Question

The following data give the number of hours 5 students spent studying and their corresponding grades on their midterm exams. 

Hours Studying 2 2 4 5 6 

Midterm Grades 64 68 86 90 96 

Step 1 of 7 :  

Calculate the sum of squared errors (SSE). Use the values above. Round your answer to three decimal places.

Step 2 of 7 :  

Calculate the estimated variance of errors, s2e. Round your answer to three decimal places.

Step 3 of 7 :  

Calculate the estimated variance of slope, s2b1. Round your answer to three decimal places.

Step 4 of 7 :  

Construct the 98% confidence interval for the slope. Round your answers to three decimal places.

Lower and upper end point:

Step 5 of 7 :  

Construct the 95% confidence interval for the slope. Round your answers to three decimal places.

Step 6 of 7 :  

Construct the 80% confidence interval for the slope. Round your answers to three decimal places.

Step 7 of 7 :  

Construct the 99% confidence interval for the slope. Round your answers to three deci

Answer 1

Given,

y^ = b0 + b1x

y^ = 51.4686 + 7.7188x

x	y	y^=51.4686+7.7188x	(y-ybar)^2
2	64	66.9062	8.446
2	68	66.9062	1.196
4	86	82.3438	13.368
5	90	90.0626	0.004
6	96	97.7814	3.173
Total	sum		26.188

a)

Sum of squared errors (SSE) = (y-ybar)^2

= 26.188

b)

Estimated variance of errors, se^2 = SSE / (n-2)

= 26.188 / (5-2)

= 8.729

c)

Estimated variance of slope, sb1^2 = se^2/(n-1)s2^2

= 8.729 / (5-1)*1.789^2

= 0.682

d)

degree of freedom = n - 2 = 5 - 2 = 3

alpha = 0.02

Critical value = t(alpha/2 ,d f) = 4.540703 = 4.54

98% confidence interval = b1 +/- t*sb1 = 7.7188 +/- 4.54*sqrt(0.682) = 7.7188 +/- 3.75 = (3.969 , 11.469)

e)

degree of freedom = n - 2 = 5 - 2 = 3

alpha = 0.05

Critical value = t(alpha/2 ,d f) = 3.182

95% confidence interval = b1 +/- t*sb1 = 7.7188 +/- 3.182*sqrt(0.682) = 7.7188 +/- 2.627 = (5.092 , 10.346)

f)

degree of freedom = n - 2 = 5 - 2 = 3

alpha = 0.2

Critical value = t(alpha/2 ,d f) = 1.64

95% confidence interval = b1 +/- t*sb1 = 7.7188 +/- 1.64*sqrt(0.682) = (6.365 , 9.073)

g)

degree of freedom = n - 2 = 5 - 2 = 3

alpha = 0.01

Critical value = t(alpha/2 ,d f) = 5.841

99% confidence interval = b1 +/- t*sb1 = 7.7188 +/- 5.841*sqrt(0.682) = (2.895, 12.543)