Elementary Statistics 2nd Edition
Elementary Statistics 2nd Edition
2nd Edition
ISBN: 9781259724275
Author: William Navidi, Barry Monk
Publisher: McGraw-Hill Education
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Chapter 12, Problem 2CS
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
    Accept601370322269147461755
    Reject3322155965234376682771
    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 applicants who were accepted by department A
    The row total is 1755, the column total is 933, and the grand total is 4526.
    17559334526361.78
    Number of applicants who were accepted by department B
    The row total is 1755, the column total is 585, and the grand total is 4526.
    17555854526226.84
    Number of applicants who were accepted by department C
    The row total is 1755, the column total is 918, and the grand total is 4526.
    17559184526355.96
    Number of applicants who were accepted by department D
    The row total is 1755, the column total is 792, and the grand total is 4526.
    17557924526307.11
    Number of applicants who were accepted by department E
    The row total is 1755, the column total is 584, and the grand total is 4526.
    17555844526226.45
    Number of applicants who were accepted by department F
    The row total is 1755, the column total is 714, and the grand total is 4526.
    17557144526276.86
    Number of applicants who were rejected by department A
    The row total is 2771, the column total is 933, and the grand total is 4526.
    27719334526571.22
    Number of applicants who were rejected by department B
    The row total is 2771, the column total is 585, and the grand total is 4526.
    27715854526358.16
    Number of applicants who were rejected by department C
    The row total is 2771, the column total is 918, and the grand total is 4526.
    27719184526562.04
    Number of applicants who were rejected by department D
    The row total is 2771, the column total is 792, and the grand total is 4526.
    27717924526484.89
    Number of applicants who were rejected by department E
    The row total is 2771, the column total is 584, and the grand total is 4526.
    27715844526357.55
    Number of applicants who were rejected by department F
    The row total is 2771, the column total is 714, and the grand total is 4526.
    27717144526437.14

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 applicants who were accepted by department A
    The observed frequency is 601 and expected frequency is 361.78
    ( 601361.78)2361.78158.18
    Number of applicants who were accepted by department B
    The observed frequency is 370 and expected frequency is 226.84
    ( 370226.84)2226.8490.35
    Number of applicants who were accepted by department C
    The observed frequency is 322 and expected frequency is 355.96
    ( 322355.96)2355.963.24
    Number of applicants who were accepted by department D
    The observed frequency is 269 and expected frequency is 307.11
    ( 269307.11)2307.114.73
    Number of applicants who were accepted by department E
    The observed frequency is 147 and expected frequency is 226.45
    ( 147226.45)2226.4527.88
    Number of applicants who were accepted by department F
    The observed frequency is 46 and expected frequency is 276.86
    ( 46276.86)2276.86192.50
    Number of applicants who were rejectedby department A
    The observed frequency is 332 and expected frequency is 571.22
    ( 332571.22)2571.22100.18
    Number of applicants who were rejected by department B
    The observed frequency is 215 and expected frequency is 358.16
    ( 215358.16)2358.1657.22
    Number of applicants who were rejectedby department C
    The observed frequency is 596 and expected frequency is 562.04
    ( 596562.04)2562.042.05
    Number of applicants who were rejectedby department D
    The observed frequency is 523 and expected frequency is 484.89
    ( 523484.89)2484.892.99
    Number of applicants who were rejectedby department E
    The observed frequency is 437 and expected frequency is 357.55
    ( 437357.55)2357.5517.66
    Number of applicants who were rejectedby department F
    The observed frequency is 668 and expected frequency is 437.14
    ( 668437.14)2437.14121.92

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

  χ2= ( OE ) 2 E}= 158.18+90.35+3.24+4.73+27.88+192.50 +100.18+57.22+2.05+2.99+17.66+121.92 χ2778.91

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: 778.91; Critical value: 15.086. χ2>CriticalValue , so we Reject H0 (null hypothesis states that acceptance to graduate school is independent of department). Weconclude that acceptance rates differ among departments.

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Chapter 12 Solutions

Elementary Statistics 2nd Edition

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