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**Unformatted text preview: **(Q: ID: MST. SLR. REE. Ell . EIDZIJf] The Image presented depicts a: negative curvilinear relationship
v' n positive curvilinear relationship
negative linear relationship positive linear relationship ((2:10: MST.SLR.REE.Dl.UDBU‘f)
Suppose the simple linear regression equation derived from a sample ls given by: A v = 3.4 + 4.9x
The sample v-Interoept is:
J a 3.4 4.9 X Y (Q=ID: MST.5LR.REE‘01.DOSOF) Select the equation that is satisfied by the total sum of squares (SST), the regression sum ofsquares (55M) and the error sum
of squares (SSE): SSM = SST + SSE
SSE = SSM + SST
v’ 6 SET = 55M + 55E
SST = 55M — SSE (Q=ID: MST.SLR.REE.Dl.DDEOf) The coefﬁcient of determination (R2) is given by: R2 = error sum of sguares
total sum of squares J a R2 = regression sum of squares
total sum of squares R2 = total sum of squares
error sum of squares R2 = regression Sle of squares
El’l’Dl’ SUITI of squares (Q=ID: MST.SLR.REE.01.00?D'F) Suppose your ﬁrm is assessing the Impact of door-to-door canvassing on total sales and has developed a regression equation
using sample data to describe this process. The error sum of squares (SSE) Is equal to 20 and the regression sum of squares (SSM) is equal to 30. The coefﬁcient of determination (R2): X o is equal to 0.40, which indicates that 40% of the variation in sales is predicted by the amount of door-to-
door canvassing undertaken is equal to 0.50, which indicates that 60% of the variation in sales is predicted by the amount of door-to-
door canvassing undertaken is equal to 0.56, which indicates that 66% of the variation in sales is predicted by the amount of door-to-
door canvassing undertaken is equal to 1.50, which indicates that 1.5% of the variation in sales is predicted by the amount of door-to-
door canvassing undertaken ((1:10: MST.SLR.ALR.01. 0030f) The difference between the observed and predicted value ofa dependant variable IS the: J - residual
standard error
deviation
variance {Q=ID: MST.SLR.Tl-‘I.01.0010f) Suppose that you have developed a simple linear regression model and have calculated the regression coefﬁcients. You conduct
a t test and ﬁnd that the population slope is not equal to zero. It is reasonable to conclude that there is: evidence of a linear relationship between the two variables
no evidence of a linear relationship between the two variables
X n evidence of autccorrelation between the two variables no evidence of autocorrelation between the two variables (Q=ID: M5T.SLR.‘IT-I.Cll.002l:lf) Suppose that you have developed a simple linear regression model and have calculated the regression coefﬁcients. You conduct
an F test and find that the population slope is not equal to zero. It is reasonable to conclude that: J - there is evidence that the y variable is significantly related to the x variable the y variable is definitely signiﬁcantly related to the x variable the y variable is definitely not significantly related to the x variable
there is no evidence that the y variable is significantly related to the x variable (Q: ID: MST.SLR.TM.DI. 0030f} In conducting a t test for the slope ([31] of a simple linear regression model, using a sample of size n, the corresponding test
statistic follows the: X - F distribution with 2 and n - 2 degrees of freedom
t distribution with n - 2 degrees of freedom
F distribution with 1 and n - 1 degrees of freedom
t distribution with n - 1 degrees of freedom :I 1 of 3 ID: MST.NDM.CC.10.IJDJ.0 Select whether the following statements regarding the ooeﬁ‘ioient of correlation (r) are true or false: a) If r :v 0 then a large value of x will cause a large value of v. ' b) If r > D then a large value ofx is related to a large value of v. - I I c) If r c Othen a large value of): will cause a small value of v. I r I :l Feedaack [3 out of 3] a) You are correct.
