MegaCon 19% [ 01:21 e-course.auca.kg 3. Employers often use standardized measure
ID: 3360512 • Letter: M
Question
MegaCon 19% [ 01:21 e-course.auca.kg 3. Employers often use standardized measures to gauge how likely it is that a new employee with little to no experience will succeed in their company. One such factor is intelligence, measured using the Intelligence Quotient (IQ). To show that this factor is related to job success, an organizational psychologist measures the IQ score and job performance (in units sold per day) in a sample of 10 new employees IQ Job performan 147 132 115 a. Compute Spearman correlation coefficient. b. According to this correlation coefficient what exactly is the relationship between IQ score and Job performance? How strong the relationship? 4. A psychologist noted that people have more difficulty sleeping in a bright light room than a dark room. She measured whether the intensity of the light could predict the time it took a sample of 12 participants to fall asleep. The data for this hypothetical study are listed in the following table Intensity of light (in Time it took to sleep watts) in minutes) 35 a. Compute the Pearson r b. According to this correlation coefficient what exactly is the relationship between intensity of light and time it took to sleep? How strong is the relationship? c. Find the regression equation. d. Interpret "a" coefficient e. Interpret "b” coefficient. t. How long will it take for a person to fall asleep if the intensity of light in the room is 50? g. Find coefficient of determination and explain what it indicates h. Find standard error of estimate and explain what it indicatesExplanation / Answer
Question 1
Average number of alcohol beverages consumption = 7
Standard deviation = 4
(i) If X is the alcohol beverage consumption of random person.
Pr(X > 13) = NORM ( X> 13; 7; 4) = 1 - Pr( X < 13; 7; 4) = 1 - (Z)
Z = (13 - 7)/4 = 1.5
Pr(X > 13) = NORM ( X> 13; 7; 4) = 1 - Pr( X < 13; 7; 4) = 1 - (Z) = 1 - (1.5)
= 1 - 0.9332 = 0.0668
(b) Pr(X < 1) = NORM ( X < 1; 7; 4) = (Z)
Z = (1 - 7)/4 = -1.5
Pr(X < 1) = NORM ( X < 1; 7; 4) = (Z) = (-1.5) = 0.0668
(c) Pr(6 < X < 12) = NORM (6 < X < 12 ; 7; 4) = Pr(X < 12) - Pr(X < 6) = (Z2) - (Z1)
Z2= (12 -7)/4 = 1.25
Z1= (6 - 7)/4 = -0.25
Pr(6 < X < 12) = NORM (6 < X < 12 ; 7; 4) = (Z2) - (Z1) = (1.25) - (-0.25)
= 0.8945 - 0.4013 = 0.4932
(d)
Pr(1 < X < 6) = NORM (1 < X < 6 ; 7; 4) = (Z2) - (Z1)
Z2= (6 -7)/4 = -0.25
Z1= (1 - 7)/4 = -1.50
Pr(1 < X < 6) = NORM (6 < X < 12 ; 7; 4) = (Z2) - (Z1) = (-0.25) - (-1.50)
= 0.4013 - 0.0668 = 0.3345
(d) Here the percentile means
Pr(X <= 10) = NORM ( X< = 10; 7; 4)
Z = (10 - 7)/4 = 0.75
Pr(X <= 10) = NORM ( X< = 10; 7; 4) = Pr(Z < 0.75) = 0.7734
so it is 77.34% percentile.
(f) Here if we say tha X0 is associated with 95% percentile.
Pr( X > X0 ; 7; 4) = 0.95
Z - vlaue = 1.645
1.645 = (X0 - 7)/4
X0 = 7 + 4 * 1.645 = 13.58