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 12, Problem 3CS
To determine

To test: the null hypothesis and show that it is rejected at the α=0.01 level of significance.

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Explanation of Solution

Given information :

    DepartmentABCDEFTotal
    Male8255603254171913732691
    Female108255933753933411835
    Total9335859187925847144526

Concept Involved:

In order to decide whether the presumed hypothesis for data sample stands accurate for the entire population or not we use the hypothesis testing.

  H0 represents null hypothesis test and Ha represents alternative hypothesis test.

  E=R×CG , where E is expected frequency, R is row total, C is column total and G is grand total for a cell

  χ2 represent value of chi square test statistics found using the formula χ2= ( OE )2E where O is observed frequency and E is expected frequency.

The value of test statistics and the critical value identified from the table help us to decide whether to reject or do not reject null hypothesis.

The critical value from Table A.4, using degrees of freedom of contingency table of any given study is provided.

If χ2CriticalValue , then reject H0 otherwise reject H0 .

The values of two qualitative variables are connected and denoted in a contingency table.

This table consists of rows and column. The variables in each row and each column of the table represent a category.

The number of rows of contingency table is represented by letter ‘r’ and number of column of contingency table is represented by letter ‘c’.

The formula to find the number of degree of freedom of contingency table is (r1)(c1) .

Calculation:

    Finding the expected frequency for the cell corresponding to:The expected frequency
    Number of male applicantsin department A
    The row total is 2691, the column total is 933, and the grand total is 4526.
    26919334526554.73
    Number of male applicantsin department B
    The row total is 2691, the column total is 585, and the grand total is 4526.
    26915854526347.82
    Number of male applicantsin department C
    The row total is 2691, the column total is 918, and the grand total is 4526.
    26919184526545.81
    Number of male applicantsin department D
    The row total is 2691, the column total is 792, and the grand total is 4526.
    26917924526470.90
    Number of male applicantsin department E
    The row total is 2691, the column total is 584, and the grand total is 4526.
    26915844526347.23
    Number of male applicantsin department F
    The row total is 2691, the column total is 714, and the grand total is 4526.
    26917144526424.52
    Number of female applicantsin department A
    The row total is 1835, the column total is 933, and the grand total is 4526.
    18359334526378.27
    Number of female applicantsin department B
    The row total is 1835, the column total is 585, and the grand total is 4526.
    18355854526237.18
    Number of female applicantsin department C
    The row total is 1835, the column total is 918, and the grand total is 4526.
    18359184526372.19
    Number of female applicantsin department D
    The row total is 1835, the column total is 792, and the grand total is 4526.
    18357924526321.10
    Number of female applicantsin department E
    The row total is 1835, the column total is 584, and the grand total is 4526.
    18355844526236.77
    Number of female applicantsin department F
    The row total is 1835, the column total is 714, and the grand total is 4526.
    18357144526289.48

All the expected frequencies are at least 5. From the results of previous part we have the below table:

    Finding the value of the chi-square corresponding to:( OE)2E
    Number of male applicantsin department A
    The observed frequency is 825 and expected frequency is 554.73
    ( 825554.73)2554.73131.68
    Number of male applicantsin department B
    The observed frequency is 560 and expected frequency is 347.82
    ( 560347.82)2347.82129.44
    Number of male applicantsin department C
    The observed frequency is 325 and expected frequency is 545.81
    ( 325545.81)2545.8189.33
    Number of male applicantsin department D
    The observed frequency is 417 and expected frequency is 470.90
    ( 417470.90)2470.906.17
    Number of male applicantsin department E
    The observed frequency is 191 and expected frequency is 347.23
    ( 191347.23)2347.2370.29
    Number of male applicantsin department F
    The observed frequency is 373 and expected frequency is 424.52
    ( 373424.52)2424.526.25
    Number of female applicantsin department A
    The observed frequency is 108 and expected frequency is 378.27
    ( 108378.27)2378.27193.11
    Number of female applicantsin department B
    The observed frequency is 25 and expected frequency is 237.18
    ( 25237.18)2237.18189.81
    Number of female applicantsin department C
    The observed frequency is 593 and expected frequency is 372.19
    ( 593372.19)2372.19131.00
    Number of female applicantsin department D
    The observed frequency is 375 and expected frequency is 321.10
    ( 375321.10)2321.109.05
    Number of female applicantsin department E
    The observed frequency is 393 and expected frequency is 236.77
    ( 393236.77)2236.77103.08
    Number of female applicantsin department F
    The observed frequency is 341 and expected frequency is 289.48
    ( 341289.48)2289.489.17

To compute the test statistics, we use the observed frequencies and expected frequency:

  χ2= ( OE ) 2 E}= 131.68+129.44+89.33+6.17+70.29+6.25 +193.11+189.81+131+9.05+103.08+9.17 χ21068.37

Here r represents the number of rows and c represents the number of columns.

Given r=2 , c=6 so the number of degree of freedom is (21)(61)=(1)(5)=5

    Degrees of freedomTable A.4 Critical Values for the chi-square Distribution
    0.9950.990.9750.950.900.100.050.0250.010.005
    10.0000.0000.0010.0040.0162.7063.8415.0246.6357.879
    20.0100.0200.0510.1030.2114.6055.9917.3789.21010.597
    30.0720.1150.2160.3520.5846.2517.8159.34811.34512.838
    40.2070.2970.4840.7111.0647.7799.48811.14313.27714.860
    50.4120.5540.8311.1451.6109.23611.07012.833 15.086 16.750

Conclusion:

Test statistic: 1068.37; Critical value: 15.086. χ2>CriticalValue ,we Reject H0 (null hypothesis states that gender is independent of department)and weconclude that gender and department are not independent.

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Chapter 12 Solutions

Loose Leaf Version For Elementary Statistics

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