Question

Engineers concerned about a tower's stability have done extensive studies of its increasing tilt. Measurements of the lean of the tower over time provide much useful information. The following table gives measurements for the years 1975 to 1987. The variable "lean" represents the difference between where a point on the tower would be if the tower were straight and where it actually is. The data are coded as tenths of a millimeter in excess of 2.9 meters, so that the 1975 lean, which was 2.9642 meters, appears in the table as 642. Only the last two digits of the year were entered into the computer.

Year	75	76	77	78	79	80	81	82	83	84	85	86	87
Lean	642	645	656	667	674	689	696	698	714	717	726	743	758

(a) Plot the data. Consider whether or not the trend in lean over time appears to be linear. (Do this on paper. Your instructor may ask you to turn in this graph.)

(b) What is the equation of the least-squares line? (Round your answers to three decimal places.)
y =  +  x

What percent of the variation in lean is explained by this line? (Round your answer to one decimal place.)
%

(c) Give a 99% confidence interval for the average rate of change (tenths of a millimeter per year) of the lean. (Round your answers to two decimal places.)
(  ,  )

Answer 1

a)

data appear to be linear

b)

X	Y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
75	642	36.000	2728.053	313.385
76	645	25.000	2423.669	246.154
77	656	16.000	1461.592	152.923
78	667	9.000	741.515	81.692
79	674	4.00	409.284	40.46
80	689	1.00	27.361	5.23
81	696	0.00	3.130	0.00
82	698	1.0000	14.2071	3.7692
83	714	4.000	390.822	39.538
84	717	9.000	518.438	68.308
85	726	16.000	1009.284	127.077
86	743	25.000	2378.438	243.846
87	758	36.000	4066.515	382.615

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	1053	9025	182.000	16172.31	1705.000
mean	81.00	694.23	SSxx	SSyy	SSxy

sample size ,   n =   13          
here, x̅ = Σx / n=   81.00   ,     ȳ = Σy/n =   694.23  
                  
SSxx =    Σ(x-x̅)² =    182.0000          
SSxy=   Σ(x-x̅)(y-ȳ) =   1705.0          
                  
estimated slope , ß1 = SSxy/SSxx =   1705.0   /   182.000   =   9.3681
                  
intercept,   ß0 = y̅-ß1* x̄ =   -64.5879          
                  
so, regression line is   Ŷ =   -64.588 +   9.368 *x

R² =    (Sxy)²/(Sx.Sy) =    0.988

98..8 percent of the variation in lean is explained by this line

c)

confidence interval for slope                  
α=   0.01              
t critical value=   t α/2 =    3.106   [excel function: =t.inv.2t(α/2,df) ]      
estimated std error of slope = Se/√Sxx =    4.26021   /√   182.00   =   0.316
                  
margin of error ,E= t*std error =    3.106   *   0.316   =   0.981
estimated slope , ß^ =    9.3681              
                  
                  
lower confidence limit = estimated slope - margin of error =   9.3681   -   0.981   =   8.39
upper confidence limit=estimated slope + margin of error =   9.3681   +   0.981   =   10.35