Question

do not round answer!

We have a dataset with n = 10 pairs of observations (Li, Yi), and 2. 683, Σ % = 813, IM:IM: 1-1 72 r} = 47, 405, Xiyi = 56,089, y = 66, 731. 1=1 i=1 What is an approximate 99% confidence interval for the slope of the line of best fit?

Answer 1

Solution-:

Given:

η = 10, t, = - 683.Σ. = 813, Σι? = 47405. Συγ = 66731 1=1 1=1 1=1 1=1
and $\sum_{i=1}^{n}x_{i}*y_{i}=56089$

We find as follows,

1,01, cou(x, y), r = corr(x, y), byr, S y

Σ1 683 T = 68.3 n 10
and $\bar{y}=\frac{\sum_{i=1}^{n}y_{i}}{n}=\frac{813}{10}=81.3$

ΣΕ 13 ση - 12 = (47405 10 68.32 = 75.61 = 8.6954 n

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

Couc, y) T corr(x,y) = 56.11 8.6954* 7.9630 0.81 01 * Oy

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

Also we find,

sands

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

$SE_{b}=\sqrt{\frac{(1-r^2)*s^2_{y}}{(n-2)*s^2_{y}}}=\sqrt{\frac{(1-0.81^2)*84.0111}{(10-2)*70.4556}} =\sqrt{\frac{28.8914}{563.6448}}=0.2264$

Here,

df = n-1= 10 – 1= 9

Table 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

(0.742143.250 +0.2264)

(0.7421 -0.7358, 0.7421 +0.7358)

(0.0063, 1.4779)

The required C.I. the slope of the line of the best fit is (0.006,1.477)