Question #187725

A researcher collected data on years of education and number of hours spent watching TV per day for a sample of 6 adults:


Years of Education (X)

17, 10, 9, 12, 14, 18


Number of Hours Spent Watching TV Per Day (Y)

1, 4, 6, 5, 3, 2


there a statistically significant correlation between years of education and the number of hours spent watching TV per day? Set α = 0.05.

 

 a. State the Null and Research Hypothesis?

b. Find the critical value?

c. Find the Pearson correlation coefficient? (r value)

e. What is the strength and direction of the correlation?

f. What is the regression equation?

 g. Predict the number of hours spent watching TV per day for someone with 11 years of education?

 h. What is the coefficient of determination?

 


1
Expert's answer
2021-05-07T11:48:03-0400

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(ρρ0)b)For the significance level α=0.05 the critical value for two tailedttest is tα2,ν=n2=t0.052,62=t0.025,4=2.78c)The Pearson correlation coefficient r=i=1n(xiyi)(i=1n(xi))(i=1n(yi))(ni=1nxi2(i=1nxi)2)(ni=1nyi2(i=1nyi)2)The number of sample points are 6, n=6i=1n(xiyi)=i=16(xiyi)(17)(1)+10(4)+9(6)+12(5)+14(3)+18(2)=17+40+54+60+42+36  =249i=16(xiyi)=249i=16(xi)=17+10+9+12+14+18=80i=16(yi)=1+4+6+5+3+2=21i=1nxi2=(17)2+(10)2+(9)2+(12)2+(14)2+(18)2=1,134i=1nyi2=(1)2+(4)2+(6)2+(5)2+(3)2+(2)2=91Substituting these values in correlation coefficient formula, we getr=249(80)(21)(6(1134)(80)2)(6(91)(21)2)=1,431(404)(105)=0.03370.034The Pearson correlation coefficient r=0.034The test statistic is t=r(n2)(1r2)t=(0.034)(62)(1(0.034)2)=0.034(4)0.998844=0.034(2.0011)=0.0680374t=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 arei=1n(yi)=ai=1n(xi)+bni=1n(xiyi)=ai=1n(xi2)+bi=1n(xi)Substituting, n=6, i=16(yi)=21, i=16(xi)=80, i=16(xi)2=1134i=16(xiyi)=249,we get21=80a+6b249=1134a+80bSolving these two equations we get a=93202=0.4604 & b=1947202=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.Years\ of \ education \ x: \quad\quad \ 17 \quad 10\quad 9\quad 12\quad 14\quad 18\\ Hrs\ spent\ watching \ tv (y):\quad 1 \quad\ 4 \quad 6 \ \quad 5 \ \quad 3\ \quad 2\\ a)\ Null \ Hypothesis \ H_{0}: Correlation \ between \ x \ and \ y\ (\rho=\rho_{0})\\ \quad Alternative \ Hypothesis\ H_{1}: No \ correlation \ between \ x \ and \ y\\ (\rho\ne\rho_{0})\\ b) For \ the \ significance \ level \ \alpha=0.05\ the \ critical \ value \ for \ two \ tailed \\ t-test\ is \ t_{\frac{\alpha}{2},\nu=n-2}=t_{\frac{0.05}{2},6-2}=t_{0.025,4}=2.78\\ c) The \ Pearson \ correlation \ coefficient \ r= \frac{\sum_{i=1}^{n}(x_iy_i)-(\sum_{i=1}^{n}(x_i))(\sum_{i=1}^{n}(y_i))}{\sqrt{(n\sum_{i=1}^{n}x_{i}^2-(\sum_{i=1}^{n}x_{i})^2)(n\sum_{i=1}^{n}y_{i}^2-(\sum_{i=1}^{n}y_{i})^2)}}\\ \because The \ number \ of \ sample \ points \ are \ 6, \ n=6\\ \sum_{i=1}^{n}(x_iy_i)=\sum_{i=1}^{6}(x_iy_i)(17)(1)+10(4)+9(6)+12(5)+14(3)+18(2)\\ \quad \quad \quad \quad=17+40+54+60+42+36\\ \quad\ \quad \ \quad=249\\ \Rightarrow \sum_{i=1}^{6}(x_iy_i)=249\\ \sum_{i=1}^{6}(x_i)=17+10+9+12+14+18=80\\ \sum_{i=1}^{6}(y_i)=1+4+6+5+3+2=21\\ \sum_{i=1}^{n}x_{i}^2=(17)^2+(10)^2+(9)^2+(12)^2+(14)^2+(18)^2=1,134\\ \sum_{i=1}^{n}y_{i}^2=(1)^2+(4)^2+(6)^2+(5)^2 +(3)^2+(2)^2=91\\ Substituting \ these\ values \ in\ correlation \ coefficient \ formula ,\ we \ get \\ r= \frac{249-(80)(21)}{\sqrt{(6(1134)-(80)^2)(6(91)-(21)^2)}}\\ \quad = \frac{-1,431}{(404)(105)}\\ \quad=-0.0337\approx-0.034\\ \therefore The \ Pearson \ correlation\ coefficient \ r= -0.034\\ The\ test \ statistic \ is \ t=r\sqrt{\frac{(n-2)}{(1-r^2)}}\\ t=(-0.034)\sqrt{\frac{(6-2)}{(1-(-0.034)^2)}}\\ =-0.034\sqrt{\frac{(4)}{0.998844}}\\ =-0.034(2.0011)\\ =-0.0680374\\ \Rightarrow t=-0.068\\ Since \ t=-0.068>-2.78, there \ is \ no \ reason \ to \ reject \ the \ null \ hypothesis.\\ Therefore, \ there \ is \ a \ correlation \ between \ y \ and x. d) Since, \ the \ correlation \ coefficient \ is \ negative, \ strength \ is \ moderate\ and \\ in \ the \ negative \ direction.\Rightarrow 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+b\\ The \ normal\ equations \ of \ regression \ line \ are \\ \sum_{i=1}^{n}(y_i)=a\sum_{i=1}^{n}(x_i)+bn\\ \sum_{i=1}^{n}(x_iy_i)=a\sum_{i=1}^{n}(x_i^2)+b\sum_{i=1}^{n}(x_i)\\ Substituting, \ n=6, \ \sum_{i=1}^{6}(y_i)=21,\ \sum_{i=1}^{6}(x_i)=80,\ \sum_{i=1}^{6}(x_i)^2=1134\\ \sum_{i=1}^{6}(x_iy_i)=249, we \ get \\ 21=80a+6b \\ 249=1134a+80b\\ Solving \ these \ two \ equations \ we \ get \ a=-\frac{93}{202} =-0.4604\ \& \ b=\frac{1947}{202}=9.6386\\ The\ regression \ line \ is \ y= -0.4604x+9.6386\\ g ) Substituting \ x=11 \ in \ the \ regression \ line\ we \ get \\ y= -0.4604(11)+9.6386= 4.5742\\ Therefore, if \ the \ number \ of \ study\ years \ is\ 11, \ the \ number \ of \ hours\ of watching \ TV \ is \ y=4.5742 hrs.\\ Strength \ of\ the\ correlation \ is\ moderate.


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