Regression analysis was applied between sales (Y in $1000) and advertising (X in $10,000), and the following estimated regression equation was obtained
Based on the above estimated regression line if advertising is $10,000, then the point estimated for sales (in dollars) is
a.503
b. 5030
c. 50,300
d.503,000
Since advertising spend is $10000, so putting x = 1, in the equation, we get,
i.e. i.e. $503000 since Y is in $10000
Hence option D.
Regression analysis was applied between sales (Y in $1000) and advertising (X in $10,000), and the...
Regression analysis was applied between sales (in $1,000) and advertising (in $100), and the following regression function was obtained. ^y = 80 + 6.2x Based on the above-estimated regression line, if advertising is $10,000, then the point estimate for sales (in dollars) is $62,080 $142,000 $700 $700,000
2. Regression analysis was applied between sales (in $1,000) and advertising (in $100), and the following regression function was obtained Y=80+6.2x Based on the above estimated regression line, if advertising is $10,000, then the point estimate for sales (in dollars) is According to my quiz the answer is 700,000 , please help me figure out how.
Regression analysis was applied between sales (in AUD1000) and advertising (in AUD100) and the following regression function was obtained: [Y = 500 + 4 X]. Based on the above estimated regression line if advertising is $10,000, then the point estimate for sales (in dollars) isSelect one:a. AUD900b. AUD505,000c. AUD40,500d. AUD900,000
Regression analysis was applied between sales data (y in $1000s) and advertising data (x in $100s) and the following information was obtained. = 12 + 1.8x n = 17 SSR = 225 SSE = 75 sb1= .2683 The point estimate of the population slope β1 is 1.8. options: True or False
A regression analysis between sales (Y in $100) and advertising (X in dollars) resulted in the following equation = 30,000 + 6 X The above equation implies that an a. increase of $1 in advertising is associated with an increase of $600 in sales b. increase of $1 in advertising is associated with an increase of $6 in sales c. increase of $1 in advertising is associated with an increase of $6,000 in sales d. None of the above
#1 In simple linear regression, r is the: a) coefficient of determination. b) mean square error. c) correlation coefficient. d) squared residual. #2 In regression analysis, with the model in the form y = β0 + β1x + ε, x is the a) estimated regression equation. b) y-intercept. c) slope. d) independent variable. #3 A regression analysis between sales (y in $1,000s) and advertising (x in dollars) resulted in the following equation. ŷ = 40,000 + 3x The above equation...
b Multiple Choice 14-019 A regression analysis between sales u in $tooo) and advertising (x in dollars) resulted in the following equation: Y = 30,000 + 4x The above equation implies that an O a. increase of $1 in advertising is associated with an increase of $4 in sales. O b. increase of $1 in advertising is associated with an increase of $4000 in sales. O c. increase of $4 in advertising is associated with an increase of $4000 in...
Question 14 A regression analysis between sales (Yin $1,000) and advertising (X in dollars) resulted in the following equation Y = 30,000+ 4X. The equation implies that an: Increase of $4 in advertising is associated with an increase of $4.000 in sales Increase of $1 in advertising is associated with an increase of $4,000 in sales Increase of $1 in advertising is associated with an increase of $34.000 in sales - Previous Next → No new data to save. Last...
A regression analysis between sales (in $1000) and price (in dollars) resulted in the following equation y = 50,000-8x The above equation implies that an a increase of $1 in price is associated with a decrease of $42,000 in sales b increase of $1 in price is associated with a decrease of $8000 in sales c increase of Ss in price is associated with an increase of $8000 in sales d increase of $1 in price is associated with a decrease of $8 in sales
NARRBEGIN: Exhibit 12-04 Exhibit 14-4 Regression analysis was applied between sales data (Y in $1,000s) and advertising data (x in $100s) and the following information was obtained. Y 12+1.8x SSR 225 SSE 75 Se-0.2683 NARREND 80. Refer to Exhibit 14-4. To perform an F test, the p-value is a. less than.01 b. ibetween.01 and.025 c. between.025 and.05 d jbetween.05 and 0.1 PTS: 1 TOP: Regression Analysis ANS: D