Question

1.) A researcher hypothesizes that people who spend more time surfing the internet will spend less time socializing in real life situations. He/she measures the relationship between internet use (X; hours per week) and social interaction (Y; hours per week) in a sample of n 10 students. The following table lists the hypothetical results of this study: Internet Use (X) Social Interaction (Y) (X-Mx) (X- MACY-My) (Y-My) 4 09 a 6 6 a 12 2 3 3 6 5 2 10 10 3 Mx My = SSx H SSY SP a.) Plot this data using the empty graph below. Be sure to label both your X and Y axis! 12 10

Answer 1

ssy= Ely-my) 2 -2.1730 42.5 V 10- n- Sy = 2.1730 2 Spa (5,-)5, ²+(127) S₂² nit12-2 2 (3.2998) = 10.8886 5-=(55x} = (3.2998) 52²= (ssy)² = (2,1730² 2.1730) = 4.7219 i Spa (10-1)+10.8886+(10-1)+4.7219 iot1o-2 , Sp- 140.4945 18 = 2.7937 i Sp=2.7937 The graph is plot on the graph papec.

Scale: an Xaxis - 1cm = 1 unit an Yaris 1cm= 1 unit 10 y Social inted action na

We have to find Peason's correlation Coefficient(?) for the given set of data. The product moment Pearson's condation coefficient is given by, Sxy Sax. Syy where ΣΥΣ Sxy=Exy Sax= {x² - (21) & Syy = zy ² (Ey)? observation table x2 2 yz SE no. 6. 4 36 2 6 16 3.6 49 36 3. 25 7 6 24 48 35 36 48 14 36 144 5 12 16 2 7 49 7 3 3 g or 5 25 30 36 4 2 20 3 100 g 345 loo 458 30 Total Go 5J 294 Calculations: Say Exy Exxy n = 294- 60x55 10 = 294-330 Say = -36

Date Page Sax= {x? (Ex) ² 458-(60) 458-(60) ² =98 Sxa=g8 Syy={y ² (EY) 345-(55) ² - 42.5 Syy = 42.5 Sxy ✓ Sxa Syy - 36 g8x42.5 -0.5 Hence there is negative correlation beta X fy i e. internet use & social interaction. We have to check significance of correlation at alpha=0.05 level. Hypothesis is, Hoi Thede is no correlation beto xf y variable.ie. internet we & social interaction ic. Hoi p=0 Hi: Thede is significant correlation beto the inteenet use & social interaction ic. Hi :P to The test statistic is, t-e n-2 1-82 dif=n-2

Date Page here idif= n-2=10-2=8 q=0.05 i cɛitical value of t if a=0105 at 8 dif is... 2.306o t = n-2 1-62 2 t = -0.5 10-2 Vi-(6ors) t = -1.6329 hece Calculated t = -10632g < 2,306o de treitical hence we do not leiect Ho at a=0.05 We accept Ho at d =0.05 hence there is no significance corelation betul inteenet we & social interaction. 2 d> coefficient of determination 82 z E = 6o.s) 8 ² = 0.25 e) We have to construct Eegression equation for the given data set here internet we is independent variable & Social intezaction is dependent Vallable

Regresion equation is, y = at bx wherei y is dependent Variable X is independent variable a is intercept & b is the slope by using least square method, the normal equations are, sy= natb Ex & Exy= aɛatbex? 2 from the above observation table if. table for second pact by wing this table we get values of Exy, {x? Exey, Hoenal equations ade, 55= lo at Gob & 294 = Go à + 458b multiplying eat by 6o feqe② by 20 , weget, 3300 - 600 a + 3600b 2940= 600 a + 4580b subst&act ege foom ② , we get, -360= grob 16= 0.3673 4 Put this value in eq 2 0, we get 55 = 10 at 6ob S5 = 10 at Got (-0,3673) lila=7.7038 hence begeesion equation is, y = 7.7038- 0,3673 x

Date Page f) we have to find predicted value of social intečaction ý give the provided inteenet we Values (X). Regression equation is, ỹ = 7,7038 -0,36732 when x= 11 9 - 77033 - 03737 || 19 = 3.6635 when x=15 = 7.7038-03 673 X 15 ŷ = 2.1943 when x=1 3 32039 - 0,373 x 1 Iŷ = 7.3365 when 0=7 9-7-7038-0.367347 9 = 5.1327 when x=0 ŷ = 707038-0,3673X ŷ= 7.7038 -

i predicted values of social interaction are Х У 11 3.6635 15 2.1943 7.3365 5.1327 7.7038 Z Z