Question

Can you show the how you worked out the problem and the formula you used?

In determining the relationship of a professor’s exams, a department checked students’ performances on the midterm and final exams in a class. Based on the following data, what can you say about the relationship of the exam scores? (a = 0.05)

H0:

H1:

a=

Decision Rule:

Student Midterm A 63

B 63 C 63 D 66 E 68 F 64 G 64 H 69

Test:

Final 110

118

103

120

125

125

120

120

Conclusion:

2b) Based on your results Midterm Exam score was 71?

above, what Final Exam score would you predict for a student whose

Answer 1

Solution :

a)

Given that =0.5

The hypothesis being tested is:

H0: There is no relationship between the test scores.  ß1=   0

Ha: There is a relationship between the test scores  ß1╪  0

Alpha = 0.05

Decision Rule: Reject H0 if p-value < 0.05.

Test:

x	y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
63	110	4.0000	58.1406	15.250
63	118	4.0000	0.1406	-0.750
63	103	4.0000	213.8906	29.250
66	120	1.0000	5.6406	2.375
68	125	9.0000	54.3906	22.125
64	125	1.0000	54.3906	-7.375
64	120	1.0000	5.6406	-2.375
69	120	16.0000	5.6406	9.500

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
Total sum	520.00	941.00	40.00	397.88	68.00
Mean	65.00	117.63	SSxx	SSyy	SSxy

sample size ,   n = 8          
here,

x̅ = Σx / n = 65.000   ,     ȳ = Σy/n = 117.625   
SSxx =    Σ(x-x̅)² =   40.0000          
SSxy=   Σ(x-x̅)(y-ȳ) =  68.0          
                  
estimated slope , ß1 = SSxy/SSxx = 68.0 / 40.000   = 1.70000

estimated std error of slope =Se(ß1) = Se/√Sxx =    6.859   /√   40.00  =  1.0845
                  
t stat = estimated slope/std error =ß1 /Se(ß1) = 1.7000 / 1.0845   = 1.5675
                  
t-critical value=    2.4469   [excel function: =T.INV.2T(α,df) ]          
Degree of freedom ,df = n-2=   6              
p-value = 0.1680              

The regression output is:

The p-value is 0.1680.

Since the p-value (0.1680) is greater than the significance level (0.05), we cannot reject the null hypothesis.

Conclusion :

Therefore, we have insufficient evidence to conclude that there is a relationship between the test scores.

b)

Final test score = 127.825.

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