Question

A research center is interested in investigating about the height and age of children who are between 5 to 9 years old. In order to do this, a sample of 15 children is selected and the data is given below.

Age (in years)	Height (inches)
7	47.3
8	48.8
5	41.3
8	50.4
8	51
7	47.1
7	46.9
7	48
9	51.2
8	51.2
5	40.3
8	48.9
6	45.2
5	41.9
8	49.6

a.. Develop a scatter chart with age as the independent variable. What does the scatter chart indicate about the relationship between the height and age of children?

b. Use the data to develop an estimated regression equation that could be used to estimate the height based on the age. What is the estimated regression model? Round the results to 3 decimal places.

c. How much of the variation in the sample values of height does the model estimated in part (c)? Explain.

d. What is the predicted value of Y (height in inches) for a 6-year old? Develop a 95% confidence interval for the predicted value.

e. Interpret the standard error for this problem

f. State the hypothesis for the significance F and test the hypothesis.

g. State the hypothesis for the significance of the independent variable and test the hypothesis.

Answer 1

a) Scatterplot:

There is a positive relationship between the height and age of children.

b)

Age, X	Height, Y	XY	X²	Y²
7	47.3	331.1	49	2237.29
8	48.8	390.4	64	2381.44
5	41.3	206.5	25	1705.69
8	50.4	403.2	64	2540.16
8	51	408	64	2601
7	47.1	329.7	49	2218.41
7	46.9	328.3	49	2199.61
7	48	336	49	2304
9	51.2	460.8	81	2621.44
8	51.2	409.6	64	2621.44
5	40.3	201.5	25	1624.09
8	48.9	391.2	64	2391.21
6	45.2	271.2	36	2043.04
5	41.9	209.5	25	1755.61
8	49.6	396.8	64	2460.16
Ʃx =	Ʃy =	Ʃxy =	Ʃx² =	Ʃy² =
106	709.1	5073.8	772	33704.59

Sample size, n =	15
x̅ = Ʃx/n = 106/15 =	7.06666667
y̅ = Ʃy/n = 709.1/15 =	47.2733333
SSxx = Ʃx² - (Ʃx)²/n = 772 - (106)²/15 =	22.9333333
SSyy = Ʃy² - (Ʃy)²/n = 33704.59 - (709.1)²/15 =	183.069333
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 5073.8 - (106)(709.1)/15 =	62.8266667

Slope, b = SSxy/SSxx = 62.82667/22.93333 = 2.73953488

y-intercept, a = y̅ -b* x̅ = 47.27333 - (2.73953)*7.06667 = 27.9139535

Regression equation :

ŷ = 27.914 + (2.740) x

c) A unit increase in age increases the height by 2.740 inches.

d) Predicted value of y at x =6

ŷ = 27.914 + (2.7395) * 6 = 44.3512

Significance level, α = 0.05

Critical value, t_c = T.INV.2T(0.05, 13) = 2.1604

Sum of Square error, SSE = SSyy -SSxy²/SSxx = 183.06933 - (62.82667)²/22.93333 = 10.9534884

Standard error, se = √(SSE/(n-2)) = √(10.95349/(15-2)) = 0.91792

95% Prediction interval :

e) Standard error, se = √(SSE/(n-2)) = √(10.95349/(15-2)) = 0.91792

f) Null and alternative hypothesis:

Ho: β₀ = β₁ = 0

H1: At least one of the β is not equal to 0.

SSR = SSxy²/SSxx = 172.1158

F =(SSR/1) / (SSE/(n-2)) = (172.1158) / (10.95349/13) = 204.273

P-value = F.DIST.RT(204.273, 1, 13) = 0.0000

p-value < α Reject the null hypothesis.

g) Null and alternative hypothesis:

Ho: β₁ = 0

Ha: β₁ ≠ 0

n=15

α = 0.05

Slope, b = 2.73953

Test statistic:

t = b /(se/√SSx) = 14.2924

df = n-2 = 13

p-value = T.DIST.2T(ABS(14.2924), 13) = 0.0000

Conclusion:

p-value < α Reject the null hypothesis.