Statistics Assignment Sample
Q1:
Answer :Introduction to Two-Sample t-Test
The two-sample t-test is used to compare the means of two independent groups to determine if there is evidence that the associated population means are significantly different. In this case, we want to test whether the mean exam scores of students taught with the traditional method differ from those taught with the innovative method.
The test involves the following steps:
- State the hypotheses.
- Calculate the sample statistics (means, standard deviations).
- Perform the t-test.
- Compare the calculated t-statistic with the critical value or use the p-value.
- Make a conclusion based on the results.
Step 1: State the Hypotheses
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Null Hypothesis (H₀): There is no significant difference in the mean exam scores between the two groups.
H0:μ1=μ2H₀: \mu_1 = \mu_2H0:μ1=μ2
Where μ1\mu_1μ1 and μ2\mu_2μ2 are the mean exam scores for Group 1 (traditional) and Group 2 (innovative), respectively. -
Alternative Hypothesis (H₁): There is a significant difference in the mean exam scores between the two groups.
H1:μ1≠μ2H₁: \mu_1 \neq \mu_2H1:μ1=μ2
We will perform the test using a two-tailed test with a significance level of α=0.05\alpha = 0.05α=0.05.
Step 2: Calculate the Sample Statistics
We need to calculate the sample means and sample standard deviations for each group.
For Group 1 (Traditional Method):
The exam scores are: 60, 62, 65, 58, 67, 72, 70, 64, 69, 61.
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Sample Mean (xˉ1\bar{x}_1xˉ1):
xˉ1=60+62+65+58+67+72+70+64+69+6110=61810=61.8\bar{x}_1 = \frac{60 + 62 + 65 + 58 + 67 + 72 + 70 + 64 + 69 + 61}{10} = \frac{618}{10} = 61.8xˉ1=1060+62+65+58+67+72+70+64+69+61=10618=61.8
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Sample Standard Deviation (s₁):
The formula for the sample standard deviation is:
s1=∑(xi−xˉ1)2n1−1s_1 = \sqrt{\frac{\sum (x_i - \bar{x}_1)^2}{n_1 - 1}}s1=n1−1∑(xi−xˉ1)2
Where xix_ixi are the individual scores, xˉ1\bar{x}_1xˉ1 is the sample mean, and n1n_1n1 is the sample size (10).
s1=(60−61.8)2+(62−61.8)2+(65−61.8)2+(58−61.8)2+(67−61.8)2+(72−61.8)2+(70−61.8)2+(64−61.8)2+(69−61.8)2+(61−61.8)29s_1 = \sqrt{\frac{(60 - 61.8)^2 + (62 - 61.8)^2 + (65 - 61.8)^2 + (58 - 61.8)^2 + (67 - 61.8)^2 + (72 - 61.8)^2 + (70 - 61.8)^2 + (64 - 61.8)^2 + (69 - 61.8)^2 + (61 - 61.8)^2}{9}}s1=9(60−61.8)2+(62−61.8)2+(65−61.8)2+(58−61.8)2+(67−61.8)2+(72−61.8)2+(70−61.8)2+(64−61.8)2+(69−61.8)2+(61−61.8)2 s1=1.44+0.04+10.24+14.44+26.44+102.44+66.24+4.84+51.84+0.649=278.609=30.96=5.57s_1 = \sqrt{\frac{1.44 + 0.04 + 10.24 + 14.44 + 26.44 + 102.44 + 66.24 + 4.84 + 51.84 + 0.64}{9}} = \sqrt{\frac{278.60}{9}} = \sqrt{30.96} = 5.57s1=91.44+0.04+10.24+14.44+26.44+102.44+66.24+4.84+51.84+0.64=9278.60=30.96=5.57
For Group 2 (Innovative Method):
The exam scores are: 75, 80, 78, 74, 82, 77, 79, 83, 75, 81.
