Determine which one is the response and which one is the explanatory variable? F

Question

Determine which one is the response and which one is the explanatory variable?

Fit a linear regression model to the data and interpret the slope coefficient.

What percent of the total variability in achievement score can be explained by fitting a linear regression line?

Test the significance of the slope at =0.05 level. Clearly state your regression model and hypotheses.

Construct a 95% confidence interval for the slope parameter.

Plot the residuals versus the X variable. From the plot, does it appear to have any issues with a linear fit?

It is assumed that achievement test scores should be correlated with student's classroom performance. One would expect that students who consistently perform well in th classroom (tests, quizes etc) would also perform well on a standardized achievement test (0-100 with 100 indicating high achievement). A teacher decides to examine this hypothesis. At the end of the academic year, she computes a correlation between the students achievement test scores (she purposefully did not look at this data until after she submitted students grades) and the overall g.p.a. foreach student computed over the entire year. The data for her class are provided below. Achievement G.PA 98 3.6 96 2.7 94 3.1 88 4.0 91 77 3.0 86 3.8 2.6 71 3.0 59 63 2.2 84 1.7 79 3.1 75 2.6 72 2.9 86 2.4 34 85 71 2.8 93 3.7 90 3.2 62

Explanation / Answer

Solution

Back-up Theory

Let X = GPA and Y = Achievement test score.

Let (xi, yi) be a pair of sample observation on (X, Y), i = 1, 2, …., n

where n = sample size.

Then, Mean X = Xbar = (1/n)sum of xiover I = 1, 2, …., n; ……………….(1)

Variance of X, V(X) = (1/n)Sxx where Sxx

= sum of (xi – Xbar)2 over i = 1, 2, …., n ………………………………..(2)

Standard Deviation of X = SDX = sq.rt of V(X). ……………………………(3)

Similarly, Mean Y = Ybar =(1/n)sum of yiover i= 1, 2, …., n;…………….(4)

Variance of Y, V(Y) = (1/n)Syy where Syy

= sum of (yi – Ybar)2 over i = 1, 2, …., n ………………………………………………(5)

Standard Deviation of Y = SDY = sq.rt of V(Y). ………………………….……………….(6)

Covariance of X and Y, Cov(X, Y)

= (1/n)Sxy where Sxy = sum of {(xi – Xbar)(yi – Ybar)} over i = 1, 2, …., n………(7)

Correlation Coefficient of X and Y = rXY

= Cov(X, Y)/(SDX.SDY) = Sxy/sq.rt(Sxx.Syy). ……………………………………………(8)

Estimated Regression of Y on X is given by: Y = a + bX, where

b = Cov(X, Y)/V(X) = Sxy/Sxx and a = Ybar – b.Xbar..…………………….(9)

Estimate of 2is given by s2= (Syy – b2Sxx)/(n - 2).

Standard Error of b is sb, where sb2 = s2/Sxx .....................................(10)

Now, to work out solution,

Part (a)

The statement, ‘one would expect that a student who consistently performs well in class room would also perform well in achievement tests’=> Achievement test score is the response variable (Y) and GPA is the explanatory variable (X). ANSWER

Part (b)

The regression model is: Y = + X +

Estimates of + are given by a and b.

Given n = 20, using Excel Functions, the following are computed using the given data.

Xbar = 81     Ybar = 2.93     Sxx = 2638    Syy = 7.762    Sxy = 75

[vide (9) under Back-up Theory], b = Cov(X, Y)/V(X) = Sxy/Sxx = 0.0284 and

a = Ybar – b.Xbar = 0.6272

So, regression fit is: y = 0.6272 + 0.0284x

Interpretation

Slope coefficient is b. b = 0.0284 => when GPA increases/decreases by 1 point, the achievement score would increase/decrease by 0.0284. ANSWER

Part (c)

Percentage of total variability explained by fitting regression is given by 100xr2 = 100 x 0.2747 = 27.47% ANSWER [vide (8) under Back-up Theory]

Part (d)

To test significance of slope at = 0.05,

Null Hypothesis H0: = 0 Vs Alternative H0: 0

Test statistic: t = (b - 0)/SE(b)

Under H0 , t ~ tn – 2

Estimate of 2 is given by s2 = (Syy – b2Sxx)/(n - 2) = 0.55932.

Standard Error of b is sb, where sb2 = s2/Sxx = 0.0109

Calculated value of r = tcal = 2.61

tcrit = Upper 2.5% (/2) point of t18 = 2.101

Since tcal > tcrit, H0 is rejected.

Conclusion: statistical evidence exists to suggest that the slope is not zero. ANSWER

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