 =259.21+26.01+620.01+26.01+259.21+123.21+222.01+47.61+26.01+141.61=1750.9= 259.21 + 26.01 + 620.01 + 26.01 + 259.21 + 123.21 + 222.01 + 47.61 + 26.01 + 141.61 = 1750.9

    Tofauti (Variance) ya Y:

    sY2=∑(Y−Yˉ)2n−1=1750.910−1=1750.99≈194.5444≈194.54 cm2s_Y^2 = \frac{\sum (Y - \bar{Y})^2}{n-1} = \frac{1750.9}{10-1} = \frac{1750.9}{9} \approx 194.5444 \approx 194.54 \text{ cm}^2

    Mkengeuko Wastani (Standard Deviation) wa Y:

    sY=sY2=194.5444≈13.9479≈13.95 cms_Y = \sqrt{s_Y^2} = \sqrt{194.5444} \approx 13.9479 \approx 13.95 \text{ cm}

    Muhtasari wa Sehemu i:

    • Maana ya urefu wa wazazi (Xˉ\bar{X}): 164 cm\boxed{\text{164 cm}}
    • Tofauti ya urefu wa wazazi (sX2s_X^2): 216 cm2\boxed{\text{216 cm}^2}
    • Mkengeuko wastani wa urefu wa wazazi (sXs_X): 14.70 cm\boxed{\text{14.70 cm}}
    • Maana ya urefu wa watoto (Yˉ\bar{Y}): 168.1 cm\boxed{\text{168.1 cm}}
    • Tofauti ya urefu wa watoto (sY2s_Y^2): 194.54 cm2\boxed{\text{194.54 cm}^2}
    • Mkengeuko wastani wa urefu wa watoto (sYs_Y): 13.95 cm\boxed{\text{13.95 cm}}

    Sehemu ii: Uwiano (Correlation) na Urejeshaji (Regression)

  5. Kokotoa ∑XY\sum XY, ∑X2\sum X^2, na ∑Y2\sum Y^2.

    Tayari tunayo ∑X=1640\sum X = 1640 na ∑Y=1681\sum Y = 1681.
    Jumla ya X2X^2:

    ∑X2=1502+1572+1882+1652+1602+1422+1702+1832+1522+1732=270904\sum X^2 = 150^2 + 157^2 + 188^2 + 165^2 + 160^2 + 142^2 + 170^2 + 183^2 + 152^2 + 173^2 = 270904

    Jumla ya Y2Y^2:

    ∑Y2=1522+1632+1932+1632+1522+1572+1832+1752+1632+1802=282327\sum Y^2 = 152^2 + 163^2 + 193^2 + 163^2 + 152^2 + 157^2 + 183^2 + 175^2 + 163^2 + 180^2 = 282327

    Jumla ya XYXY:

    ∑XY=(150×152)+(157×163)+(188×193)+(165×163)+(160×152)+(142×157)+(170×183)+(183×175)+(152×163)+(173×180)\sum XY = (150 \times 152) + (157 \times 163) + (188 \times 193) + (165 \times 163) + (160 \times 152) + (142 \times 157) + (170 \times 183) + (183 \times 175) + (152 \times 163) + (173 \times 180) =22800+25571+36284+26955+24320+22294+31110+32025+24776+31140=277275= 22800 + 25571 + 36284 + 26955 + 24320 + 22294 + 31110 + 32025 + 24776 + 31140 = 277275
  6. Kokotoa mgawo wa uwiano (correlation coefficient), rr.

    Tumia fomula:

    r=n∑XY−(∑X)(∑Y)[n∑X2−(∑X)2][n∑Y2−(∑Y)2]r = \frac{n \sum XY - (\sum X)(\sum Y)}{\sqrt{[n \sum X^2 - (\sum X)^2][n \sum Y^2 - (\sum Y)^2]}}

    Kwanza, kokotoa nambari (numerator):

    n∑XY−(∑X)(∑Y)=10(277275)−(1640)(1681)n \sum XY - (\sum X)(\sum Y) = 10(277275) - (1640)(1681) =2772750−2756840=15910= 2772750 - 2756840 = 15910

    Kisha, kokotoa sehemu ya kwanza ya denominator:

    n∑X2−(∑X)2=10(270904)−(1640)2n \sum X^2 - (\sum X)^2 = 10(270904) - (1640)^2 =2709040−2689600=19440= 2709040 - 2689600 = 19440

    Na sehemu ya pili ya denominator:

    n∑Y2−(∑Y)2=10(282327)−(1681)2n \sum Y^2 - (\sum Y)^2 = 10(282327) - (1681)^2 =2823270−2825761=−2491= 2823270 - 2825761 = -2491

    Kuna tatizo hapa. n∑Y2−(∑Y)2n \sum Y^2 - (\sum Y)^2 haiwezi kuwa hasi. Hii inaashiria kuwa kuna kosa katika data ya Y au hesabu ya ∑Y2\sum Y^2 au ∑Y\sum Y.
    Hebu turekebishe ∑Y2\sum Y^2 kwa kutumia ∑(Y−Yˉ)2=1750.9\sum (Y - \bar{Y})^2 = 1750.9:
    Tunajua ∑(Y−Yˉ)2=∑Y2−(∑Y)2n\sum (Y - \bar{Y})^2 = \sum Y^2 - \frac{(\sum Y)^2}{n}.
    Hivyo, 1750.9=∑Y2−16812101750.9 = \sum Y^2 - \frac{1681^2}{10}.
    1750.9=∑Y2−2825761101750.9 = \sum Y^2 - \frac{2825761}{10}.
    1750.9=∑Y2−282576.11750.9 = \sum Y^2 - 282576.1.
    ∑Y2=282576.1+1750.9=284327\sum Y^2 = 282576.1 + 1750.9 = 284327.
    Hivyo, ∑Y2\sum Y^2 sahihi ni 284327.

    Sasa turudie hesabu ya sehemu ya pili ya denominator na ∑Y2=284327\sum Y^2 = 284327:

    n∑Y2−(∑Y)2=10(284327)−(1681)2n \sum Y^2 - (\sum Y)^2 = 10(284327) - (1681)^2 =2843270−2825761=17509= 2843270 - 2825761 = 17509

    Sasa tunaweza kuendelea na rr:

    r=15910(19440)(17509)=15910340375000=1591018440.98≈0.8627r = \frac{15910}{\sqrt{(19440)(17509)}} = \frac{15910}{\sqrt{340375000}} = \frac{15910}{18440.98} \approx 0.8627

    Mgawo wa uwiano (correlation coefficient): 0.86\boxed{0.86}

  7. Kokotoa mlinganyo wa urejeshaji (regression equation), Y=a+bXY = a + bX.

    Kwanza, kokotoa mteremko (slope), bb:

    b=n∑XY−(∑X)(∑Y)n∑X2−(∑X)2b = \frac{n \sum XY - (\sum X)(\sum Y)}{n \sum X^2 - (\sum X)^2} b=1591019440≈0.8184b = \frac{15910}{19440} \approx 0.8184

    Kisha, kokotoa sehemu ya kukatiza (y-intercept), aa:

    a=Yˉ−bXˉa = \bar{Y} - b\bar{X} a=168.1−(0.8184)(164)a = 168.1 - (0.8184)(164) a=168.1−134.1976=33.9024a = 168.1 - 134.1976 = 33.9024

    Mlinganyo wa urejeshaji (regression equation):

    Y=33.90+0.82X\boxed{Y = 33.90 + 0.82X}
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