Years of education x: 17109121418Hrs spent watching tv(y):1 46 5 3 2a) Null Hypothesis H0:Correlation between x and y (ρ=ρ0)Alternative Hypothesis H1:No correlation between x and y(ρ=ρ0)b)For the significance level α=0.05 the critical value for two tailedt−test is t2α,ν=n−2=t20.05,6−2=t0.025,4=2.78c)The Pearson correlation coefficient r=(n∑i=1nxi2−(∑i=1nxi)2)(n∑i=1nyi2−(∑i=1nyi)2)∑i=1n(xiyi)−(∑i=1n(xi))(∑i=1n(yi))∵The number of sample points are 6, n=6∑i=1n(xiyi)=∑i=16(xiyi)(17)(1)+10(4)+9(6)+12(5)+14(3)+18(2)=17+40+54+60+42+36 =249⇒∑i=16(xiyi)=249∑i=16(xi)=17+10+9+12+14+18=80∑i=16(yi)=1+4+6+5+3+2=21∑i=1nxi2=(17)2+(10)2+(9)2+(12)2+(14)2+(18)2=1,134∑i=1nyi2=(1)2+(4)2+(6)2+(5)2+(3)2+(2)2=91Substituting these values in correlation coefficient formula, we getr=(6(1134)−(80)2)(6(91)−(21)2)249−(80)(21)=(404)(105)−1,431=−0.0337≈−0.034∴The Pearson correlation coefficient r=−0.034The test statistic is t=r(1−r2)(n−2)t=(−0.034)(1−(−0.034)2)(6−2)=−0.0340.998844(4)=−0.034(2.0011)=−0.0680374⇒t=−0.068Since t=−0.068>−2.78,there is no reason to reject the null hypothesis.Therefore, there is a correlation between y andx.d)Since, the correlation coefficient is negative, strength is moderate andin the negative direction.⇒If y increases x decreases.e)The regression equation is obtained by method of least squares.Let the least squares regresion line be y=ax+bThe normal equations of regression line are∑i=1n(yi)=a∑i=1n(xi)+bn∑i=1n(xiyi)=a∑i=1n(xi2)+b∑i=1n(xi)Substituting, n=6, ∑i=16(yi)=21, ∑i=16(xi)=80, ∑i=16(xi)2=1134∑i=16(xiyi)=249,we get21=80a+6b249=1134a+80bSolving these two equations we get a=−20293=−0.4604 & b=2021947=9.6386The regression line is y=−0.4604x+9.6386g)Substituting x=11 in the regression line we gety=−0.4604(11)+9.6386=4.5742Therefore,if the number of study years is 11, the number of hours ofwatching TV is y=4.5742hrs.Strength of the correlation is moderate.
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