Question

imports and exports a random sample of countries shows the value of u.s. exports and the u.s. imports for each country for a recent hundreds of millions of dollars. assume the independent variable is exports.
country | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11
exports | 34 | 23 | 54 | 67 | 35 | 32 | 66 | 7 | 44 | 39 | 18
imports | 40 | 24 | 10 | 33 | 45 | 63 | 48 | 35 | 43 | 12 | 25
send data to excel
the correlation coefficient for the data is r=0.026 and α=0.05. should regression analysis be done?
the regression analysis should not be done.
the regression analysis should be done.
find the equation of the regression line. round the coefficients to at least three decimal places.
y=a+bx
a =
b =
find y when x=20 hundred million dollars. round to one decimal place.
y =

Explanation:

Step1: Check regression validity

Given r=0.026 (weak correlation), α=0.05. Critical r for n=11 is ~0.602. |r| < critical r → no linear relationship. So regression should not be done.

Step2: Calculate mean of x and y

x (Exports): 34+23+54+67+35+32+66+7+44+39+18 = 419 → $\bar{x} = 419/11 ≈ 38.0909$
y (Imports): 40+24+10+33+45+63+48+35+43+12+25 = 378 → $\bar{y} = 378/11 ≈ 34.3636$

Step3: Compute slope b

$b = r \times \frac{s_y}{s_x}$. Using data, $s_x ≈ 19.44$, $s_y ≈ 17.53$ → $b = 0.026 \times (17.53/19.44) ≈ 0.013$

Step4: Compute intercept a

$a = \bar{y} - b\bar{x} ≈ 34.3636 - (0.013×38.0909) ≈ 30.735$

Step5: Predict y' at x=20

$y' = 30.735 + (0.013×20) = 30.735 + 0.26 ≈ 30.9$

Answer:

The regression analysis should not be done.
a = 30.735
b = 0.013
y' = 30.9