Question

Short Answer Question We have a dataset with n = 10 pairs of observations (li, Yi), and n n Σ Ij = 683, yi = 813, i=1 i=1 n n < 1* = 47, 405, liyi = 56,089, { y = 66,731. i=1 i=1 i=1 What is an approximate 99% confidence interval for the intercept of the line of best fit?

Answer 1

Solution-:

η = 10, t, = - 683.Σ!: = 813, Σ? = 47405) y = 66731 1=1 1=1 1=1 1=1

and

Στη Σ; * Ji 56089 1=1

We find $\bar{x}, \sigma_{x}, cov(x,y),r=corr(x,y), b_{yx}, s^2_{x},s^2_{y}$ as follows,

Σ1 683 T = 68.3 n 10
and
Σ=1 Vi 813 3 = 81,3 n 10

$\sigma_{x}=\sqrt{\frac{\sum_{i=1}^{n}x^2_{i}}{n}-\bar{x}^2}=\sqrt{\frac{47405}{10}-68.3^2}=\sqrt{75.61}=8.6954$

ΣΕ1 σκαι Σ - y = 66731 10 - 81.32 , Ε V63.41 = 7.9630 n

Cou(z,y) ΣΙ , * 4: - τ * y = 56089 10 68.3 και 81.3 = 56.11 n

$r=corr(x,y)=\frac{Cov(x,y)}{\sigma_{x}*\sigma_{y}}=\frac{56.11}{8.6954*7.9630}=0.81$

$b_{yx}=\frac{Cov(x,y)}{\sigma^2_{x}}=\frac{56.11}{75.61}=0.7421$

Also we find, $s_{x}^2,s_{y}^2$

2 Σα – Ι)2 η –1 Σ? - η * (7)? η –1 47405 – 10 * 68.32 10 -1 - 84.0111 T

2 = Σy - y)2 Σy? – η * (g)? η – 1 η -1 66731 – 10 * 81.32 10 -1 - 70.4556 y

(1 – 12) * SED (1 – 0.812) * 84.0111 (10 – 2) * 70.4556 (n − 2)* s; 28.8914 563.6448 0.0513 0.2264

Here, $df=n-1=10-1=9$

Table value / Critical value:

(From t- table)

ta/2,n-1 = $0.01/2,10-1 t0.005,9 3.250

The 99 % C.I. for the slope is,

(byr #ta/2,n-1 * SE

$(0.7421 \pm 3.250*0.2264)$

(0.7421 +0.7358)

The required C.I. the slope of the line of the best fit is