Explain whether it is possible that the acceptance to graduate school is independent of department at the level of significance of

Answer to Problem 2CS
There is not enough evidence to conclude that the acceptance rate does not differ among departments.
Explanation of Solution
Calculation:
The
Contingency table:
A contingency table is obtained as using two qualitative variables. One of the qualitative variable is row variable that has one category for each row of the table another is column variable has one category for each column of the table.
Step 1:
The hypotheses are:
Null Hypothesis:
That is, the acceptance rate does not differ among departments.
Alternate Hypothesis:
That is, the acceptance rate differs among departments.
Step 2:
Now, it is obtained that,
Department | A | B | C | D | E | F | Row Total |
Accept | 601 | 370 | 322 | 269 | 147 | 46 | 1,755 |
Reject | 332 | 215 | 596 | 523 | 437 | 668 | 2,771 |
Column Total | 933 | 585 | 918 | 792 | 584 | 714 | 4,526 |
Step 3:
Expected frequencies:
The expected frequencies in case of contingency table is obtained as,
Now, using the formula of expected frequency it is found that the expected frequency for the accepted applicants in department A is obtained as,
Hence, in similar way the expected frequencies are obtained as,
Department | A | B | C |
Accept | |||
Reject |
Department | D | E | F |
Accept | |||
Reject |
Step 4:
Level of significance:
The level of significance is given as 0.01.
Chi-Square statistic:
The chi-square statistic is obtained as
The classification of department can be rewritten as,
A | 1 |
B | 2 |
C | 3 |
D | 4 |
E | 5 |
F | 6 |
Test Statistic:
Software procedure:
Step -by-step software procedure to obtain test statistic using MINITAB software is as follows:
- Select Stat > Table > Cross Tabulation and Chi-Square.
- Check the box of Raw data (categorical variables).
- Under For rows enter Row.
- Under For columns enter Department.
- Check the box of Count under Display.
- Under Chi-Square, click the box of Chi-Square test.
- Select OK.
- Output using MINITAB software is given below:
Thus, the value of chi-square statistic is 778.907.
It is known that under the null hypothesis
In the given question there are 2 rows and 6 columns.
Hence, the degrees of freedom is
Thus, the degree of freedom is 5.
It is known that when the null hypothesis
It is found that all the expected frequencies corresponding to all rows and columns of the given contingency table are more than 5.
Hence, the test of independence is appropriate.
Step 5:
Critical value:
In a test of hypotheses the critical value is the point by which one can reject or accept the null hypothesis.
Software procedure:
Step-by-step software procedure to obtain critical value using MINITAB software is as follows:
- Select Graph > Probability distribution plot > view probability
- Select Chi -Square under distribution.
- In Degrees of freedom, enter 5.
- Choose Probability Value and Right Tail for the region of the curve to shade.
- Enter the Probability value as 0.01 under shaded area.
- Select OK.
- Output using MINITAB software is given below:
Hence, the critical value at
Rejection rule:
If the
Step 6:
Conclusion:
Here, the
That is,
Thus, the decision is “reject the null hypothesis”.
Thus, there is not enough evidence to conclude that the acceptance rate does not differ among departments.
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Chapter 10 Solutions
ALEKS 360 ESSENT. STAT ACCESS CARD
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