Let x = boiler steam pressure in 100 lb/ in 2 and let y = critical sheer strengt
ID: 3250546 • Letter: L
Question
Let x = boiler steam pressure in 100 lb/ in2 and let y = critical sheer strength of boiler plate steel joints in tons/ in2. We have the following data for a series of factory boilers.
(a) Make the logarithmic transformations x' = log (x) and y' = log (y). Then make a scatter plot of the (x', y') values. Does a linear equation seem to be a good fit to this plot?
The transformed data does not fit a straight line well. The data seem to explode as x increases.The transformed data does not fit a straight line well. The data seem to have a parabolic shape. The transformed data does not fit a straight line well. The data seem to explode as x decreases.The transformed data fit a straight line well.
(b) Use the (x', y') data points and a calculator with regression keys to find the least-squares equation y' = a + bx'. What is the correlation coefficient? (Use 3 decimal places.)
(c) Use the results of part (b) to find estimates for and in the power law y = x. Write the power equation for the relationship between steam pressure and sheer strength of boiler plate steel. (Use 3 decimal places.)
Explanation / Answer
a) The transformed data fit a straight line well.
Find XY and X2 as it was done in the table below.
Step 2: Find the sum of every column:
X=3.98 , Y=4.11 , XY=3.432 , X2=3.2684
Step 3: Use the following equations to find a and b:
a=(YX2XXY)/(nX2(X)2)= (4.113.26843.983.432)/(53.26843.982)0.451
b=(nXYXY)/(nX2(X)2)=(53.4323.984.11)/(53.2684(3.98)2) 1.599
Step 4: Substitute a and b in regression equation formula
y = a + bx
y = 0.451 + 1.599x
correlation coefficient
r = (nXYXY)/[nX2(X)2][nY2(Y)2]
= 53.4323.984.11)/[53.26843.982][53.64134.112] 0.988
X Y X' = logx Y' = log y 4 3.4 0.60206 0.531479 5 4.2 0.69897 0.623249 6 6.3 0.778151 0.799341 8 10.9 0.90309 1.037426 10 13.3 1 1.123852Related Questions
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