In) You are boned. c) You are correct. The coen'icient of correlation is a measure of the linear relationship between two variables. A positive coefﬁcient of correlation means that the two variables tend
to move In the same olrectlon.1'hls means that if one varlable (x) ls large then the other varlable (v) also tends to be large. A negatlve ooeﬁldent of correlation
means that the two variables tend to move in opposite directions. 111Is means that if one vanable (x) is large then the other variable (3:) tends to be small. It is important to remember that the ooeﬁident of correlation only oesmbes the observed relationship between x and v. 1t om not describe a cause and elfect
relationship. It is possible for x and v to be positiver or negatively correlated simpiv bv coincidence. For two positively correlated variables It is incorrect to state
that a large value of x causes a large value of v. The toned: Eatement is to sag.r that a large value of x is related to a large value of v. 2| 2 of 3 m: MST.NDM.LP.po.oom Regarding the analysis oi a linear relationship between two variables, It Is true that the: covariance is always between -1 and 1
covariance is Independent of the size of the data being considered
- coefficient of correlation ls always between —1 and 1 :| Feedback [1 out of 1] You are correct. Discussion Given two variables x and v. the formula for the covariance is: shmadabies = Elxi ' Elm ' ti mum) n _ 1 The oovarianoe gives a measure or the extent to which two variablﬁ are linearly related. In qualitative terms, it does this by measuring the variation in both
variables and to what extent these variations occur together. Data sets with larger values will tend to have larger absolute variation. For this reason, the
disadvantage of covariance is that It is dependent on the size of the data being considered. In general, the covariance between two variables can have any
numerical value. The formula for the coefﬁcient of correlation is: M
r 2 mix! sxs, The coefﬁcient of correlation is a measure that solves the disadvantage of covariance. It is calculated by divining the covariance by the standard deviations of
both variables. As a result, if a data set has a lag variation, this does not aﬁect the ooeﬁ'ident of correlation. The coeiﬁclent oi oorrelation takes values between
-1 and 1. If r is close to 1. the variabla have a ebony positive linear relationship. if r is close to -l. the variables have a strong negative linear relationship. If r is
close to 0, there is no identiﬁable linear relationship between the variables. T 3 or 3 ID: usr.nou.cc.os.oono An investigation has been conducted to determine whether there is a relationship between the salary paid to a Chief
Executive Ofﬁcer (CEO) and the productivity oi that CED as measured by the change In proﬁts from the time the CED
was employed. The scatter plot plots the salary paid 0:) against the «ﬂange in profits (y) of a sample of CEOs. Without doing any
calculations and according only to this scatter diagram. a reasonable ooelﬁcient oi correlation {r} between x and y would CWhmlﬂ be: .
. r = 0.39 " - ' .
r = 43.68 '
r = -2.45 '. ' r"
a is
r = 2_4iiI Salary vtﬂtx)
I Feedback [1 outof 1] You are correct. Discussion The apparent positive linear dope of the scatter plot reveals that larger values or x are related to larger values oi y. This means that the coeiﬁcient of correlation
will be positive. It is also important to recall that the coefﬁcient of correlation must be between -1 and 1. The onlyr option that satisﬁes the requirements or being
positive reveals that larger values oi it are related to larger values of y. This means that the coeﬁicient of correlation will be positive. It is also important to recall
that the coefﬁcient of correlation must be between -1 and 1.111e only option that satisﬁes the requirements of being positive and between -:I. and 1 is r = 0.39. This could imply that the higher the salary paid to a CEO. the more value they add ho a company. It is Important to remember that the cceﬁ‘lolent of correlation
only describes the observed relationship between x and v. It does not describe a cause and effect relationship. It Is possible for x and y to be positively or
negatively correlated simply Ur coincidence. It is Incorrect no state that a large value of x causes a large value of v. The correct statEmem is to say that a large
value of x is related to a large value of y. :I 3 of 3 1o: MST.NoM.cc.o5.do1o The following sample statistirs have been calculated for a set of data: Sample statistics
covariance between x and v: cov(x,v) = 29,177,777.7B
standard deviation of x: 5x = 15,167.22
standard deviation of v: 3,,r = 15,43?.69 Use these statistics to calculate the coefﬁcient of correlation (r) between x and v. Give your answer to 3 decimal places. r = 0.125 3| Feedback [1 out of 1] You are correct. calculation 111E ooefﬁcient of correlation can be calculated using the following formula: ML:
, _ M
sxsy = 29.177.777.73
15, 167.22 x 15,437.69 0.12461316...
0.125 Rounded as last step ;| 2 of 3 ID: Msr.NoM.cc.o9.ddlo You are told that the ooeﬁicient of correlation between the variable x and the variable y is 4112. According to this correlation coefﬁcient, larger value: of x bend to be related to m values of v. :I Feedback [1 out of 1] You are correct. Discussion The coefﬁcient of correlation is a measure or the linear refationship between two variables. A negative coefﬁcient of correration means that the two varieties
tend to move in opooﬂte directions. In other words. if one variabte 0:) Is large then the other variable (v) tends to be small. Conversely, If x is small then if tends
to be large. It is important to remember that the coefﬁcient oi correlation only deseribes the observed relationship between x and y. It does not describe a cause and effect
relationship. It Is possible for v and v to be positiver or negativer correlated slmplvI bv oolncloenoe. It Is lnoorrect to state that a large value of x causes a small
value ofv.1'neocrrect statement is to sayr that a Large value cfx is related to a small value of v. 2| 3 of 3 m: r-rsr.uon.cc.oe.oom An Investigation has been conducted to determine whether there is a relationship between the salary paid to a Chief ‘
Executive Oﬂioer (CEO) and the productivity of that CEO as measured by the change in proﬁts from the time the CEO .
was employed. The scatter plot plots the salaryI paid (it) against the change In proﬁts (v) of a sample of CEOs. Without doing any
calculations and according only to this scatter diagramr a reasonable coefficient of correlation (r) between x and y would Clem III will (1] be: , . .
