'm having some difficulty with the following question:
The following data shows expenditures (in millions of dollars) and case (sales in millions) for 7 major soft drink brands. Calculate the correlation coefficient. Is this significant at the 5% level (ie, α=.05)? Construct the regression equation for this data. Decide how much RC Cola would make if they spend 643.8. Do the same for Canada Dry if they spend 13.8. Is it a good idea to use this regression equation for RC and Canada Dry? Why?
Brand Spending Sales
Coke 131.3 1929.2
Pepsi 92.4 1384.6
Diet Coke 60.4 811.4
Sprite 55.7 541.5
Dr Pepper 40.2 536.9
Mt Dew 29.0 535.6
7-Up 11.6 219.5
I made my table & came up with:
Ex=420.6
Ey=5958.7
Exy=500073.09
Exsq=35119.7
Eysq=7213830.96
I'm having trouble getting the correlation coefficient. I used the formula in my textbook:
r=nΣxy-(Σx)( Σy) / √n(Σx²)-(Σx)² * √n(Σy²)-(Σy)²
but I keep coming up with different answers (I've done it by hand & with my TI-84 plus (LinRegTTest) several times each). I've figured out b1 (14.42378282) for the slope equation but for b0, my answer keeps changing (I get -15.something & the "something" is what keeps changing every time I work the problem out).
Please help!
I used an online calculator and came up with this:
y = a + bx where:
a= -15.4
b= 14.4
r = 0.978
Your formula looks correct. It's easy to make errors with the calculations when doing these kinds of problems.
y = -15.4 + 14.4x
the correlation coefficient is correct because it is geomatric mean of two regression coefficient the second regression coefficient is 0.066428.
To calculate the correlation coefficient, you need to use the formula you mentioned:
r = (nΣxy - ΣxΣy) / √[(nΣx² - (Σx)²)(nΣy² - (Σy)²)]
Let's go through the steps together to calculate the correlation coefficient:
1. First, calculate the values needed for the formula:
- n: the number of data points (in this case, 7)
- Σxy: the sum of all the products of each data point's x-value multiplied by its y-value (in this case, Σxy = 500073.09)
- Σx: the sum of all x-values (in this case, Σx = 420.6)
- Σy: the sum of all y-values (in this case, Σy = 5958.7)
- Σx²: the sum of all squared x-values (in this case, Σx² = 35119.7)
- Σy²: the sum of all squared y-values (in this case, Σy² = 7213830.96)
2. Now, substitute these values into the formula and calculate the numerator and denominators separately:
Numerator = nΣxy - ΣxΣy = (7 * 500073.09) - (420.6 * 5958.7)
Denominator = √[(nΣx² - (Σx)²)(nΣy² - (Σy)²)] = √[(7 * 35119.7 - (420.6)²)(7 * 7213830.96 - (5958.7)²)]
3. Calculate each part of the denominator separately:
nΣx² - (Σx)² = 7 * 35119.7 - (420.6)²
nΣy² - (Σy)² = 7 * 7213830.96 - (5958.7)²
4. Substitute the calculated values back into the denominator and calculate the square root of the result:
Denominator = √[(nΣx² - (Σx)²)(nΣy² - (Σy)²)]
5. Finally, calculate the correlation coefficient by dividing the numerator by the denominator:
r = Numerator / Denominator
This should give you the correct value for the correlation coefficient. If you are still getting different answers, please double-check your calculations and make sure you are using the correct numbers for each step.
Once you have the correlation coefficient, you can check if it is significant at the 5% level (i.e., α = 0.05) by comparing it to the critical value for a two-tailed t-test with n-2 degrees of freedom.
As for constructing the regression equation, you already calculated the slope (b1) as 14.42378282. To find the intercept (b0), you need to use the following formula:
b0 = (Σy - b1Σx) / n
Substitute the values you already calculated, and you should get the correct value for b0.
To determine how much RC Cola and Canada Dry would make if they spend specific amounts, you can use the regression equation:
Y = b0 + b1X
Where Y represents sales and X represents spending. Plug in the specific spending amounts for each brand, and you will get the predicted sales value.
Whether it is a good idea to use this regression equation for RC Cola and Canada Dry depends on factors such as the significance of the correlation coefficient, the range of the data, and the assumptions of linear regression. If the correlation coefficient is significant and the data meets the assumptions of linear regression, then it may be reasonable to use the regression equation for predictions. However, further analysis and validation should be done before making any decisions based on the regression equation.