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 100 third-grade classes, estimates the OLS regression Testscore= 515.1960-5.7618CS, R2-.08, SER= 11.4 (20.1960) (2.3426) (a) Construct a 95% confidence interval for A , the regression slope coefficient Lower bound: -10.35 Upper bound: -1.17 (b) The t-statistic for the two-sided test of the null hypothesis Ho : 0 is Round your responses to two decimal places.) (Round your responses to two decimal places.) 2.46 . (Round your response to two decimal places.) (c) The p-value for the two-sided test of the null hypothesis H0 : 1-0 is 0.016 Round your response to three decimal places.) Do you reject the null hypothesis at the 1% level(yes/no)? yes The p-value for the two-sided test of the null hypothesis Ho : A =-5.5 is 0.112 . (Round your response to three decimal places.) Without doing any additional calculations, is-5.5 is contained in the 95% confidence interval for 31 (yes/no)? (d) Construct a 99% confidence interval for A30 Lower bound: 463.2 Upper bound: 567.2 yes Round your responses to one decimal place.) (Round your responses to one decimal place.)

Explanation / Answer

Answers to the question is as follows:

a. Lower and upper bound of B1 is given by :

Lower bound = B1 - Z*SE = - 5.7618 - 1.96*2.3426 = -10.35
Upper bound = B1 + Z*SE = - 5.7618 + 1.96*2.3426 = -1.17

b. t = (B-0)/SE = -5.7618/2.3426 = -2.46

c. p-value is The P-Value is .015627 or .016 rounded off the 2 decimals

No. we will not reject null hypothesis. since .01>.016

d. t = (B-0)/SE = -5.5/2.3426 = -2.35
The P-Value is .0208.

Doesn the 95% interval we created in parta have -5.5 between upper and lower bounds. If yes, then Yes is the answer. Since our lower and upper bounds are -1.17 and -10.35 and -5.5 is between the two, we can say that
b1 lies in the 95% interval.

So, Yes

d. The 99% interval is  

Upper bound = B1 + Z*SE = - 5.7618 + 2.56*2.3426 = 0.2

Lower bound = B1 - Z*SE = - 5.7618 - 2.56*2.3426 = -11.8

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