Question

Modern medical practice tells us not to encourage babies to become too fat. Is there a positive correlation between the weight x of a 1-year old baby and the weight y of the mature adult (30 years old)? A random sample of medical files produced the following information for 14 females.

x (lb) 23 27 22 26 20 15 25 21 17 24 26 22 18 19

y (lb) 127 124 117 125 130 120 145 130 130 130 130 140 110 115

In this setting we have Σx = 305, Σy = 1773, Σx2 = 6819, Σy2 = 225,669, and Σxy = 38,803. (

a) Find x, y, b, and the equation of the least-squares line. (Round your answers for x and y to two decimal places. Round your answers for least-squares estimates to four decimal places.) x = y = b = ŷ = + x

(b) Draw a scatter diagram displaying the data. Graph the least-squares line on your scatter diagram. Be sure to plot the point (x, y). Selection Tool Line Ray Segment Circle Vertical Parabola Horizontal Parabola Point No Solution Help Clear Graph Delete Layer Fill WebAssign Graphing Tool Graph LayersToggle Open/Closed After you add an object to the graph you can use Graph Layers to view and edit its properties.

(c) Find the sample correlation coefficient r and the coefficient of determination. (Round your answers to three decimal places.) r = r2 = What percentage of variation in y is explained by the least-squares model? (Round your answer to one decimal place.) %

(d) Test the claim that the population correlation coefficient ρ is positive at the 1% level of significance. (Round your test statistic to three decimal places and your P-value to four decimal places.) t = P-value = Conclusion Reject the null hypothesis. There is sufficient evidence that ρ > 0. Reject the null hypothesis. There is insufficient evidence that ρ > 0. Fail to reject the null hypothesis. There is sufficient evidence that ρ > 0. Fail to reject the null hypothesis. There is insufficient evidence that ρ > 0.

(e) If a female baby weighs 15 pounds at 1 year, what do you predict she will weigh at 30 years of age? (Round your answer to two decimal places.) lb

(f) Find Se. (Round your answer to two decimal places.) Se =

(g) Find a 99% confidence interval for weight at age 30 of a female who weighed 15 pounds at 1 year of age. (Round your answers to two decimal places.) lower limit lb upper limit lb

(h) Test the claim that the slope β of the population least-squares line is positive at the 1% level of significance. (Round your test statistic to three decimal places and your P-value to four decimal places.) t = P-value = Conclusion Reject the null hypothesis. There is sufficient evidence that β > 0. Reject the null hypothesis. There is insufficient evidence that β > 0. Fail to reject the null hypothesis. There is sufficient evidence that β > 0. Fail to reject the null hypothesis. There is insufficient evidence that β > 0.

(i) Find a 99% confidence interval for β and interpret its meaning. (Round your answers to three decimal places.) lower limit upper limit Interpretation For each pound less a female infant weighs at 1 year, the adult weight increases by an amount that falls within the confidence interval. For each pound more a female infant weighs at 1 year, the adult weight increases by an amount that falls within the confidence interval. For each pound less a female infant weighs at 1 year, the adult weight increases by an amount that falls outside the confidence interval. For each pound more a female infant weighs at 1 year, the adult weight increases by an amount that falls outside the confidence interval.

Answer 1

X	Y	XY	X²	Y²
23	127	2921	529	16129
27	124	3348	729	15376
22	117	2574	484	13689
26	125	3250	676	15625
20	130	2600	400	16900
15	120	1800	225	14400
25	145	3625	625	21025
21	130	2730	441	16900
17	130	2210	289	16900
24	130	3120	576	16900
26	130	3380	676	16900
22	140	3080	484	19600
18	110	1980	324	12100
19	115	2185	361	13225
Ʃx =	Ʃy =	Ʃxy =	Ʃx² =	Ʃy² =
305	1773	38803	6819	225669

Sample size, n =	14
x̅ = Ʃx/n = 305/14 =	21.78571429
y̅ = Ʃy/n = 1773/14 =	126.6428571
SSxx = Ʃx² - (Ʃx)²/n = 6819 - (305)²/14 =	174.3571429
SSyy = Ʃy² - (Ʃy)²/n = 225669 - (1773)²/14 =	1131.214286
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 38803 - (305)(1773)/14 =	176.9285714

a)

x̅ = Ʃx/n = 305/14 = 21.79

y̅ = Ʃy/n = 1773/14 = 126.64

Slope, b = SSxy/SSxx = 176.92857/174.35714 = 1.0147481

y-intercept, a = y̅ -b* x̅ = 126.64286 - (1.01475)*21.78571 = 104.53585

Regression equation :

ŷ = 104.5358 + (1.0147) x

b) Scatter plot:

c)

Correlation coefficient, r = SSxy/√(SSxx*SSyy) = 176.92857/√(174.35714*1131.21429) = 0.398

Coefficient of determination, r² = (SSxy)²/(SSxx*SSyy) = (176.92857)²/(174.35714*1131.21429) = 0.159

15.9% variation in y is explained by the least squares model.

d)

Null and alternative hypothesis:

Ho: ρ = 0 ; Ha: ρ > 0

Test statistic :  

t = r*√(n-2)/√(1-r²) = 0.3984 *√(14 - 2)/√(1 - 0.3984²) = 1.505

df = n-2 = 12

p-value = T.DIST.RT(1.5046, 12) = 0.0791

Conclusion:

Fail to reject the null hypothesis. There is insufficient evidence that ρ > 0.

e)

Predicted value of y at x = 15

ŷ = 104.5358 + (1.0147) * 15 = 119.76

f)

Sum of Square error, SSE = SSyy -SSxy²/SSxx = 1131.21429 - (176.92857)²/174.35714 = 951.6763621

Standard error, se = √(SSE/(n-2)) = √(951.67636/(14-2)) = 8.90541 = 8.91

g)

Critical value, t_c = T.INV.2T(0.01, 12) = 3.0545

99% Confidence interval :

Lower limit = ŷ - tc*se*√((1/n) + ((x-x̅)²/(SSxx))) = 119.7571 - 3.0545*8.9054*√((1/14) + ((15 - 21.7857)²/(174.3571))) = 104.00

Upper limit = ŷ + tc*se*√((1/n) + ((x-x̅)²/(SSxx))) = 119.7571 + 3.0545*8.9054*√((1/14) + ((15 - 21.7857)²/(174.3571))) = 135.51

h)

Test statistic:

t = b/(se/√SSxx) = 1.505

df = n-2 = 12

p-value = T.DIST.RT(1.5046, 12) = 0.0791

Conclusion:

Fail to reject the null hypothesis. There is insufficient evidence that β > 0.

i)

Critical value, t_c = T.INV.2T(0.01, 12) = 3.0545

99% Confidence interval for slope:

Lower limit = β₁ - tc*se/√SSxx = 1.0147 - 3.0545*8.9054/√174.3571 = -1.045

Upper limit = β₁ + tc*se/√SSxx = 1.0147 + 3.0545*8.9054/√174.3571 = 3.075