Suppose that a researcher, using data on class size (CS) and average test scores

Question

Suppose that a researcher, using data on class size (CS) and average test scores from 50 third-grade classes, estimates the OLS regression TestScore = 640.3-4.93 × CS, R2 = 0.11, SER = 8.7 (23.5) (2.02) a. Construct a 95% confidence interval for A, the regression slope coefficient. b. Calculate p-value for the two-sided test of the null hypothesisHo: A = 0. Do you reject the null hypothesis (i) at the 5% level (ii) at the 1% level? C. Calculate the p-value for the two-sided test of the null hypothesisHo: ,--5. Without any doing any additional calculations, determine whether IS contained in the 95% confidence level for 1 d. Construct a 90% confidence level. e. Construct a 99% confidence level.

Explanation / Answer

a)95% confidence interval for 1 : 1 ± 1.96* standard error

(since the critical value for t at 95% is 1.96)

95% confidence interval for 1 : -4.93 ± 1.96*2.02 = -4.93 ± 3.9592

95% confidence interval for 1 : (-8.8892 , -0.9708)

b)t = 1 / SE(1) = -4.93/2.02 = -2.4405

p-value = 2 (- |t|) = 2 (-2.4405) = 2*0.0073 (from the z table for z=-2.44)

p-value =0.0146

Interpretation for 5% level: Since the p-value is less than 0.05, we reject the null hypothesis and conclude that 1 0.

Interpretation for 1% level: Since the p-value is greater than 0.01, we accept the null hypothesis and conclude that 1 = 0.

c)t = 1-(-5) / SE(1) = 0.07/2.02 = 0.0347

p-value = 2 (- |t|) = 2 (-0.0347) = 2*0.4880 (from the z table for z=0.03)

p-value =0.976

Since the 95% confidence interval for 1 is (-8.8892 , -0.9708), it is evident that -5 is contained in the confidence interval. Therefore 1 = -5 is not rejected at 5% level of significance.

d)90% confidence interval for 1 : 1 ± 1.65* standard error

(since the critical value for t at 90% is 1.65)

90% confidence interval for 1 : -4.93 ± 1.65*2.02 = -4.93 ± 3.333

90% confidence interval for 1 : (-8.263 , -1.597)

e)99% confidence interval for 1 : 1 ± 2.58* standard error

(since the critical value for t at 99% is 2.58)

99% confidence interval for 1 : -4.93 ± 2.58*2.02 = -4.93 ± 5.2116

99% confidence interval for 1 : (-10.1416 , 0.2816)

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