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Download DocumentSTAT200Introduction to StatisticsName______________________________
Final Examination: Spring 2016 OL2Instructor __________________________
Answer Sheet
Instructions:
This is an open-book exam. You may refer to your text and other course materials as you work on the exam, and you may use a calculator.
Record your answers and work in this document.
Answer all 20 questions. Make sure your answers are as complete as possible. Show all of your work and reasoning. In particular, when there are calculations involved, you must show how you come up with your answers with critical work and/or necessary tables. Answers that come straight from calculators, programs or software packages will not be accepted. If you need to use software (for example, Excel)
H1: The data are not consistent with the heart beat.(b)20
(c) 0.05 to 0.03
(d).No
Work for (b) and (c):
(40-80)2/80 =20
test statistic 20 and d f 39 p-value lies between 0.05 to 0.03
18
Answer:
(a) Here chi-square test for goodness of fit will be used.
H0: The data are consistent with the five colors.
H1: The data are not consistent with five colors.
(b) 33.3
(c) 0.05 to 0.025
(d) No
Work for (b) and (c):
b. Test statistic
Expected value=(1/5) 100=20
Χ2 = Σ [ (x - yi)2 / yi ]
Where xi is the observed frequency count yi is the expected frequency count
test statistic 33.3 and d f 19 p-value lies between 0.05 to 0.025
19
Answer:
(a) Y = 1.97 +0.89x
(b) 4.64
Work for (a) and (b):
Y= a + bx
N = 4
Using the formula for slope (b): QUOTE
QUOTE = 0.89
Incorporate the intercept (a) formula: QUOTE
QUOTE = 1.97
Y= 1.97 + 0.89x
If x= 3
Incorporating into the equation:
Y= 1.97 + 0.89 (3)
Y = 4.64
20
Answer :
(a)
(b) 0.405
(c) 0.810
(d) No, p-value is much greater than the level of significance.
Work for (a), (b) and (c)
Sum of squares (error-within)
x+42.36 = 1100.76
x=1100.76 -42.36
x=1058.40
Degree of freedom
Factor (between): 50 -1 = 49
Error (within): 500 – 1 = 499
Mean Squares
Factor (between) = 42.36/49
=0.86
Error (within) =1058.40/499
= 2.12
F-test is used
MS(a)/MS(b0 = 0.86 /2.12
=0.405
p-value= P(z < -1) + P(z > 1)
= 0.405+ 0.405= 0.810
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