PRACT STAT W/ ACCESS 6MO LOOSELEAF
PRACT STAT W/ ACCESS 6MO LOOSELEAF
4th Edition
ISBN: 9781319215361
Author: BALDI
Publisher: Macmillan Higher Education
Question
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Chapter 18, Problem 18.21E

(a)

To determine

To explain:

The difference in mean SRSs score for the populations of the given sample.

(a)

Expert Solution
Check Mark

Answer to Problem 18.21E

  t=6.2993

Explanation of Solution

Given:

    Group nˉxs
    Delayed clamping2165.309527.704
    Early clamping2081.9511.55

Concept used:

Formula

Standard error

  s21n1+s22n2

Degree of freedom

  =[s21n1+s22n2]2(s21n1)2n11+(s21n1)2n21

Significance level

  α=1confidence

Calculation:

  n1=21,¯x1=65.30952,s1=7.704668n2=20,¯x2=81.95,s2=11.55183

Standard error

  =s21n1+s22n2=(21.6)2130+(27.6)2115=3.196

Degree of freedom

  =[s21n1+s22n2]2(s21n1)2n11+(s21n1)2n21=39

At 95% confidence interval for the proportion p

Critical z value

  =zα/2=2.0227

Margin of error

  t=zα/2SEt=1.96(3.971)t=6.2993

(b)

To determine

To explain:

The difference in mean for the populations of the given sample.

(b)

Expert Solution
Check Mark

Answer to Problem 18.21E

  (22.8,10.5)

Explanation of Solution

Given:

    Group nˉxs
    Delayed clamping2165.309527.704
    Early clamping2081.9511.55

Concept used:

Formula

Standard error

  s21n1+s22n2

Degree of freedom

  =[s21n1+s22n2]2(s21n1)2n11+(s21n1)2n21

Significance level

  α=1confidence

Calculation:

  n1=21,¯x1=65.30952,s1=7.704668n2=20,¯x2=81.95,s2=11.55183

Standard error

  =s21n1+s22n2=(21.6)2130+(27.6)2115=3.196

Degree of freedom

  =[s21n1+s22n2]2(s21n1)2n11+(s21n1)2n21=39

At 95% confidence interval for the proportion p

Critical z value

  =zα/2=2.0227

Margin of error

  E=zα/2SEE=1.96(3.971)E=6.2993

The 95% confidence interval for difference

  =¯x1¯x2±zα/2SE=(65.3095281.95)±6.1744=16.6405±6.1744=(22.8,10.5)

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