Question

A study analyzing the relationship between the age in years of particular species of fish and its weight in pounds produces the following data set.

For problem 5 a) Find the linear correlation coefficient b) Find the least square regression line c) Find the coefficient of determination d) Find the sample standard deviation of errors se e) Construct a 90% confidence interval for the slope of the population regression line, Pi. 5. A study analyzing the relationship between the age in years of particular species of fish and its weight in pounds produces the following data set. Age : x 2 3 4 Weight: y66 n=7 Ex = 30 2y = 66 Zxy = 313 Ex+= 144 Zy? = 692 SS– 14 – 39 – 1543 ss = 692 – 60 69,71 SS = 313– 30X06 30.14

Answer 1

Page From the above information na 7 Ex= 30 Ey= 66 Exy = 310, E2²=144 212= 692 ssaaz 144 & 302 8 15:43 Soyy = 692 – 662 N 69.71 Bio IMS AM Sony = 313 - 30066 N 3049 ALPHA comeration coefficien (5 - 05:ฯ 1 - 2 174 Inexil-Exig2 ney; 2 (eyir E = ALC d/dx : Sdx Jabl r x² hyp (-) 60999 STO - ENG ( RCL harat 8= & 74313 - 30x66 17x144-(30)2 57 x 692 - (6632 7 г МАТ1 и 8 VCThm IS-SUM FIS-VAR PA Rand Rnd I r= 0.97 The correlation coefficient is o.gr age The relation between a por and y is possitive . If one the variable is increases the other variable is alro increasen or Niseversa.

it feast square regression line 4= Bot Box Po Exiyi- Exizyi Exi2 - exin - 313 - 30x66 144 - - (30)2 7 Bo = 1.056 Biad J-BoT COAT = g. 4285-1.056% 4.2857 B = 1.954 The least sequan regnerrin fine is . 9 = Bot BiX 1 - 1.056 + 1.9548

determination The coefficient of is 22 P2= SSR SST = Elfi - 72 Elyint = 58.8902 69. 7142 . coefficient of determination The 84 % variation ofy explained by the X . standar devication = leur of enor Se seie Elyi-g2 n-2 = 10.8249 - 52.1648 Fra 1.47 sample standard deviation

Page for intewal for go% confidence a Slopelo p1 + taal, on see Bi z t212, 0-2) = (t + x2 SXX dig 54 to 10/2, 7-2) X1:47 +4.2857) 17 15:43 1954 I 2.015*1.47 1.1546 1954 1 3.4201 Peze [-1.47, 5.38 go % confidence internal for Bi ir 1.47 5.387

I will provide R codes also

x=c(2,3,4,4,5,5,7)
> y=c(6,6,8,9,9,13,15)
> x
[1] 2 3 4 4 5 5 7
> y
[1] 6 6 8 9 9 13 15
> model=lm(y~x)
> model

Call:
lm(formula = y ~ x)

Coefficients:
(Intercept) x  
1.056 1.954  

> summary(model)

Call:
lm(formula = y ~ x)

Residuals:
1 2 3 4 5 6 7
1.0370 -0.9167 -0.8704 0.1296 -1.8241 2.1759 0.2685

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 1.0556 1.6989 0.621 0.56163
x 1.9537 0.3746 5.216 0.00342 **
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 1.471 on 5 degrees of freedom
Multiple R-squared: 0.8447, Adjusted R-squared: 0.8137
F-statistic: 27.2 on 1 and 5 DF, p-value: 0.003422
cor(x,y)
[1] 0.9190956