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. 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 a = 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. 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:
1. Almond: 19
2. Cashew: 1.59
3. Macadamia: 2.5g
4. Peanut: 0.59

Answer 1

Check the explanation

Explanation:

p=12/20=0.6

H_0:p=0.4,H_a:p>0.4

`t^*`=
`\frac{} \frac{0.6-0.4}/{\sqrt{(0.6*0.4)/(20)}`=1.83

pvalue=P(t>1.83,19dof)=0.042

Hence, we reject H0 and say the proportion of peanuts is more than 40%

b)E[weight]=0.2*1+0.3*1.5+0.1*2.5+0.4*0.5=1.1

V[weight]=0.2(1-1.1)^2+0.3(1.5-1.1)^2+0.1(2.5-1.1)^2+0.4(0.5-1.1)^2=0.39

SD[weight]=sqrt(0.39)=0.6245

`P(\bar{X}<1)=P(Z<(1-1.1)/(0.6245/√(100))`=-1.6013)=0.0547=5.47\%