Question

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

Answer 1

Solution-:

Given:
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 as follows,

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

Στι 683 ΣΕ 9: 813 1=. T = - 68.3andy : = 81.3 10 n 10

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

$\sigma_{y}=\sqrt{\frac{\sum_{i=1}^{n}y^2_{i}}{n}-\bar{y}^2}=\sqrt{\frac{66731}{10}-81.3^2}=\sqrt{63.41}=7.9630$

$Cov(x,y)=\frac{\sum_{i=1}^{10}x_{i}*y_{i}}{n}-\bar{x}*\bar{y} =\frac{56089}{10}-68.3*81.3=56.11$

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,

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

Critical value:

$t_{\alpha/2,n-1}=t_{0.01/2,10-1}=t_{0.005,9}=3.250$ (From t- table)

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

$(b_{yx} \pm t_{\alpha/2,n-1}*SE_{b})$

$(0.7421 \pm 3.250*0.2264)$

(0.7421 -0.7358, 0.7421 +0.7358)

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