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.
Since the equation is Y= 80+6.2 *X where X is the advertising cost, and since the equation which is given was calculated from the sample of which advertising was in $100. So the advertising value will be $100k here.
So, now the sales is calculated as:
Y= 80+6.2*100 = $700k.
Now since the sales is given in $1000 hence the correct sale will be 700*1000 = 700,100.
2. Regression analysis was applied between sales (in $1,000) and advertising (in $100), and the following...
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
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 Ÿ=500+ 3x
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
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 (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
Exhibit 14-3 Regression analysis was applied between sales data (in $1000s) and advertising data (in $100s), and the following information was obtained. 121.8x n 17 SSR 225 SSE 75 Sb 0.2683 Refer to Exhibit 14-3. The critical F value at a .05 is O a. 4.54 b. 3.68 О с. 3.59 d. 4.45
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