Question

Question 16 2 pts As people age they begin experience hearing loss. A study was done on a random sample of 8 subjects to determine "comfort level of sound and age in people. The data are given here: Age, x(years) 15 25 35 45 55 65 75 85 Sound level y (decibels) 56 57 a a 68 74 78 85 xy = 29010, n= 7 x = 400, z = 24200, Ey=546, y = 37986, 550=4200 SSxy=1710, SSyy=721.5 SSE=25.2857 Se=2.052872 Assume the existence of a linear relationship between comfortable sound level and age, find the equation of the regression line relating y tox. what is slope B, Question 17 2 pts Assume the existence of a linear relationship between comfortable sound level and age, find the equation of the regression line relating y tox. what is y intercept point B = y - . . * Find =? Question 18 4 pts Find the predicted value of sound for a 60 year old person. Question 19 4 pts Construct a 90% prediction interval for the comfortable level of sound for a 60 year old person. Put only the margin of error. format is (X.XXX)

Answer 1

Answer:

Given That:-

As people age they begin experience hearing loss. A study was done on a random sample of 8 subsets to determine "confirmed" comfort level" of sound and again people.

	X	Y	X * Y	X2	Ŷ
	15	56	840	225	54
	25	57	1425	625	58.07143
	35	64	2240	1225	62.14286
	45	64	2880	2025	66.21429
	55	68	3740	3025	70.28571
	65	74	4810	4225	74.35714
	75	78	5850	5625	78.42857
	85	85	7225	7225	82.5
Total	400	546	29010	24200	3631.049

(16) Assume the existence of a linear relationship between comfortable sound level and age, find the equation of the regresion line realting y to x.

Equation of regression line is Ŷ = a + bX
b = ( n Σ(XY) - (ΣX* ΣY) ) / ( n Σ X2 - (ΣX)2 )
b = ( 8 29010 - 400 546 ) / ( 8 * 24200 - ( 400 )2)
b = 0.4071

a =( ΣY - ( b * ΣX ) ) / n
a =( 546 - ( 0.4071 * 400 ) ) / 8
a = 47.8929
Equation of regression line becomes Ŷ = 47.8929 + 0.4071 X

(18) Find the predicted value of sound for a 60 year old person.

Ŷ = 47.8929 + 0.4071X

Ŷ = 47.8929 + 0.4071 (60)
Ŷ = 72.3189

(19) Construct a 90% prediction interval for the comfortable levle of sound for a 60 year ole person.

Estimated Error Variance $(\hat{\sigma}_{2})$

= 52 = (Sy – b * Szy)/n - 2

S2 = ( 721.5 - 0.4071 * 1710 ) / 8 - 2
S2 = 4.2265
S = 2.0558

Predictive Confidence Interval of  

09=xlA

Ŷ = 47.8929 + 0.4071X
Ŷ = 72.3189

1 se[yo] ta/2 x 1/[1+ - + n 20-0 SE

t(4.2265/2) = t(0.1/2) = 1.943

X̅ = (Xi / n ) = 400/8 = 50

se yol = Y = 72.318972.3189 + 1.943 x 1 [1++ (60) - 50) 4200 -] x 4.2265 A
90% Predictive confidence interval is  
(68.0375 < MY X=60 < 76.6003)

ISSE SS 4 R = 0.9852 .- R502857 1710 21) jan(exy)-({x)(84) - [1823(53)?][654? czyjej • 8(2900) -(400) (546) f8f242009640233343437989=654632 r=0.9823

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