Question

A pediatrician wants to determine the relation that may exist between a​ child's height and head circumference. She randomly selects 5 children and measures their height and head circumference. The data are summarized below. Complete parts​ (a) through​ (f) below.

Height_(inches)_-_x   Head_Circumference_(inches)_-_y
25   16.9
27   17.5
27.75   17.6
27.5   17.5
26.5   17.3

a) Treating height as the explanatory​ variable, x, use technology to determine the estimates of

beta 0β0

and

beta 1β1.

beta 0β0almost equals≈b 0b0equals=nothing

​(Round to four decimal places as​ needed.)

beta 1β1almost equals≈b 1b1equals=nothing

​(Round to four decimal places as​ needed.)

​(b) Use technology to compute the standard error of the​ estimate,

s Subscript ese.

s Subscript eseequals=nothing

​(Round to four decimal places as​ needed.)

​(c) A normal probability plot suggests that the residuals are normally distributed. Use technology to determine

s Subscript b 1sb1.

s Subscript b 1sb1equals=nothing  

​(Round to four decimal places as​ needed.)

​(d) A normal probability plot suggests that the residuals are normally distributed. Test whether a linear relation exists between height and head circumference at the

alphaαequals=0.010.01

level of significance. State the null and alternative hypotheses for this test.

Choose the correct answer below.

A.

Upper H 0H0​:

beta 0β0equals=0

Upper H 1H1​:

beta 0β0greater than>0

B.

Upper H 0H0​:

beta 0β0equals=0

Upper H 1H1​:

beta 0β0not equals≠0

C.

Upper H 0H0​:

beta 1β1equals=0

Upper H 1H1​:

beta 1β1not equals≠0

D.

Upper H 0H0​:

beta 1β1equals=0

Upper H 1H1​:

beta 1β1greater than>0

Determine the​ P-value for this hypothesis test.

​P-valueequals=nothing

​(Round to three decimal places as​ needed.)

What is the conclusion that can be​ drawn?

A.

Do not rejectDo not reject

Upper H 0H0

and conclude that a linear relation

does not existdoes not exist

between a​ child's height and head circumference at the level of significance

alphaαequals=0.010.01.

B.

Do not rejectDo not reject

Upper H 0H0

and conclude that a linear relation

existsexists

between a​ child's height and head circumference at the level of significance

alphaαequals=0.010.01.

C.

RejectReject

Upper H 0H0

and conclude that a linear relation

existsexists

between a​ child's height and head circumference at the level of significance

alphaαequals=0.010.01.

D.

RejectReject

Upper H 0H0

and conclude that a linear relation

does not existdoes not exist

between a​ child's height and head circumference at the level of significance

alphaαequals=0.010.01.

​(e) Use technology to

construct

a​ 95% confidence interval about the slope of the true​ least-squares regression line.

Lower​ bound: nothing

Upper​ bound: nothing

​(Round to three decimal places as​ needed.)

​(f) Suppose a child has a height of 26.5 inches. What would be a good guess for the​ child's head​ circumference?

A good estimate of the​ child's head circumference would be

nothing

inches.

​(Round to two decimal places as​ needed.)

Answer 1

X	Y	XY	X²	Y²
25	16.9	422.5	625	285.61
27	17.5	472.5	729	306.25
27.75	17.6	488.4	770.0625	309.76
27.5	17.5	481.25	756.25	306.25
26.5	17.3	458.45	702.25	299.29

Ʃx =	133.75
Ʃy =	86.8
Ʃxy =	2323.1
Ʃx² =	3582.5625
Ʃy² =	1507.16
Sample size, n =	5
x̅ = Ʃx/n = 133.75/5 =	26.75
y̅ = Ʃy/n = 86.8/5 =	17.36
SSxx = Ʃx² - (Ʃx)²/n = 3582.5625 - (133.75)²/5 =	4.75
SSyy = Ʃy² - (Ʃy)²/n = 1507.16 - (86.8)²/5 =	0.312
SSxy = Ʃxy - (Ʃx)(Ʃy)/n = 2323.1 - (133.75)(86.8)/5 =	1.2

a)

Slope, b1 = SSxy/SSxx = 1.2/4.75 =    0.252631579 = 0.2526

y-intercept, b0 = y̅ -b1* x̅ = 17.36 - (0.25263)*26.75 =    10.60210526 = 10.6021

Regression equation :   

ŷ = 10.6021 + (0.2526) x  

b)

Sum of Square error, SSE = SSyy -SSxy²/SSxx = 0.312 - (1.2)²/4.75 =    0.00884

Standard error, se = √(SSE/(n-2)) = √(0.00884/(5-2)) = 0.0543

c)

Standard error for slope, se(b1) = se/√SSxx = 0.0543/√4.75 = 0.0249

d)

Null and alternative hypothesis:  

Ho: β₁ = 0  

Ha: β₁ ≠ 0  

Test statistic:  

t = b1/se(b1) = 0.2526/0.0249 =    10.1419

df = n-2 =    3

p-value = T.DIST.2T(ABS(10.1419), 3) =    0.002

Conclusion:  

Reject H0 and conclude that a linear relation exists between a​ child's height and head circumference at the level of significance α = 0.01.

e)

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

95% Confidence interval for slope:  

Lower limit = b1 - tc*se(b1) = 0.2526 - 3.1824*0.0249 =    0.173

Upper limit = b1 + tc*se(b1) = 0.2526 + 3.1824*0.0249 =    0.332

f)

Predicted value of y at x =    26.5

ŷ = 10.6021 + (0.2526) * 26.5 = 17.3