First, let's find the correlation coefficient (r) to determine if there's a linear relationship between the price and quality score. We can do this using the formula:
r = (nΣ(xy) - ΣxΣy) / √[(nΣ(x^2) - (Σx)^2) (nΣ(y^2) - (Σy)^2)]
Where x represents the price, y represents the quality score, and n is the number of data points.
Calculating the necessary values:
Σx = 17100
Σy = 528
Σ(xy) = 1202700
Σ(x^2) = 29123000
Σ(y^2) = 35424
n = 8
r = (8 * 1202700 - 17100 * 528) / √[(8 * 29123000 - 17100^2) * (8 * 35424 - 528^2)]
r = (9621600 - 9029440) / √[(232984000 - 292410000) * (283392 - 278784)]
r = (592160) / √[(-59626000) * (4608)]
r ≈ -0.91
Since r ≈ -0.91, which is close to -1, we can conclude that there is a strong negative linear relationship between the price and quality score.
Now we must find the line of regression (y = a + bx), using the formulas:
b = Σ(xy) - [(Σx)(Σy)]/n / (Σ(x^2)) - ((Σx)^2)/n
a = (Σy - bΣx) / n
Calculating b:
b = (1202700 - (17100 * 528)/8) / (29123000 - (17100)^2/8)
b = (1202700 - 225315) / (29123000 - 36610500/8)
b = (978385) / (29123000 - 4576312.5)
b ≈ -0.0448
Calculating a:
a = (Σy - bΣx) / n
a = (528 - (-0.0448 * 17100)) / 8
a = (528 + 765.84) / 8
a ≈ 64.23
So the line of regression is y = 64.23 - 0.0448x.
Now we can use this line to predict the quality score of a TV set with a price of $1900:
y = 64.23 - 0.0448 * 1900
y ≈ 64.23 - 85.12
y ≈ 57
Therefore, the predicted quality score for a TV set with a price of $1900 is approximately 57.
use the table of values to find the line of regression and, justified at the 0.05 significance level, use it to find the predicted quality score of a TV set with a price od $1900. if the data does not suggest linear correlation, the average quality score as a prediction. PRICE Q. Score
2300 74
1800 73
2500 70
2700 66
2000 63
1700 62
1500 52
2700 68
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