Statistics Assignment Sample
Q1:
Answer :Introduction to ANOVA
Analysis of Variance (ANOVA) is a statistical technique used to compare the means of three or more groups to see if there are any statistically significant differences between them. In a one-way ANOVA, we are interested in determining whether the mean scores for the three different teaching methods differ from each other.
The steps involved in performing a one-way ANOVA are as follows:
- State the null and alternative hypotheses.
- Calculate the means of the groups and the overall mean.
- Calculate the between-group variability (Mean Square Between, MSB) and within-group variability (Mean Square Within, MSW).
- Calculate the F-statistic and compare it to the critical value to make a decision.
Step 1: State the Hypotheses
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Null Hypothesis (H₀): There is no significant difference in the mean performance scores between the three teaching methods.
H0:μA=μB=μCH₀: \mu_A = \mu_B = \mu_CH0:μA=μB=μC
Where μA\mu_AμA, μB\mu_BμB, and μC\mu_CμC represent the mean performance scores of the students for teaching methods A, B, and C, respectively. -
Alternative Hypothesis (H₁): At least one of the teaching methods leads to a significantly different mean performance score.
H1:At least one mean is different.H₁: \text{At least one mean is different.}H1:At least one mean is different.
We will use a significance level of α=0.05\alpha = 0.05α=0.05 to test the hypothesis.
Step 2: Calculate the Group Means and Overall Mean
First, we need to calculate the mean of each teaching method group and the overall mean of all students’ scores combined.
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Mean of Method A:
Mean of A=85+88+90+92+865=4415=88.2\text{Mean of A} = \frac{85 + 88 + 90 + 92 + 86}{5} = \frac{441}{5} = 88.2Mean of A=585+88+90+92+86=5441=88.2
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Mean of Method B:
Mean of B=78+82+81+80+855=4065=81.2\text{Mean of B} = \frac{78 + 82 + 81 + 80 + 85}{5} = \frac{406}{5} = 81.2Mean of B=578+82+81+80+85=5406=81.2
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Mean of Method C:
Mean of C=92+94+96+93+915=4665=93.2\text{Mean of C} = \frac{92 + 94 + 96 + 93 + 91}{5} = \frac{466}{5} = 93.2Mean of C=592+94+96+93+91=5466=93.2
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Overall Mean (grand mean):
Overall Mean=85+88+90+92+86+78+82+81+80+85+92+94+96+93+9115=130415=86.93\text{Overall Mean} = \frac{85 + 88 + 90 + 92 + 86 + 78 + 82 + 81 + 80 + 85 + 92 + 94 + 96 + 93 + 91}{15} = \frac{1304}{15} = 86.93Overall Mean=1585+88+90+92+86+78+82+81+80+85+92+94+96+93+91=151304=86.93
Step 3: Calculate the Between-Group Variability (MSB) and Within-Group Variability (MSW)
Between-Group Variability (MSB)
The between-group sum of squares (SSB) measures the variability between the group means and the overall mean. We can compute the formula for SSB as follows:
SSB=n[(XˉA−Xˉoverall)2+(XˉB−Xˉoverall)2+(XˉC−Xˉoverall)2]SSB = n \left[ (\bar{X}_A - \bar{X}_{\text{overall}})^2 + (\bar{X}_B - \bar{X}_{\text{overall}})^2 + (\bar{X}_C - \bar{X}_{\text{overall}})^2 \right]SSB=n[(XˉA−Xˉoverall)2+(XˉB−Xˉoverall)2+(XˉC−Xˉoverall)2]
Where nnn is the number of observations in each group (5 in this case). Using the group means and the overall mean:
SSB=5[(88.2−86.93)2+(81.2−86.93)2+(93.2−86.93)2]SSB = 5 \left[ (88.2 - 86.93)^2 + (81.2 - 86.93)^2 + (93.2 - 86.93)^2 \right]SSB=5[(88.2−86.93)2+(81.2−86.93)2+(93.2−86.93)2] SSB=5[(1.27)2+(−5.73)2+(6.27)2]SSB = 5 \left[ (1.27)^2 + (-5.73)^2 + (6.27)^2 \right]SSB=5[(1.27)2+(−5.73)2+(6.27)2] SSB=5[1.6129+32.8329+39.3129]SSB = 5 \left[ 1.6129 + 32.8329 + 39.3129 \right]SSB=5[1.6129+32.8329+39.3129] SSB=5×73.7587=368.7935SSB = 5 \times 73.7587 = 368.7935SSB=5×73.7587=368.7935
Next, we calculate the Mean Square Between (MSB):
MSB=SSBdf between=368.79353−1=368.79352=184.3968MSB = \frac{SSB}{\text{df between}} = \frac{368.7935}{3 - 1} = \frac{368.7935}{2} = 184.3968MSB=df betweenSSB=3−1368.7935=2368.7935=184.3968
Within-Group Variability (MSW)
The within-group sum of squares (SSW) measures the variability within each group. The formula for SSW is:
SSW=∑i=13∑j=1n(Xij−Xˉi)2SSW = \sum_{i=1}^{3} \sum_{j=1}^{n} (X_{ij} - \bar{X}_i)^2SSW=i=1∑3j=1∑n(Xij−Xˉi)2
Where XijX_{ij}Xij represents each individual score and Xˉi\bar{X}_iXˉi represents the mean of group iii.
