please use rstudio and provide all code 1.4 Broadway Box Office Let X and Y deno
ID: 3177040 • Letter: P
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please use rstudio and provide all code
1.4 Broadway Box Office Let X and Y denote the weekly reports on the box office ticket sales for plays on Broadway in New York for two consecutive weeks, respectively, in October 2004. (You can actually download similar data from www.playbill.com.) The regression output for this data set is shown in the table below: Variable Coefficient t-value p-value S.e 0.685 9929 0.503 Intercept 6805 0.9821 68.071 2 x 10 16 0.01443 n 18 R 0.9966 s 18007.56 Suppose that the model satisfies the usual SLR model assumptions, and that the SST for Y is 1.507773 x 10 (i) What were the degrees of freedom used in calculating se? What are the SSE and SSR? (ii) Compute the sample variance for Y S and sample correlation between X and Y (rxY) (iii) Suppose that the ticket sales in the first week for a particular play was $822,000. What is the expected sales for the same play in the following week? (iv) Suppose further that X 622186.6 and sx 302724.5. Construct the 95% forecast interval for the estimate in (iii) (v) Construct the 95% confidence interval for the slope of the true regression line B1Explanation / Answer
Answer to the question)
Answer to part (i)
The sample size n = 18
thus degree of freedom (df) = n - 1
df = 18 - 1
df = 17
.
We know that :
R square = SSR / SST
we got :
R square = 0.9966
SST = 1.507773 x 10^12
.
On pluggging these values in the formula we get:
0.9966 = SSR / 1.507773 x 10^12
SSR = 0.9966 * 1.507773 x 10^12
SSR = 1.5026 x 10^12
.
SSE = SST - SSR
SSE = 1.507773 x 10^12 - 1.5026 x 10^12
SSE = (1.507773 - 1.5026) x 10^12
SSE = 0.005173 x 10^12
SSE = 5.173 x 10^9
.
Answer to part (ii)
Sample Variance of Y (Syy) = SST / n
Sample Variance = 1.507773 x 10^12 / 18
Sample Variance = 8.3765 x 10^10
.
Sample correlation r = Square root of (R square)
Sample correlation r = 0.9966
Sample correlation r = 0.998299
.
Answer to part (iii)
The regression equation is :
Y =6805 + 0.9821 * x
Where:
6805 is the intercept value obtained from the "coefficient column"
0.9821 is the slope value obtained from the "coefficient column"
Y = sales of following week
X = sales of previous week
.
We got x = 822000
Thus we plug in the value of x, to get the value of Y
Y = 6805 + 0.9821 * 822000
Y = 6805 + 807286.2
Y = 814091.2
.
Answer to part (iv)
x bar = 622186.6
Sx = 302724.5
Confidence level = 95%
df = n -2
df = 17
T value is = 2.110
.
Thus confidence interval is: x bar - t * SE , x bar + t*SE
on plugging the values we get
622186.6 - 2.11*302724.5 , 622186.6 + 2.11*302724.5
-16562.1 , 1260935
.
Answer to part (v)
The formula of slope's confidence interval is:
Slope - t* SE , Slope +t *SE
t for 95% confidence interval , df = 17, is 2.110
Thus t = 2.110
.
Slope = 0.9821
SE = 0.01443
.
On plugging in the values we get:
0.9821 - 2.11 *0.01443 , 0.9821 + 2.11*0.01443
0.9821 - 0.0304473 , 0.9821 +0.0304473
0.9516527 , 1.0125473
.
Answer to part (vi)
No we cannot assume the two weeks to have equal sales, as per the model that fits in the data in the best possible manner
we got r square = 0.9966 which shows that the two values are not same , but strongly associated
.
Answer to part (vii)
Even if Y and X are revered the value of R square would remain the same
The value of r or R square is same for (x and y) and (y and x)
Thus R square = 0.9966
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