Loose Leaf Version For Elementary Statistics
Loose Leaf Version For Elementary Statistics
3rd Edition
ISBN: 9781260373523
Author: William Navidi Prof., Barry Monk Professor
Publisher: McGraw-Hill Education
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Chapter 15.2, Problem 11E

a.

To determine

To find:The hypothesis for the test statistic.

a.

Expert Solution
Check Mark

Answer to Problem 11E

The hypothesis are H0:m1=m2 and H1:m1<m2 .

Explanation of Solution

Given information:

The level of significance is 0.05 and the given data is,

    Compact
    34.533.726.128.527.4
    30.631.128.033.033.0
    32.828.525.532.134.9
    Midsize
    21.621.129. 124.828.5
    28.121.922.520.526.1

Calculations:

The hypothesis are,

  H0: The median mileage is not less for compact car than for the midsize car.

  H1: The median mileage is less for compact car than for the midsize car.

Therefore, the hypothesis are H0:m1=m2 and H1:m1<m2 .

b.

To determine

To find:The value of test statistic.

b.

Expert Solution
Check Mark

Answer to Problem 11E

The value of test statistic is 3.24 .

Explanation of Solution

Given information:

The level of significance is 0.05 and the given data is,

    Compact
    34.533.726.128.527.4
    30.631.128.033.033.0
    32.828.525.532.134.9
    Midsize
    21.621.129. 124.828.5
    28.121.922.520.526.1

Calculations:

The rank table is shown below.

    Gas mileagesSampleRank
    20.5Midsize1
    21.1Midsize2
    21.6Midsize3
    21.9Midsize4
    22.5Compact5
    24.8Midsize6
    25.5Compact7
    26.1Midsize8.5
    26.1Compact8.5
    27.4Compact10
    28Compact11
    28.1Midsize12
    28.5Midsize14
    28.5Compact14
    28.5Compact14
    29.1Midsize16
    30.6Compact17
    31.1Compact18
    32.1Compact19
    32.8Compact20
    33Compact21.5
    33Compact21.5
    33.7Compact23
    34.5Compact24
    34.9Compact25

The sum of rank of compact car is 253.5 and the rank of midsize car is 71.5 .

The sum of smaller rank is S=71.5 and the sample size are n1=10 and n2=15 .

The value of μs is,

  μs=n1(n1+n2+1)2

Substitute the values in above equation.

  μs=( 10)( 10+15+1)2=2602=130

The value of μs is,

  σs=n1n2( n 1 + n 2 +1)12

Substitute the values in above equation.

  σs= ( 15 )( 10 )( 10+15+1 ) 12= 3900 12=18.028

The z- value is,

  z=Sμsσs

Substitute the values in above equation.

  z=71.513018.028=3.24

Therefore, the value of test statistics is 3.24 .

c.

To determine

To find:The p-value.

c.

Expert Solution
Check Mark

Answer to Problem 11E

The p-valueis 0.0006 .

Explanation of Solution

Given information:

The level of significance is 0.05 and the given data is,

    Compact
    34.533.726.128.527.4
    30.631.128.033.033.0
    32.828.525.532.134.9
    Midsize
    21.621.129. 124.828.5
    28.121.922.520.526.1

Calculations:

From the table of cumulative normal distribution the p-value corresponding to z-value is 0.0006 .

Therefore, the p-value is 0.0006 .

d.

To determine

To find:The conclusion for the test.

d.

Expert Solution
Check Mark

Answer to Problem 11E

The median mileage is less for compact car than for the midsize car.

Explanation of Solution

Given information:

The level of significance is 0.05 and the given data is,

    Compact
    34.533.726.128.527.4
    30.631.128.033.033.0
    32.828.525.532.134.9
    Midsize
    21.621.129. 124.828.5
    28.121.922.520.526.1

Calculations:

Since, the p-value is 0.0006 which less than 0.05 .

Thus, the hypothesis H0 is rejected.

Therefore, the median mileage is less for compact car than for the midsize car.

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Chapter 15 Solutions

Loose Leaf Version For Elementary Statistics

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