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To find:
To show that there is a relationship between grades and the combination of subject and gender
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Answer to Problem 13E
Solution:
There is a relationship between grades and the combination of particular subject and gender for student in the state.
Explanation of Solution
The following table represents the random sample of relationship between grades and the combination of particular subject and gender for student in the state is,
Observed Sample of 440 Students | ||||||
A | B | C | D | F | Total | |
Math, Female | 12 | 30 | 44 | 15 | 9 | 110 |
Math, Male | 5 | 24 | 37 | 36 | 8 | 110 |
English, Female | 16 | 41 | 39 | 10 | 4 | 110 |
English, Male | 10 | 40 | 35 | 19 | 6 | 110 |
Total | 43 | 135 | 155 | 80 | 27 | 440 |
Step 1: To state the null and alternative hypotheses.
Let the null hypothesis be that grades and combination of subjects and gender are independent of one another.
Null hypothesis:
Alternative hypothesis:
Step 2: To determine which distribution to use for the test statistic, and state the level of significance.
To determine there is a relationship between grades and the combination of particular subject and gender for student in the state. Since we assume that the necessary condition that have been met. So, use the chi-square test for this association and the level of siginificance is
Step 3: To calculate the test statistic.
First calculate the expected value for each cell in the
Formula for calculating expected value of a frequency in a contingency table:
The expected value of the frequency for the ith possible outcome in a contingency table is given by
Where n is the
From the above table represent the value of n is 146.
The
The expected values of for the Grade B and the combination of subjects and gender is given below,
The expected values of for the Grade C and the combination of subjects and gender is given below,
The expected values of for the Grade D and the combination of subjects and gender is given below,
The expected values of for the Grade A and the combination of subjects and gender is given below,
The table represents the contingency table of expected values by using the above formula is given below,
Expected values | ||||||
A | B | C | D | F | Total | |
Math, Female | 10.75 | 33.75 | 38.75 | 20 | 6.75 | 110 |
Math, Male | 10.75 | 33.75 | 38.75 | 20 | 6.75 | 110 |
English, Female | 10.75 | 33.75 | 38.75 | 20 | 6.75 | 110 |
English, Male | 10.75 | 33.75 | 38.75 | 20 | 6.75 | 110 |
Total | 43 | 135 | 155 | 80 | 27 | 440 |
Formula for calculating test statistic for a Chi-Square test for association:
The test statistic for a chi-square test for association is given below,
Where Oi, is the observed frequency for the ith possible outcome and Ei is the expected frequency for the ith possible outcome and n is the sample size.
From the above table represent the value of n is 440.
The test statistic,
The following table represents the chi square test statistic value s given below,
Observed | Expected | |
|
|
12 | 10.75 | 1.25 | 1.5625 | 0.14535 |
5 | 10.75 | -5.75 | 33.0625 | 3.07558 |
16 | 10.75 | 5.25 | 27.5625 | 2.56395 |
10 | 10.75 | -0.75 | 0.5625 | 0.05233 |
30 | 33.75 | -3.75 | 14.0625 | 0.41667 |
24 | 33.75 | -9.75 | 95.0625 | 2.81667 |
41 | 33.75 | 7.25 | 52.5625 | 1.55741 |
40 | 33.75 | 6.25 | 39.0625 | 1.15741 |
44 | 38.75 | 5.25 | 27.5625 | 0.71129 |
37 | 38.75 | -1.75 | 3.0625 | 0.07903 |
39 | 38.75 | 0.25 | 0.0625 | 0.00161 |
35 | 38.75 | -3.75 | 14.0625 | 0.3629 |
15 | 20 | -5 | 25 | 1.25 |
36 | 20 | 16 | 256 | 12.8 |
10 | 20 | -10 | 100 | 5 |
19 | 20 | -1 | 1 | 0.05 |
9 | 6.75 | 2.25 | 5.0625 | 0.75 |
8 | 6.75 | 1.25 | 1.5625 | 0.23148 |
4 | 6.75 | -2.75 | 7.5625 | 1.12037 |
6 | 6.75 | -0.75 | 0.5625 | 0.08333 |
So, the value of test statistic,
Step 4:
Draw a conclusion and interpret the decision.
Degrees of freedom in a Chi-square test for association:
In a chi-square test for association the number of degrees of freedom for the chi-square distribution of the test statistic is given by,
Where R is the number of rows of the data in the contingency table (not including the row total) and C is the number of columns of data in the contingency table (not including the column totals).
Rejection Region for Chi-Square Test for Association:
Reject the null hypothesis,
From the given information R = 4 and C = 5.
Substitute the above values in the formula of degrees of freedom to get the following,
The number of degrees of freedom for this test is 12.
Use the level of significance of
Use the “Area to the right of the critical value
The test statistic value is
Comparing the test statistic and critical value to get the following,
So, there is a relationship between grades and the combination of particular subject and gender for student in the state.
Final statement:
There is a relationship between grades and the combination of particular subject and gender for student in the state.
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Chapter 10 Solutions
Beginning Statistics, 2nd Edition
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