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
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.