Question

Chloe opens another smaller package from the same company, which is also supposedly 20% almonds (A), 30% cashews (C), 10% macadamias (M), and 40% peanuts. Out of 20 nuts, she find that there are 12 peanuts.

a. (0.5 points) Chloe wants to directly calculate the exact p-value of this occurring! (i.e. she does not want to do a proportion test). Please perform this test for Chloe at α = 0.05. Show/explain how to set up this problem, and you may use a table or calculator to get the answer.

Note: We can not use normal or chi squared approximate tests to calculate exact p-values

b. (1.5 point) Suppose the company’s claim about its distribution of nuts is true. Given the following weights, use the central limit theorem to calculate the (approximate) probability that a random sample of 100 nuts has a sample mean that weighs less than 1g:

Almond: 1g

Cashew: 1.5g

Macadamia: 2.5g

Peanut: 0.5g

Answer 1

(a) Here number of nuts = n = 20

P(Peanuts) = 0.40

Here if x is the number of peanuts out of 20.

x ~ BINOMDIST(n = 20, p = 0.40)

P- value = P(x 12) = 1 - BINOMDIST(11, 20, 0.4, true) = 1- 0.9435 = 0.0565

so exact p value = 0.0565

(b) Here as given

Almond: 1g

Cashew: 1.5g

Macadamia: 2.5g

Peanut: 0.5g

20% almonds (A), 30% cashews (C), 10% macadamias (M), and 40% peanuts.

Average of 100 nuts = 1 * 0.2 + 1.5 * 0.3 + 2.5 * 0.1 + 0.5 * 0.4 = 1.1 gm

Variance of 1 random nut = E[x2] - E[x]2

E[x2] = 1 * 1 * 0.2 + 1.5 * 1.5 * 0.3 + 2.5 * 2.5 * 0.1 + 0.5 * 0.5 * 0.4 = 1.6

VaR[x] = 1.6 - 1.1 * 1.1 = 0.39

Variance of 100 nuts weight = 0.39 * 100 = 39

variance of average of 100 nuts = 0.06245

so here we have to find

P( < 1.0 g) = NORMDIST( < 1.0 g; 1.1 gm; 0.06245 g)

z = (1 - 1.1)/0.06245 = -1.6013

P( < 1.0 g) = NORMSDIST(-1.6013) = 0.0547

probability that a random sample of 100 nuts has a sample mean that weighs less than 1g is 0.0547