Connect Hosted by ALEKS Access Card or Elementary Statistics
Connect Hosted by ALEKS Access Card or Elementary Statistics
3rd Edition
ISBN: 9781260373752
Author: William Navidi Prof., Barry Monk Professor
Publisher: McGraw-Hill Education
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Chapter 12, Problem 4CS
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 :

  Connect Hosted by ALEKS Access Card or Elementary Statistics, Chapter 12, Problem 4CS

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 Harepresents 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:

For department A:

    Department A
    GenderAcceptRejectTotal
    Male512313825
    Female8919108
    Total601332933
    Finding the expected frequency for the cell corresponding to:The expected frequency
    Number of male applicants acceptedin department A
    The row total is 825, the column total is 601, and the grand total is 933.
    825601933531.43
    Number of male applicants rejectedin department A
    The row total is 825, the column total is 332, and the grand total is 933.
    825332933293.57
    Number of female applicantsaccepted in department A
    The row total is 108, the column total is 601, and the grand total is 933.
    10860193369.57
    Number of female applicantsrejected in department A
    The row total is 108, the column total is 332, and the grand total is 933.
    10833293338.43
    Finding the value of the chi-square corresponding to:( OE)2E
    Number of male applicants acceptedin department A
    The observed frequency is 512 and expected frequency is 531.43
    ( 512531.43)2531.430.71
    Number of male applicants rejectedin department A
    The observed frequency is 313 and expected frequency is 293.57
    ( 313293.57)2293.571.29
    Number of female applicantsaccepted in department A
    The observed frequency is 89 and expected frequency is 69.57
    ( 8969.57)269.575.43
    Number of female applicantsrejected in department A
    The observed frequency is 19 and expected frequency is 38.43
    ( 1938.43)238.439.82

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

  χ2= ( OE ) 2E}=0.71+1.29+5.43+9.82χ217.25

In department A, % of male accepted = 512625=62.1% and % of female accepted = 89108=82.4%

For department B:

    Department B
    GenderAcceptRejectTotal
    Male353207560
    Female17825
    Total370215585
    Finding the expected frequency for the cell corresponding to:The expected frequency
    Number of male applicants acceptedin department B
    The row total is 560, the column total is 370, and the grand total is 585.
    560370585354.19
    Number of male applicants rejectedin department B
    The row total is 560, the column total is 215, and the grand total is 585.
    560215585205.81
    Number of female applicantsaccepted in department B
    The row total is 25, the column total is 370, and the grand total is 585.
    2537058515.81
    Number of female applicantsrejected in department B
    The row total is 25, the column total is 215, and the grand total is 585.
    252155859.19
    Finding the value of the chi-square corresponding to:( OE)2E
    Number of male applicants acceptedin department B
    The observed frequency is 353 and expected frequency is 354.19
    ( 353354.19)2354.190.00
    Number of male applicants rejectedin department B
    The observed frequency is 207 and expected frequency is 205.81
    ( 207205.81)2205.810.01
    Number of female applicantsaccepted in department B
    The observed frequency is 17 and expected frequency is 15.81
    ( 1715.81)215.810.09
    Number of female applicantsrejected in department B
    The observed frequency is 8 and expected frequency is 9.19
    ( 89.19)29.190.15

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

  χ2= ( OE ) 2 E}=0.00+0.01+0.09+0.15χ20.25

In department B, % of male accepted = 353560=63.0% and % of female accepted = 1725=68.0%

For department C:

    Department C
    GenderAcceptRejectTotal
    Male120205325
    Female202391593
    Total322596918
    Finding the expected frequency for the cell corresponding to:The expected frequency
    Number of male applicants acceptedin department C
    The row total is 325, the column total is 322, and the grand total is 918.
    325322918114.00
    Number of male applicants rejectedin department C
    The row total is 325, the column total is 596, and the grand total is918.
    325596918211.00
    Number of female applicantsaccepted in department C
    The row total is 593, the column total is 322, and the grand total is 918.
    593322918208.00
    Number of female applicantsrejected in department C
    The row total is 593, the column total is 596, and the grand total is 918.
    593596918385.00
    Finding the value of the chi-square corresponding to:( OE)2E
    Number of male applicants acceptedin department C
    The observed frequency is 120 and expected frequency is 114
    ( 120114)21140.32
    Number of male applicants rejectedin department C
    The observed frequency is 205 and expected frequency is 211
    ( 205211)22110.17
    Number of female applicantsaccepted in department C
    The observed frequency is 202 and expected frequency is 208
    ( 202208)22080.17
    Number of female applicantsrejected in department C
    The observed frequency is 391 and expected frequency is 385
    ( 391385)23850.09

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

  χ2= ( OE ) 2 E}=0.32+0.17+0.17+0.09χ20.75

In department C, % of male accepted = 120325=36.9% and % of female accepted = 202593=34.1%

For department D:

