A beverage company has developed a new low-calorie sports drink and is conducting taste tests to gauge consumer acceptance. Historically, 60% of consumers have favoured the company's sports drinks in taste tests. Due to the change in formula to reduce calories, there is concern that this new drink may not be as well-received as previous versions.

A study is conducted with 100 participants, which observed 56 that favoured the new low-calorie sports drink.

Hypotheses:
H0: The acceptance rate for the new low-calorie sports drink is 60%

H1: The acceptance rate for the new low-calorie sports drink is less than 60%

The box model is below:

box_model = c(1, 1, 1, 1, 1, 1, 0, 0, 0, 0)

What R code produces the EV and SE of the sample sum?

Question

A beverage company has developed a new low-calorie sports drink and is conducting taste tests to gauge consumer acceptance. Historically, 60% of consumers have favoured the company's sports drinks in taste tests. Due to the change in formula to reduce calories, there is concern that this new drink may not be as well-received as previous versions.

A study is conducted with 100 participants, which observed 56 that favoured the new low-calorie sports drink.

Hypotheses:
H0: The acceptance rate for the new low-calorie sports drink is 60%

H1​: The acceptance rate for the new low-calorie sports drink is less than 60%

The box model is below:

box_model = c(1, 1, 1, 1, 1, 1, 0, 0, 0, 0)

What R code produces the EV and SE of the sample sum?

BlackTom AI · Accepted Answer

EV of a sum of n draws equals n times the box mean, and the SE of the sum equals sqrt(n) times the population standard deviation of the box (assuming independent draws / sampling with replacement).

Relevant concepts:
- Box model: values {1,1,1,1,1,1,0,0,0,0} has mean μ and population variance σ^2 computed over the box.
- EV(sum of n draws) = n * μ.
- Var(sum of n independent draws) = n * σ^2, so SE(sum) = sqrt(n) * σ.
- In R, use mean() for μ and popsd() (or compute the population sd manually) for σ; sd() computes the sample sd (dividing by n-1) and is not appropriate for a population box.

Step-by-step calculations:
1. Compute the box mean:
   mean(box_model) = (6 ones + 4 zeros) / 10 = 6/10 = 0.6.
2. Compute the population variance:
   For the box, sum of squared deviations = 6*(1-0.6)^2 + 4*(0-0.6)^2
                                 = 6*(0.4)^2 + 4*(0.6)^2
                                 = 6*0.16 + 4*0.36 = 0.96 + 1.44 = 2.40.
   Population variance σ^2 = 2.40 / 10 = 0.24.
   Population sd σ = sqrt(0.24) ≈ 0.4898979.
3. EV of the sample sum for n = 100:
   EV = 100 * 0.6 = 60.
4. SE of the sample sum:
   SE = sqrt(100) * 0.4898979 = 10 * 0.4898979 ≈ 4.898979.

Why the provided R code is correct:
- EV = 100 * mean(box_model) implements n * μ.
- SE = sqrt(100) * popsd(box_model) implements sqrt(n) * σ using the population sd.

Why alternatives are incorrect:
- Using sd(box_model) instead of popsd(box_model) yields the sample sd (dividing by n-1), producing a slightly larger SE (≈5.164) which misstates the box-model population variability.
- Omitting sqrt(n) or using n * popsd would not match Var(sum)=nσ^2.

Final confirmation: The correct R code is
EV = 100 * mean(box_model)
SE = sqrt(100) * popsd(box_model)
which matches the provided answer.

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