Question

a grass seed company conducts a study to determine the relationship between the density of seeds planted and the quality of the resulting lawn. eight similar plots of land are selected and each is planted with a particular density of seed. one month later the quality of each lawn is rated on a scale of 0 to 100. Using a 5% significance level determine whether the data provide sufficient evidence to conclude that seed density and lawn quality are positively linearly correlated.

A grass seed company conducts a study to determine the relationship between the density of seeds planted (in pounds per 500 sq ft) and the quality of the resulting lawn. Eight similar plots of land are selected and each is planted with a particular density of seed. One month later the quality of each lawn is rated on a scale of 0 to 100. The sample data are given below. Seed Density, x Lawn Quality, y 1 30 1 40 2 40 3 40 3 50 3 65 4 SO 5 50 Using a 5% significance level, determine whether the data provide sufficient evidence to conclude that seed density and lawn quality are positively linearly correlated

Answer 1

Solution: correlation coefficient is given by r= cov(x,y) 1 EXyX ſox² oy ² 43x=**)# 281-9") Y x? yoo - 1 1 1 Seed Density Lawn Quality X, 30 40 2 40 3 ЧО 3 3 65 4 50 5 50 SIX=22 Sy = 365 2 xy 30 40 1600 80 4 1600 120 9 1600 150 9 2500 195 9 4225 200 16 2500 250 25 2500 EXY = 1065 | Ex?=74 SE7425 50 n=8 EX= 22 EY= 365 EXY= 1065 EX²=74 EY2 = 17425 X: ΣΧ n X = 2.75 22 8 2.75

y= sy n 365 8 = 15.625 Y= 45.625 DR 1x1065 2.75X 45.625 x+4-(25) x17425-4 -(पड.625) K - 133.125 125.469 9.25 - 7.5625] [2178.125 – 2081• ६५ ।] 7.656 7.656 16875x96.484 162.81615 ४ 7.656 12.7599 0.60 correlation (7) = 0.60

Hypothesis testing Ho; p=0 Hi pro ; seed density and lawn quality are positivelyo linearly Córrelated. This is one tailed (Right tailed test) test statistic Under Ho ta rv (0-2) ver2 t = 0.60 /(8-2). 0.60 X16 J-0.60) th-0.36) ta 1.4697 vooby 1.4697 0.8 1-8372 test statistic t = 1.8372 Significance level oto 1% = 0.01 degree of freedom = n-2 af = 8-2 = 6 Idf=6 Critical value of t at 0.01 level of significance for 6 degree of freedom for Right failed test torp (6) = 3.143 Ifrom t-tables Right tailed

Calculated t < critical t 1.8372 < 3.143 So, we fail to reject Ho at 1% level of Significance. conclusion: Do not Reject Her At the 1% Significance level, the data do not provide Sufficient evidence to conclude that seed density and lawn qu quality are positively linearly correlated.