    Department D
    GenderAcceptRejectTotal
    Male138279417
    Female131244375
    Total269523792
    Finding the expected frequency for the cell corresponding to:The expected frequency
    Number of male applicants acceptedin department D
    The row total is 417, the column total is 269, and the grand total is 792.
    417269792141.63
    Number of male applicants rejectedin department D
    The row total is 417, the column total is 523, and the grand total is792.
    417523792275.37
    Number of female applicantsaccepted in department D
    The row total is 375, the column total is 269, and the grand total is 792.
    375269792127.37
    Number of female applicantsrejected in department D
    The row total is 375, the column total is 523, and the grand total is 792.
    375523792247.63
    Finding the value of the chi-square corresponding to:( OE)2E
    Number of male applicants acceptedin department D
    The observed frequency is 138 and expected frequency is 141.63
    ( 138141.63)2141.630.09
    Number of male applicants rejectedin department D
    The observed frequency is 279 and expected frequency is 275.37
    ( 279275.37)2275.370.05
    Number of female applicantsaccepted in department D
    The observed frequency is 131 and expected frequency is 127.37
    ( 131127.37)2127.370.10
    Number of female applicantsrejected in department D
    The observed frequency is 244 and expected frequency is 247.63
    ( 244247.63)2247.630.05

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

  χ2= ( OE ) 2 E}=0.09+0.05+0.10+0.05χ20.30

In department D, % of male accepted = 138417=33.1% and % of female accepted = 131375=34.9%

For department E:

    Department E
    GenderAcceptRejectTotal
    Male53138191
    Female94299393
    Total147437584
    Finding the expected frequency for the cell corresponding to:The expected frequency
    Number of male applicants acceptedin department E
    The row total is 191, the column total is 147, and the grand total is 584.
    19114758448.08
    Number of male applicants rejectedin department E
    The row total is 191, the column total is 437, and the grand total is584.
    191437584142.92
    Number of female applicantsaccepted in department E
    The row total is 393, the column total is 147, and the grand total is 584.
    39314758498.92
    Number of female applicantsrejected in department E
    The row total is 393, the column total is 437, and the grand total is 584.
    393437584294.08
    Finding the value of the chi-square corresponding to:( OE)2E
    Number of male applicants acceptedin department E
    The observed frequency is 53 and expected frequency is 48.08
    ( 5348.08)248.080.50
    Number of male applicants rejectedin department E
    The observed frequency is 138 and expected frequency is 142.92
    ( 138142.92)2142.920.17
    Number of female applicantsaccepted in department E
    The observed frequency is 94 and expected frequency is 98.92
    ( 9498.92)298.920.24
    Number of female applicantsrejected in department E
    The observed frequency is 299 and expected frequency is 294.08
    ( 299294.08)2294.080.08

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

  χ2= ( OE ) 2 E}=0.50+0.17+0.25+0.08χ21.00

In department E, % of male accepted = 53191=27.7% and % of female accepted = 94393=23.9%

For department F:

    Department F
    GenderAcceptRejectTotal
    Male22351373
    Female24317341
    Total46668714
    Finding the expected frequency for the cell corresponding to:The expected frequency
    Number of male applicants acceptedin department F
    The row total is 373, the column total is 46, and the grand total is 714.
    3734671424.03
    Number of male applicants rejectedin department F
    The row total is 373, the column total is 668, and the grand total is714.
    373668714348.97
    Number of female applicantsaccepted in department F
    The row total is 341, the column total is 46, and the grand total is 714.
    3414671421.97
    Number of female applicantsrejected in department F
    The row total is 341, the column total is 668, and the grand total is 714.
    341668714319.03
    Finding the value of the chi-square corresponding to:( OE)2E
    Number of male applicants acceptedin department F
    The observed frequency is 22 and expected frequency is 24.03
    ( 2224.03)224.030.17
    Number of male applicants rejectedin department F
    The observed frequency is 351 and expected frequency is 348.97
    ( 351348.97)2348.970.01
    Number of female applicantsaccepted in department F
    The observed frequency is 24 and expected frequency is 21.97
    ( 2421.97)221.970.19
    Number of female applicantsrejected in department F
    The observed frequency is 317 and expected frequency is 319.03
    ( 317319.03)2319.030.01

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

  χ2= ( OE ) 2 E}=0.17+0.01+0.19+0.01χ20.39

In department E, % of male accepted = 22373=5.9% and % of female accepted = 24341=7.0%

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

For all the contingency table r=2 , c=2 so the number of degree of freedom is (21)(21)=(1)(1)=1

    Degrees of freedomTable A.4 Critical Values for the chi-square Distribution
    0.9950.990.9750.950.900.100.050.0250.010.005
    10.0000.0000.0010.0040.0162.7063.8415.024 6.635 7.879
    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
    50.4120.5540.8311.1451.6109.23611.07012.83315.08616.750

The critical value is same for all the contingency table.

Conclusion:

For department A:

Test statistic: 17.25; Critical value: 6.635. χ2>CriticalValue ,we Reject H0

For department B:

Test statistic: 0.25; Critical value: 6.635. χ2<CriticalValue , we do notreject H0

For department C:

Test statistic: 0.75; Critical value: 6.635. χ2<CriticalValue , we do notreject H0

For department D:

Test statistic: 0.30; Critical value: 6.635. χ2<CriticalValue , we do notreject H0

For department E:

Test statistic: 1.00; Critical value: 6.635. χ2<CriticalValue , we do notreject H0

For department F:

Test statistic: 0.39; Critical value: 6.635. χ2<CriticalValue , we do notreject H0

In departmentA, 82.4% of the women were accepted, but only 62.1% of themen were accepted.

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Connect Hosted by ALEKS Access Card or Elementary Statistics

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