. r = 41.62 ‘ ' r = 0.62 _ 3‘.
r = L42 ' ' . ., , r = —1.56 Salary namm [1 outof 1] YOU are UOI'I'ECt. Discussion The apparent negative linear slope oi the scatter plot reveals that larger values of x are related to smaller values of v. This means that the ooeﬁ'lclent of
correlation will be negative. It is also important to recall that the coeﬂ’lclent of correlation must be between '1. and 1. The onlyr option that satisﬁes the
requirements of being negative reveals that larger values of ‘x are related to larger values oi v. This means that the ooeﬁ‘lclent of oorrelatlon will be positive. It is
also important to recall that the melﬁcient of correlation must be between -1 and 1. The only option that satisﬁa the requirements of being positive and
between —1 and 1 is r = $.52. This could imply that the higher the salary paid to a CEO, the less value they add to a company. It is important to remember that the coefﬁcient of correlation
onhr descrﬂzies the observed relationship between K and y. It does not describe a cause and effect relationship. It is posible for x and y to be positively or
negativer correlated simplyr by oolncldenoe. It Is incorrect to state that a large value of x causes a small value of v. The correct statement is to say that a large
value of x is related to a small value of v. :I 1 of3 ID: MST.SLR.REE.oe.ou1o You are part ofan emnomim research team within Eastern Mining Company. Your team has developed a linear regression model for predicting the amount of copper
produced based on the amount of material unearthed. You have been passed the following information relating to the model: The regression sum ofsquares (55M) and the error sum of squares (SSE) for the
copper production model have been calculated: 55M = 57.64
SSE = 26.01 Calculate the coefﬁcient of determination (R2). Give your answer to 2 decimal plaoes. R2 : 0.69 3 Feedback [1 out of 1] You are correct. Calculation The coefﬁcient of determination (R2) can be calculated using the following formula: show variables 55M
SST 57.64
83.65 (168906157...
0.69 Rounded as last step R2 3 2 of 3 ID: MsT.SLR.REE.oa.do:m A study Is conducted to determine the simple linear regresdon relation (If any) between average temperature over a week and ice-cream sales In a particular city.
The following table provides the regression data required to construct the model: Average Temperature and Ice-cream Sales Average Temperature Weekly Ice-cream Sales cvera Week (°F) (3'0005)
7B 244. l 13
53 194.1 13
55 186.9}?
92 279.?11
33 131.933
6? 218.091
40 149.81?
38 145.353
44 159. 125
95 236.23?
76 239.699
38 145.803 Calculate the slope (b1) and intercept (ha) of the simple regression equation using the data provided. Give your answers to 2 decimal places.
(a) Slope = bl = 2.49 (b) Intercept = be = 50.35 (a) You are correct. (b) You are correct. Calculation Using a software package, one may obtain the following statistics:
(a) bl = 2.49 (b) II.) = 50.35 Calculating the sample slope and intercept (hide) a) Without using a software package. the sample slope can be calculated as follows: hide variables b1 sample slope
SSXY sum of squares for x and y
55X = sum of squares for x SSXY b=
1 ssx Where: hide variables b = ssxv
1 ssx where: hide variables mean of the response variable = 198.41433333...
mean of the explanatory variable = 59.5 21‘
ll im observation of the explanatory variable
i"h observation of the response variable
size of the sample = 12 ‘5.
ll U1
U1
X
.<
II E [m - i)(Vi ‘ V)]
i = 1 13.052.314- ll 55X n _
Z (Xi ' X)2
i = 1
= 5,257 Using the above: = 13,082.314
5,257 2.43855127...
= 2.49 Rounded as fast step '31 h) The sample intercept can be calculated using the following formula: n Z [M - i)(ll'i ' V”
i = 1 13,082.314 SS XY n _
z (xi-X)2
i = 1
3257 SSX Using the above: b _ 13,082.314
1 5257 2.48855127...