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Method A (SSW for A):
SSWA=(85−88.2)2+(88−88.2)2+(90−88.2)2+(92−88.2)2+(86−88.2)2SSW_A = (85 - 88.2)^2 + (88 - 88.2)^2 + (90 - 88.2)^2 + (92 - 88.2)^2 + (86 - 88.2)^2SSWA=(85−88.2)2+(88−88.2)2+(90−88.2)2+(92−88.2)2+(86−88.2)2 SSWA=(−3.2)2+(−0.2)2+(1.8)2+(3.8)2+(−2.2)2=10.24+0.04+3.24+14.44+4.84=32.80SSW_A = (-3.2)^2 + (-0.2)^2 + (1.8)^2 + (3.8)^2 + (-2.2)^2 = 10.24 + 0.04 + 3.24 + 14.44 + 4.84 = 32.80SSWA=(−3.2)2+(−0.2)2+(1.8)2+(3.8)2+(−2.2)2=10.24+0.04+3.24+14.44+4.84=32.80
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Method B (SSW for B):
SSWB=(78−81.2)2+(82−81.2)2+(81−81.2)2+(80−81.2)2+(85−81.2)2SSW_B = (78 - 81.2)^2 + (82 - 81.2)^2 + (81 - 81.2)^2 + (80 - 81.2)^2 + (85 - 81.2)^2SSWB=(78−81.2)2+(82−81.2)2+(81−81.2)2+(80−81.2)2+(85−81.2)2 SSWB=(−3.2)2+(0.8)2+(−0.2)2+(−1.2)2+(3.8)2=10.24+0.64+0.04+1.44+14.44=26.80SSW_B = (-3.2)^2 + (0.8)^2 + (-0.2)^2 + (-1.2)^2 + (3.8)^2 = 10.24 + 0.64 + 0.04 + 1.44 + 14.44 = 26.80SSWB=(−3.2)2+(0.8)2+(−0.2)2+(−1.2)2+(3.8)2=10.24+0.64+0.04+1.44+14.44=26.80
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Method C (SSW for C):
SSWC=(92−93.2)2+(94−93.2)2+(96−93.2)2+(93−93.2)2+(91−93.2)2SSW_C = (92 - 93.2)^2 + (94 - 93.2)^2 + (96 - 93.2)^2 + (93 - 93.2)^2 + (91 - 93.2)^2SSWC=(92−93.2)2+(94−93.2)2+(96−93.2)2+(93−93.2)2+(91−93.2)2 SSWC=(−1.2)2+(0.8)2+(2.8)2+(−0.2)2+(−2.2)2=1.44+0.64+7.84+0.04+4.84=14.80SSW_C = (-1.2)^2 + (0.8)^2 + (2.8)^2 + (-0.2)^2 + (-2.2)^2 = 1.44 + 0.64 + 7.84 + 0.04 + 4.84 = 14.80SSWC=(−1.2)2+(0.8)2+(2.8)2+(−0.2)2+(−2.2)2=1.44+0.64+7.84+0.04+4.84=14.80
Thus, the total within-group sum of squares is:
SSW=SSWA+SSWB+SSWC=32.80+26.80+14.80=74.40SSW = SSW_A + SSW_B + SSW_C = 32.80 + 26.80 + 14.80 = 74.40SSW=SSWA+SSWB+SSWC=32.80+26.80+14.80=74.40
Now, calculate the Mean Square Within (MSW):
MSW=SSWdf within=74.4015−3=74.4012=6.20MSW = \frac{SSW}{\text{df within}} = \frac{74.40}{15 - 3} = \frac{74.40}{12} = 6.20MSW=df within SSW=15−374.40=1274.40=6.20
Step 4: Calculate the F- statistic
The F-statistic is calculated as the ratio of the between-group variability to the within-group variability:
F=MSBMSW=184.39686.20=29.7F = \frac{MSB}{MSW} = \frac{184.3968}{6.20} = 29.7F=MSWMSB=6.20184.3968=29.7
Step 5: Compare the F- statistic with the Critical Value
To determine if the observed F-statistic is statistically significant, we compare it with the critical value from the F-distribution table. For α=0.05\alpha = 0.05α=0.05, the degrees of freedom for between-groups (df between) is 2, and for within-groups (df within) is 12.
Using an F-distribution table or statistical software, the critical value for F at α=0.05\alpha = 0.05α=0.05, df1 = 2, and df2 = 12 is approximately 3.89.
Since the calculated F-statistic (29.7) is much greater than the critical value (3.89), we reject the null hypothesis.
Conclusion
There is enough evidence to conclude that there is a statistically significant difference in the mean performance scores between the three teaching methods (A, B, and C) at the 5% significance level. Therefore, at least one teaching method leads to a significantly different mean performance score compared to the others.