Question

A statistician took a walk down the street at noon during a day when schools were in session. She recorded the age and measured the height of the first children of different ages she encounter on this street. The oldest child was a six-year old.

The statistician made sure that all the children were unrelated to each other.

Here are the measurements

Child	1	2	3	4	5
Age (years)	2	3	4	5	6
Height (cm)	87	90	102	112	110

SSx=10, SSy=516.8, SSxy=68, ∑x=20, ∑y=501, ∑xy=2072

1. Find the equation describing the linear relationship between age and height for 2– to 6-year old children on that street.

2. Conduct a test to determine if age, x, is useful linear predictor of height change. Assume s=4.2 and use a = 0.01.

Find the 95% confidence limits on the expected height of 2-year-old child on the street.

Answer 1

X	Y	XY	X²	Y²
2	87	174	4	7569
3	90	270	9	8100
4	102	408	16	10404
5	112	560	25	12544
6	110	660	36	12100
Ʃx =	Ʃy =	Ʃxy =	Ʃx² =	Ʃy² =
20	501	2072	90	50717

Sample size, n =	5
x̅ = Ʃx/n = 20/5 =	4
y̅ = Ʃy/n = 501/5 =	100.2
SSxx = Ʃx² - (Ʃx)²/n = 90 - (20)²/5 =	10
SSyy = Ʃy² - (Ʃy)²/n = 50717 - (501)²/5 =	516.8
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 2072 - (20)(501)/5 =	68

a)

Slope, b = SSxy/SSxx = 68/10 = 6.8

y-intercept, a = y̅ -b* x̅ = 100.2 - (6.8)*4 = 73

Regression equation :

ŷ = 73 + (6.8) x

b)

Null and alternative hypothesis:

Ho: β₁ = 0

Ha: β₁ ≠ 0

n = 5

α = 0.01

Slope, b = 6.8

Standard error of slope, se(b1) = 1.3466

Test statistic:

t = b/(s/√SSxx) = 6.8/(4.2/√10) = 5.12

df = n-2 = 3

p-value = T.DIST.2T(ABS(5.12), 3) = 0.0144

Conclusion:

p-value > α , Fail to reject the null hypothesis.

There is enough evidence to conclude that x is useful linear predictor of height change at 0.01 significance level.

--

Predicted value of y at x = 2

ŷ = 73 + (6.8) * 2 = 86.6

Critical value, t_c = T.INV.2T(0.05, 3) = 3.1824

95% confidence interval on the expected height of 2-year-old child on the street :

Lower limit = ŷ - tc*se*√(1 + (1/n) + ((x-x̅)²/(SSxx)))

= 86.6 - 3.1824*4.2*√(1 + (1/5) + ((2 - 4)²/(10))) = 69.69

Upper limit = ŷ + tc*se*√(1 + (1/n) + ((x-x̅)²/(SSxx)))

= 86.6 + 3.1824*4.2*√(1 + (1/5) + ((2 - 4)²/(10))) = 103.51