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Sample Mean (xˉ2\bar{x}_2xˉ2):
xˉ2=75+80+78+74+82+77+79+83+75+8110=80410=80.4\bar{x}_2 = \frac{75 + 80 + 78 + 74 + 82 + 77 + 79 + 83 + 75 + 81}{10} = \frac{804}{10} = 80.4xˉ2=1075+80+78+74+82+77+79+83+75+81=10804=80.4
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Sample Standard Deviation (s₂):
s2=∑(xi−xˉ2)2n2−1s_2 = \sqrt{\frac{\sum (x_i - \bar{x}_2)^2}{n_2 - 1}}s2=n2−1∑(xi−xˉ2)2 s2=(75−80.4)2+(80−80.4)2+(78−80.4)2+(74−80.4)2+(82−80.4)2+(77−80.4)2+(79−80.4)2+(83−80.4)2+(75−80.4)2+(81−80.4)29s_2 = \sqrt{\frac{(75 - 80.4)^2 + (80 - 80.4)^2 + (78 - 80.4)^2 + (74 - 80.4)^2 + (82 - 80.4)^2 + (77 - 80.4)^2 + (79 - 80.4)^2 + (83 - 80.4)^2 + (75 - 80.4)^2 + (81 - 80.4)^2}{9}}s2=9(75−80.4)2+(80−80.4)2+(78−80.4)2+(74−80.4)2+(82−80.4)2+(77−80.4)2+(79−80.4)2+(83−80.4)2+(75−80.4)2+(81−80.4)2 s2=28.96+0.16+5.76+40.96+2.56+11.56+1.96+6.76+28.96+0.369=127.049=14.11=3.76s_2 = \sqrt{\frac{28.96 + 0.16 + 5.76 + 40.96 + 2.56 + 11.56 + 1.96 + 6.76 + 28.96 + 0.36}{9}} = \sqrt{\frac{127.04}{9}} = \sqrt{14.11} = 3.76s2=928.96+0.16+5.76+40.96+2.56+11.56+1.96+6.76+28.96+0.36=9127.04=14.11=3.76
Step 3: Perform the t-Test
Now that we have the sample statistics, we can calculate the t-statistic using the formula for the two-sample t-test:
t=(xˉ1−xˉ2)s12n1+s22n2t = \frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}t=n1s12+n2s22(xˉ1−xˉ2)
Where:
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xˉ1=61.8\bar{x}_1 = 61.8xˉ1=61.8, xˉ2=80.4\bar{x}_2 = 80.4xˉ2=80.4
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s1=5.57s_1 = 5.57s1=5.57, s2=3.76s_2 = 3.76s2=3.76
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n1=n2=10n_1 = n_2 = 10n1=n2=10
t=61.8−80.45.57210+3.76210t = \frac{61.8 - 80.4}{\sqrt{\frac{5.57^2}{10} + \frac{3.76^2}{10}}}t=105.572+103.76261.8−80.4 t=−18.631.0410+14.1110=−18.63.104+1.411=−18.64.515=−18.62.13=−8.73t = \frac{-18.6}{\sqrt{\frac{31.04}{10} + \frac{14.11}{10}}} = \frac{-18.6}{\sqrt{3.104 + 1.411}} = \frac{-18.6}{\sqrt{4.515}} = \frac{-18.6}{2.13} = -8.73t=1031.04+1014.11−18.6=3.104+1.411−18.6=4.515−18.6=2.13−18.6=−8.73
Step 4: Compare the t-statistic with the Critical Value
For a two-tailed test at α=0.05\alpha = 0.05α=0.05, the degrees of freedom (df) are calculated using the formula:
df=(s12n1+s22n2)2(s12n1)2n1−1+(s22n2)2n2−1df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}}df=n1−1(n1s12)2+n2−1(n2s22)2(n1s12+n2s22)2
Using statistical software or a t-distribution table, the critical t-value for α=0.05\alpha = 0.05α=0.05 (two-tailed test) and df ≈ 17 is approximately 2.109.
Since the calculated t-statistic (-8.73) is much less than -2.109 (in absolute value), we reject the null hypothesis.
Step 5: Conclusion
Since the absolute value of the calculated t-statistic (8.73) is greater than the critical value (2.109), we reject the null hypothesis at the 5% significance level. Therefore, there is sufficient evidence to conclude that there is a statistically significant difference in the mean exam scores between students who received the traditional teaching method and those who received the innovative teaching method.
The innovative teaching method seems to result in significantly higher exam scores compared to the traditional method.