Question

In: Statistics and Probability

A pediatrician wants to determine the relation that may exist between a​ child's height and head...

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

25

27

26.5

27.5

25.5

Head Circumference​ (inches), y

16.9

17.5

17.3

17.5

17.1

(a) Treating height as the explanatory variable, X, use technology to determine the estimates of β0 and β1.

β0 ≈b0 = ____ ​(Round to two decimal places as​ needed.)

β1 ≈b1 = ____ (Round to two decimal places as​ needed.)

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

se= ______ ( Rounding to four decimal places)

​(c)  A normal probability plot suggests that the residuals are normally distributed . Use technology to determine sb1 .

Sb1= _____ ( Rounding to four decimal places)

(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 α=0.01 level of significance.

The first step is to set up a hypothesis test. The parameter being tested is the​ slope, β1​, of the linear regression line. If there is no linear relation between the response and explanatory​ variables, the slope of the true regression line will be zero.

The​ P-value for this test is _____ (rounding to three decimal places.)

(e) Use technology to construct a​ 95% confidence interval about the slope of the true​ least-squares regression line.

Lower bound: ____

Upper bound : _____

(f) Suppose a child has a height of 27 inches. What would be a good guess for the​ child's head​circumference?

Substitute the given​ height, 27​ inches, for x in the regression line and​ simplify, rounding to two decimal places.

Y= ______

Solutions

Expert Solution

a)

bo=10.777

b1=0.247

b)se =0.0597

c)sb1=0.0288

d) ​ P-value for this test is 0.003

e) Lower bound: =0.155

upper bound: =0.338

f)

Y =10.777+0.247*27=17.45 ( please try 17.43 if this comes wrong)


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