Answer to Question #187725 in Quantitative Methods for Olivia Padula

Question

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?

 

Answer 1

$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.$