Happy Lodge Ski Resorts tries to forecast monthly attendance. The management has
ID: 3300886 • Letter: H
Question
Happy Lodge Ski Resorts tries to forecast monthly attendance. The management has noticed a direct relationship between the average monthly temperature and attendance.
Month
Average
Temperature
Resort Attendance
(in thousands)
5
40
28
Given five months of average monthly temperatures and corresponding monthly attendance, compute a linear regression equation of the relationship between the two. Round your answer to 2 decimal places, the tolerance is +/-0.01. For negative amounts use a negative sign preceding the number eg -45.
Resort attendance =
+
(average temperature)
If next month’s average temperature is forecast to be 45 degrees, use your linear regression equation to develop a forecast. (To achieve this answer do not round your interim calculations. Round your answer to 1 decimal place, the tolerance is +/-0.1.)
Resort attendance =
thousand attendees.
Compute a correlation coefficient for the data and determine the strength of the linear relationship between average temperature and attendance. How good a predictor is temperature for attendance? (Round your answer to 2 decimal places, the tolerance is +/-0.01.)
The correlation coefficient is
It indicates that the average temperature
predictor of resort attendance. (Do not round your intermediate computations to answer this question.)
Month
Average
Temperature
Resort Attendance
(in thousands)
5
40
28
Explanation / Answer
Q1. Line of Regression Y on X i.e Y = bo + b1 X
calculation procedure for regression
mean of X = X / n = 31.6
mean of Y = Y / n = 35.6
(Xi - Mean)^2 = 153.2
(Yi - Mean)^2 = 155.2
(Xi-Mean)*(Yi-Mean) = -142.8
b1 = (Xi-Mean)*(Yi-Mean) / (Xi - Mean)^2
= -142.8 / 153.2
= -0.93
bo = Y / n - b1 * X / n
bo = 35.6 - -0.93*31.6 = 65.05
value of regression equation is, Y = bo + b1 X
Y'=65.05-0.93* X
Resort attendance = 65.05 - 0.93(average temperature)
Q2.
when next month’s average temperature is forecast to be 45 degrees
=> Y'=65.05-0.93* (45) = 23.2
Q3.
calculation procedure for correlation
sum of (x) = x = 158
sum of (y) = y = 178
sum of (x^2)= x^2 = 5146
sum of (y^2)= y^2 = 6492
sum of (x*y)= x*y = 5482
to caluclate value of r( x,y) = covariance ( x,y ) / sd (x) * sd (y)
covariance ( x,y ) = [ x*y - N *(x/N) * (y/N) ]/n-1
= 5482 - [ 5 * (158/5) * (178/5) ]/5- 1
= -28.56
and now to calculate r( x,y) = -28.56/ (SQRT(1/5*5482-(1/5*158)^2) ) * ( SQRT(1/5*5482-(1/5*178)^2)
=-28.56 / (5.5353*5.5714)
=-0.9261
value of correlation is =-0.9261
coeffcient of determination = r^2 = 0.8576
properties of correlation
1. If r = 1 Corrlation is called Perfect Positive Corrlelation
2. If r = -1 Correlation is called Perfect Negative Correlation
3. If r = 0 Correlation is called Zero Correlation
& with above we conclude that correlation ( r ) is = -0.9262< 0, strong nagative correlation
X Y (Xi - Mean)^2 (Yi - Mean)^2 (Xi-Mean)*(Yi-Mean) 24 45 57.76 88.36 -71.44 28 35 12.96 0.36 2.16 31 37 0.36 1.96 -0.84 35 33 11.56 6.76 -8.84 40 28 70.56 57.76 -63.84Related Questions
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