Question

The Blue Diamond Company advertises that their nut mix contains (by weight) 40% cashews, 15% Brazil nuts, 20% almonds and only 25% peanuts. The truth-in-advertising investigators took a random sample (of size 20 lbs) of the nut mix and found the distribution to be as follows:
5 lbs of Cashews,
6 lbs of Brazil nuts,
6 lbs of Almonds and 3 lbs of Peanuts.
A. At the 0.05 level of significance, is the claim made by Blue Diamond true?

Answer 1

Answer:

At the 0.05 level of significance,  the claim made by Blue Diamond is true

Step-by-step explanation:

Given that the Blue Diamond Company advertises that their nut mix contains (by weight) 40% cashews, 15% Brazil nuts, 20% almonds and only 25% peanuts.

$H_0: The mix are as per percents given\\H_a:Atleast two differ from the above percent.$

(Two tailed chi square test at 5% significance level)

Chi square is calculated as (Obs_exp)^2/Expected

Cashews Brazil nuts Almonds Peanuts Total

Observed 5 6 6 3 20

Expected  40% 15% 20% 25% 1

Expected  8 3 4 5 20

     

Chi square 1.125 3 1 0.8 5.925

df = 3

p value = 0.1153

Since p value >0.0 we accept null hypothesis.