You can put this solution on YOUR website! what are you comparing? means or variances?
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student states the following
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The sample covariance between the mid-term and final exam grades is given as 68.9. I need to test the null hypothesis that the within-student average grades for the mid-term and final exams are the same against a one-sided alternative that the within-student average grade for the final exam is lower than the average grade for the mid-term exam.
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The sample covariance between the mid-term and final exam grades is given as 68.9
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This means the final exam grade increases as the mid-term exam grade increases
because the sample covariance is greater than zero.
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Ho: mid-term average = final grade average
H1: mid-term average > final grade average
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Standard Error(SE) = square root[(s1^2/n1) + (s2^2/n2)]
SE = square root[(14.4^2/176) + (14.5^2/176)] = 1.5404
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Degrees of Freedom (DF) = (s1^2/n1 + s2^2/n2)^2 / { [ (s1^2/ n1)^2 / (n1 - 1) ] + [ (s2^2 / n2)^2 / (n2 - 1) ] }
DF = (14.4^2/176 + 14.5^2/176)^2 / {[(14.4^2/176)^2 / 175] + [(14.5^2/176)^2 / 175]} = 349.98 approx 350
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t statistic = (x1 - x2) / SE = (72.5 - 68.2) / 1.5404 = 2.79
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Probability ( t < or = 2.79) is 0.9972
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Since the P-value (0.9972) is greater than the significance level (0.05), we cannot reject the null hypothesis.
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