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Dmitry_Shevchenko [17]
3 years ago
9

The data is given as follow. xi 2 6 9 13 20 yi 7 18 9 26 23 The estimated regression equation for these data is = 7.6 + .9x. Com

pute SSE, SST, and SSR (to 1 decimal). SSE SST SSR What percentage of the total sum of squares can be accounted for by the estimated regression equation (to 1 decimal)? % What is the value of the sample correlation coefficient (to 3 decimals)?

Mathematics
1 answer:
Tanzania [10]3 years ago
7 0

Answer:

✓SSE = 127.3

✓SST = 281.2

✓SSR = 153.9

✓percentage of the total sum of squares = 54.73%

✓The value of the sample correlation coefficient= 0.7398

Step-by-step explanation:

Given:

xi :2 6 9 13 20

yi:7 18 9 26 23

The given regression equation = 7.6 + 0.9 x.

we need to calculate the required SSE, SST and SSR , after this The percentage of total sum of squares can be computed. To do this we need to find the square difference between the actual value of y and average value of y, also the difference between actual value and predicted y value.

CHECK THE ATTACHMENT FOR DETAILED EXPLANATION.

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Derivative of the denominator:
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Hmm our numerator is 2x+7. Ok this let's us know that a simple u-substitution is NOT going to work. But let's apply some clever Algebra to the numerator splitting it up into two separate fractions. Split the +7 into +2 and +5.

\rm \int \dfrac{x^2+2+5}{x^2+2x+5}~dx

and then split the fraction,

\rm \int \dfrac{x^2+2}{x^2+2x+5}~dx+\int\dfrac{5}{x^2+2x+5}~dx

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\rm \int \dfrac{1}{u}~du+\int\dfrac{5}{x^2+2x+5}~dx

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\rm ln|x^2+2x+5|+\int\dfrac{5}{x^2+2x+5}~dx

Other one is a little tricky. We'll need to complete the square on the denominator. After that it will look very similar to our arctangent integral so perhaps we can just match it up to the identity.

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So we have this going on,

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Let's factor the 5 out of the intergral,
and the 4 from the denominator,

\rm ln|x^2+2x+5|+\frac54\int\dfrac{1}{\frac{(x+1)^2}{2^2}+1}~dx

Bringing all that stuff together as a single square,

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simplying a lil bit,

\rm ln|x^2+2x+5|+\frac52\int\dfrac{1}{u^2+1}~du

and hopefully from this point you recognize your arctangent integral,

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undo your substitution as a final step,
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Hope that helps!
Lemme know if any steps were too confusing.

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