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?

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.

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

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

CashewsBrazil nutsAlmondsPeanutsTotal

Observed566320

Expected  40%15%20%25%1

Expected  834520

Chi square1.125310.85.925

df = 3

p value = 0.1153

Since p value >0.0 we accept null hypothesis.

step-by-step explanation:

f

-7/8

step-by-step explanation:

instead of subtracting you add the -6/8 and -1/8