Question

A researcher studies text messages and sleep in teenagers. Consider the following data for text messages sent per day and average hours of sleep for five teenagers. Texts per day Hours of sleep 4.2 132 528 77 115 209 6 5.2 3.2 (a) Find the sample mean and standard deviation of texts per day and hours of sleep. (b) Find the correlation coefficient r between texts per day and hours of sleep. Describe what your value of r means.

Answer 1

flow of sleep 115 20 = (a) Sample mean my n texto per day 132527115 + 901 585 1124 Sample men ua af hours of sleep = 42.+ 9 + 6 45.2 +32 = 56-6.5 32 -Uz Ilerdandvatten Xiu)? Texts per day = (132-1177+ (52-1179?+ (77-1173'+ (115-1177) + (209-115? = 29036 Howrs y oleep = (4-2-5-32)° +18-5-32)+(6-5-32) + (5-2-5-32)+(3.2-532)2 5 . -2.6816 (6) Correlation coefficient ar=? It's a stastical oneasure of the strength of the relationship between the relative morements of two 1. variables. The values ronge between I tot It si-l or r> 1 ; ensor in correlation measurement Gr=0 y no linear relatio

Ifra1.0 perfeeſ 20 » perfect negative correlahu * perfeul poilne lomelation Py = E(X;-)lyi 9) 5 =(x,-)'>(i :D)' Here xu y = 1 hellys 117, l, 5'32 2 texts per day of students Yi hour of sleep of studente xil S Y (34;=7) (477) -16.8 1.12 2.68 - 65 -17.4.2 -40 0.68 -9 -27.2 0.12 2.12 0:24 -195.04 fim - 413 E(X-4,7 = 14518 Z(X ) = Elyi-T)' = E(Yi -4,)2 = 13.408 Txy = _-413 578 X 13.408 - 4113 441199991 =-0.936 08364 negative correlation coeffierent Higher chances of getting dess hout of sleep it texts per day are more

Answer 2

Persons Hours of sleep Texts per day 132 52 77 115 209 5.2 3 .2 na no of data points a) Sample mean For texts per day x = 3 X = X = 132+52477 +)151209 = 585 lă = 117 Sample mean For Hours of sleep ý = sy AW: - 4.2+8+6+5.2 +3.2 5 - 266 19 25.32 texts per day standard deviation for Ž (xi-x) 2 Sx=v no) Xi = each value of the data n= no. of data points X= mean of xi

this Texts per day (-7)e (24- ) (x-117) 15 -65 132 117 52 77 115 225 4225 16oo 117 -40 -2 209 117 92 8464 {= 14518 Total= using Sx Ŝ (-)2 n-) where n = 5 V 14 518 5-1 114518 V 4 = √3629.5 Sx = 60. 245 Standard deviation of texts per day = 60.245 Now, we will do sume for hours of sleep

For this 1 Hours of sleep ý Cyi-9) ² Table(II) 4.2 5.32 (2 7 ) (y- 5.32) -1.12 2.68 0.68 -0.12 5.32 5.32 5.32 1.2544 7.1824 0.462 4 0.0144 4.4944 5.2 - 2 .12 Total (E) 13.408 using Sy= 2 (4 ;-})? whese nas 13.408 V 5-1 13.408 sy = 3.6616 Standaod deviation of hours of sleep = 3.6616
i have used a better correction coefficient formula and also explained why .you will get it read carefully. Hope you got your answer
13225 b) Now correlation coefficient & between texts per Tably day and hours of sleep Text perday Hours or sleep 2² 1 y ² xy 132 4.2 17424 17.64 554.4 52 2704 64 77 416 115 5929 36 462 209 27.04 593 436 81 110:24 668.8 Total a 82963 154.92 2699.2 8= {(x-7)(y-7) Shere Ely-y) = comes out - (9-3)*xEcy-g)2 to bezero ) 1 See Table (I) =(4-3) last wro) so we will use a better simplified way new formula for calculation purpose. r= {xy-năý ý -5.32 Sex-n22) (E4²_n92) r = 2699.2 - 5 x (117)x(5:32) x= 117 (froma) 182963-15x (117)) (154.92 -(5x (5-32)2) r= - 413 (82963-68445) 154.32-141.512) r= - 413 -413 14518 x 13408 J 19 4 657.344 - 413 4.41.1991 r = -0.93608 these is high negative correlation between text perday. and hours of skep. here & means