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.

类似问题

Emily tosses a fair die three times. The probability she gets an even number at least 2 times is

State-wide surveys indicates that 14.5% of toddlers in Ney York are obese. Let the random variable X be the number of toddlers who are obese in a random sample of 20 toddlers from the state of New York. Is this a binomial scenario?

已知 G(n, p) 模型，观察到 m 个边的概率呈： Given the G(n, p) model, the probability of observing m edges the

Let 𝑋 equal the number of tomato seeds out of 𝑛 = 48 seeds that will germinate when 𝑝 = 0.75 is the probability that a particular seed germinates. Approximate 𝑃 ( 35 ≤ 𝑋 ≤ 40 ) . Choose the correct answer below.

What does the following code calculate? dbinom(4, 10, 0.5)

中奖的概率是0.000001（即百万分之一）。你每周买一张票。即使你坚持十年，也有可能永远赢不了？

The proportion of individuals that regularly smoke in a population is known to be 0.07. What is the probability that a random sample of 100 individuals drawn from the population will have 10 or less smokers?

更多留学生实用工具

智能学习助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单