Question

In: Statistics and Probability

A statistician took a walk down the street at noon during a day when schools were...

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.
  3. Find the 95% confidence limits on the expected height of 4.5-year-old child on the street.

Solutions

Expert Solution

X Y XY
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 = 4.5

ŷ = 73 + (6.8) * 4.5 = 103.6

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

c) 95% confidence limits on the expected height of 4.5-year-old child on the street:

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

= 103.6 - 3.1824*4.2*√(1 + (1/5) + ((4.5 - 4)²/(10))) = 88.81

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

= 103.6 + 3.1824*4.2*√(1 + (1/5) + ((4.5 - 4)²/(10))) = 118.39


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