2.49 Rounded as fast step b) The sample intercept can be calculated using the following formula: to ? - b1?
198.41433333... — (2.48855127...)(59.5)
50.34553307... 50.35 Rounded as last step :| 3 or 3 1o: nsr.sm.n:e.us.uozo The following table shows the average petrol price and the number of online shopping orders over a given month: Petrol Price and Online Shopping Average Petrol Price Number of Onllne
per gallon over a monﬂl [$) Shopping Orders 3.0825 3,218 2.925 2.95? 3.?4 4,760 4.2 5, 103 4.7325 4,6?0 3.2??5 3,487 2. 185 2,121 2.?1 2.?95 3.3825 3,883 3.?4 3,829 The relationship between the average petrol price and the number of onllne shopping orders In a given month ls proposed to follow the simple regression equation
below: show variables Q = be + ng
Calculate the proportion of variability in the number of onllne shopping orders that Is not explained by the average petrol prloe. Give your answer as a percentage to
1 decimal plaoe. Proportion - 20 % :| Feedback [u out of 1] 111ls is not correct.
Proportion = 14.6% Calculation The statistic R2 represents the proportion of variability in the observed response variable (y) that is explained by the explanatory variable (x) in the regression
model. If R2 is very high, the relationship between the response variables and the explanatory variables is very strong and vice versa. 111erefore, the proportion of variability that is not explained by the price of petrol in the model is the complement of Ft2 and can be mlculatecl by subtracting R2
from 1 (or 100%). Using a soltware package, R1 = 035415439... Calculating the Coefﬁcient of Determination {hide} The ooeﬁicient oi determination (R2) an be calmlated using the following formula: hide variables coefﬁcient of determination
regression sum of squares
SST = total sum of squares ﬂ
3 7E:
II II SSM
2 = —
R SST where: hide variable; = SSM 2 7
R SST Where: hide variables ‘ it"I observation of y — ith predicted value for y
- mean of the response
number of observations = 10 = 7,022,866.25359069... n
SST = z (yi-VF
i = 1
8,221,914.1 Therefore, SSM
SST 7,022,866.25359059. ..
B,221,914.1 = 0.85416439... R2: Therefore, n A
SSM = E (Yi- I'I
SST = z (yi-Y)2 g Therefore, 2 SSM
R SST 7[022[866.25359069. .. 3,221,914.1
0.85416439... Therefore, 1 - R2 1 — 0.85416439... = 0.14583561... Multiply by 100 to convert to a percentage
= 1458356086... 14.6% Rounded as a last step Proportion 2| 1 of :1 ID: usr.stn.e£E.o4.oo1o
111e pricing department of Vopstra needs to determine how much it should charge for its new smell phones in order to maximize sales revenue. A dmple linear regesslon equation has been oonstruaed co redraent the relationship between the price of a mobilenphone (x) and the annual sales revenue earned from those phones {v}. The simple linear regression equation was found
to be: vi = 15,000 + 500xi. This equation was based on a sample ofsize 500 where the minimum price in the sample was $30 and the maximum price in the sample was $300.
Select whether the following conduslons drawn from this simple linear regression Equation are valid or invalid: Valid Invalid a) If the pricing department decides to sell the smell phones For $150 each, Vopstra's predicted annual revenue is $90,000. of - b) If the pricing department decides to sell me smell phones for $800 each, Vopstra's predicted annual revenue is $415,000. J > c) If the pricing department decides to give away the smell phones for free, Uopstra's predicted annual revenue is $15,000. J - Tl Feedback [3 out of 3] Legend
J You are correct. Simple linear regression allows for the prediction of values of a dependent variable based on values of an independent variable. These predctlons are made using
a regrsﬁion equation whid’l summarizes the relationship between the two variables. In a simple linear reqresa'on model the relationship between the two
variables is represented using a stralgit line. The simple linear regression model used or vopstra involved the relationship between the price of mobile phones 0:) and the annual revenues that would be
received from those phonﬁ (v). This relationship was estimated by the equation: vi = 15,000 + 500xi. It is possible to prelid: values of v by substituting Vllll‘l values of XI into this equation. A valid value of xi is one which rails within the range of values in the random sample that was used to construct the reqmsion
equation. The regesslon equatlon describing the relationship between R and P was ccnstrurxed based on sample data with values of xi between $30 and $300.
Therefore, a valid prediction of v can only be obtained using a value of xi between $30 and $300. Faedaaclr [3 out of 3] Legend
J You are correct. D's: ussinn Simple linear regression allows for the prediction or values of a dependent variable based on values or an independent variable. These predictions are made using
a regression equation which summarizes the relationship between the two variables. In a simple linear regression model the relationship between the two variables is represented using a straight line. The simple linear regression model used by Vopstra involved the relationship between the price or mobile ohona [x] and the annual revenua that would be
received from those phones (v). ‘lhis relationship was estimated by the equation: vi - 15,000 + 500m. It is possible to predict values of v by substituting valid
values of x. into this equation. A valid value of x. is one which falls within the range of values in the random sample that was used to construct the regression
equation. The regression equation describing the relationship between R and P was constructed based on sample data with values oi xi between $30 and $300. Therefore, a valid predi...

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- Fall '13
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