Question

You are working with the marketing team for a FMCG firm that produces shaving cream. The team believes that sales of some of the products are closely related to sales of other products. They want you to explore this in a little more depth for two products, SKU 123 and SKU 456. Unfortunately, all of the base sales data for these products has been destroyed. All that you have is the weekly summary data:
Data Mean Standard Deviation
SKU123 721 176
SKU456 1059 266
Part 1.
The marketing team believes the correlation of these items is 0.79.
What would the covariance need to be for the marketing team to be correct?
Write your answer as a number with at least 2 decimal places.
Part 2.
Now the marketing team wants to understand the potential weekly sales for these two products. Let the sales price for the two SKUs be 12.50, 7.75, respectively.
What is the expected weekly revenue?
Write your answer as a number with at least 2 decimal places.
Assume that marketing is correct and the correlation = 0.79.
What is the standard deviation of the weekly revenue?
Write your answer as a number with at least 2 decimal places.
Part 3.
Assuming the marketing team’s correlation of 0.79 is correct. What is the probability that weekly sales will be between 10,000 and 20,000 dollars?
Part 4.
Which of the following statements are true:__________.
Choose the correct answer.
The population mean is always greater than the sample mean.
Everything else being equal, the Confidence Interval for a sample increases with the number of observations (n) in the sample.
The t-distribution is used for small sample sizes instead of the Normal.
Holding all else equal, reducing the probability of a Type I error actually increases the probability of a Type II error.

Answer 1

Answer:

Part 1. The covariance needs to be 36,984.64 for the marketing team to be correct

Part 2. The expected weekly revenue = 17,219.75 Dollars

The standard deviation of the weekly revenue = 3014.93

Part 3. The probability that that the sales will be between 10,000 and 20,000 dollars is 0.8128

Part 4 The correct option is;

The t-distribution is used for small sample sizes instead of the Normal distribution

Explanation:

Part 1. The relationship between correlation and covariance is given as follows;

$Corr(x, y) = r_{xy}=\dfrac{Cov \left (x,y \right )}{S_{x}S_{y}}$

Where:

Corr(x, y) = The correlation between the items = 0.79

$S_x$ = The standard deviation of SKU123 = 176

$S_y$ = The standard deviation of SKU456 = 266

Cov (x, y) = Covariance between SKU123 and SKU456

Therefore we have;

$Cov(x, y) =S_x \times S_y \times CORR(X, Y)$

Which gives;

Cov (x, y) = 176 × 266 × 0.79 = 36,984.64

The covariance needs to be 36,984.64 for the marketing team to be correct

Part 2. Where we have the expected sales of the two SKUs are given as follows;

                    Mean sale  Price

SKU123         721             12.50

SKU456        1059           7.75

The expected weekly revenue = 721 × 12.50 + 1059 × 7.75 = $17,219.75

The standard deviation SD (aX + bY) of the weekly revenue is given by the equation

SD (aX + bY) = √(12.5²× 176² + 7.75²×266²) = √(9089782.25) = 3014.93

Part 3. Given the correlation of 0.79, therefore, we have

The z value at 10,000 is given as follows;

$Z=\dfrac{x-E(X) }{SD(aX + bY) }$

Where:

x = The observed test value = $10,000

E(X) = The expected sales = $17219.75

Therefore, we have;

Z = (10000 - 17219.75)/3014.93 = -2.395

From the z table, we have p:(z = -2.395) = 0.0084

Similarly, when x = $20,000 we have;

Z = (20000 - 17219.75)/3014.93 = 0.9222 which gives p:(z = 0.9222) = 0.8212

Therefore, at a correlation of 0.79, the probability that that the sales will be between 10,000 and 20,000 dollars is 0.8212 - 0.0084 = 0.8128 or 81.28%

Part 4 The Student's t-distribution is a probability statistics distribution  used for small sample size, therefore, the correct option is that the t-distribution is used for small sample sizes instead of the Normal distribution