Statistics Assignment Sample
Q1:
Answer :Introduction to the Chi-Square Test for Independence
The Chi-square test for independence is used to determine whether two categorical variables are independent or whether there is an association between them. In this case, we are testing whether there is an association between gender and music preference.
The steps for performing the Chi-square test are as follows:
- State the hypotheses.
- Calculate the expected frequencies.
- Calculate the Chi-square statistic.
- Determine the degrees of freedom.
- Compare the calculated Chi-square statistic with the critical value.
- Make a conclusion.
Step 1: State the Hypotheses
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Null Hypothesis (H₀): There is no association between gender and music preference (i.e., the two variables are independent).
H0:Gender and Music Preference are independent.H₀: \text{Gender and Music Preference are independent.}H0:Gender and Music Preference are independent. -
Alternative Hypothesis (H₁): There is an association between gender and music preference (i.e., the two variables are dependent).
H1:Gender and Music Preference are dependent.H₁: \text{Gender and Music Preference are dependent.}H1:Gender and Music Preference are dependent.
We will perform the test at a significance level of α=0.05\alpha = 0.05α=0.05.
Step 2: Calculate the Expected Frequencies
The expected frequency for each cell in the contingency table is calculated using the formula:
Eij=(Row Totali)×(Column Totalj)Grand TotalE_{ij} = \frac{(Row \, Total_i) \times (Column \, Total_j)}{Grand \, Total}Eij=GrandTotal(RowTotali)×(ColumnTotalj)
Where:
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EijE_{ij}Eij is the expected frequency for the cell in the iii-th row and jjj-th column.
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Row TotaliRow \, Total_iRowTotali is the total number of observations in row iii.
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Column TotaljColumn \, Total_jColumnTotalj is the total number of observations in column jjj.
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Grand TotalGrand \, TotalGrandTotal is the total number of observations (in this case, 200).
Let’s calculate the expected frequencies for each cell in the table.
For Classical Music:
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Expected frequency for Male (M):
ECM,M=(80×120)200=9600200=48E_{CM, M} = \frac{(80 \times 120)}{200} = \frac{9600}{200} = 48ECM,M=200(80×120)=2009600=48
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Expected frequency for Female (F):
ECM,F=(80×80)200=6400200=32E_{CM, F} = \frac{(80 \times 80)}{200} = \frac{6400}{200} = 32ECM,F=200(80×80)=2006400=32
For Pop Music:
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Expected frequency for Male (M):
EPM,M=(70×120)200=8400200=42E_{PM, M} = \frac{(70 \times 120)}{200} = \frac{8400}{200} = 42EPM,M=200(70×120)=2008400=42
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Expected frequency for Female (F):
EPM,F=(70×80)200=5600200=28E_{PM, F} = \frac{(70 \times 80)}{200} = \frac{5600}{200} = 28EPM,F=200(70×80)=2005600=28
For Rock Music:
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Expected frequency for Male (M):
ERM,M=(80×120)200=9600200=48E_{RM, M} = \frac{(80 \times 120)}{200} = \frac{9600}{200} = 48ERM,M=200(80×120)=2009600=48
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Expected frequency for Female (F):
ERM,F=(80×80)200=6400200=32E_{RM, F} = \frac{(80 \times 80)}{200} = \frac{6400}{200} = 32ERM,F=200(80×80)=2006400=32
Step 3: Calculate the Chi-Square Statistic
The Chi-square statistic is calculated using the formula:
χ2=∑(Oij−Eij)2Eij\chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}}χ2=∑Eij(Oij−Eij)2
Where OijO_{ij}Oij is the observed frequency in cell i,ji,ji,j, and EijE_{ij}Eij is the expected frequency for that cell.
Now, let’s calculate the contributions to the Chi-square statistic for each cell:
For Classical Music:
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Male (M):
(30−48)248=(−18)248=32448=6.75\frac{(30 - 48)^2}{48} = \frac{(-18)^2}{48} = \frac{324}{48} = 6.7548(30−48)2=48(−18)2=48324=6.75
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Female (F):
(50−32)232=(18)232=32432=10.125\frac{(50 - 32)^2}{32} = \frac{(18)^2}{32} = \frac{324}{32} = 10.12532(50−32)2=32(18)2=32324=10.125
For Pop Music:
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Male (M):
(40−42)242=(−2)242=442=0.0952\frac{(40 - 42)^2}{42} = \frac{(-2)^2}{42} = \frac{4}{42} = 0.095242(40−42)2=42(−2)2=424=0.0952
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Female (F):
(30−28)228=(2)228=428=0.1429\frac{(30 - 28)^2}{28} = \frac{(2)^2}{28} = \frac{4}{28} = 0.142928(30−28)2=28(2)2=284=0.1429
For Rock Music:
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Male (M):
(50−48)248=(2)248=448=0.0833\frac{(50 - 48)^2}{48} = \frac{(2)^2}{48} = \frac{4}{48} = 0.083348(50−48)2=48(2)2=484=0.0833
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Female (F):
(30−32)232=(−2)232=432=0.125\frac{(30 - 32)^2}{32} = \frac{(-2)^2}{32} = \frac{4}{32} = 0.12532(30−32)2=32(−2)2=324=0.125
Now, sum these values to get the Chi-square statistic:
χ2=6.75+10.125+0.0952+0.1429+0.0833+0.125=17.2224\chi^2 = 6.75 + 10.125 + 0.0952 + 0.1429 + 0.0833 + 0.125 = 17.2224χ2=6.75+10.125+0.0952+0.1429+0.0833+0.125=17.2224
Step 4: Determine the Degrees of Freedom
The degrees of freedom (df) for a Chi-square test of independence is given by:
df=(r−1)(c−1)df = (r - 1)(c - 1)df=(r−1)(c−1)
Where rrr is the number of rows and ccc is the number of columns. In this case, there are 3 rows (Music Preferences) and 2 columns (Gender), so:
df=(3−1)(2−1)=2×1=2df = (3 - 1)(2 - 1) = 2 \times 1 = 2df=(3−1)(2−1)=2×1=2
Step 5: Compare the Chi-Square Statistic with the Critical Value
Using a Chi-square distribution table or statistical software, we find the critical value for df=2df = 2df=2 and α=0.05\alpha = 0.05α=0.05 is 5.991.
The calculated Chi-square statistic is 17.22, which is greater than the critical value of 5.991.
Step 6: Conclusion
Since the calculated Chi-square statistic (17.22) is greater than the critical value (5.991), we reject the null hypothesis at the 5% significance level.
Interpretation:
There is sufficient evidence to conclude that there is a significant association between gender and music preference. This means that the preference for different types of music (classical, pop, rock) differs significantly between males and females.
Summary of Results:
- Null Hypothesis (H₀): Gender and Music Preference are independent.
- Alternative Hypothesis (H₁): Gender and Music Preference are dependent.
- Chi-Square Statistic: 17.22
- Degrees of Freedom (df): 2
- Critical Value: 5.991
- Conclusion: Reject H₀. There is a significant association between gender and music preference