The Blue Diamond Company advertises that their nut mix contains (by weight) 40%

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, 5 lbs of Brazil nuts, 8 lbs of Almonds and 2 lbs of Peanuts. At the 0.05 level of significance, is the claim made by Blue Diamond true?
Select the [p-value, Decision to Reject (RH0) or Failure to Reject (FRH0)].  

a) [p-value = 0.004, RH0]

b) [p-value = 0.020, RH0]

c) [p-value = 0.041, RH0]

d) [p-value = 0.004, FRH0]

e) [p-value = 0.041, FRH0]

Explanation / Answer

Solving the above chi-squared test in R:

chisq.test(c(5,5,8,2),p=c(.40,.15,.20,.25))

Chi-squared test for given probabilities

data: c(5, 5, 8, 2)

X-squared = 8.2583, df = 3, p-value = 0.04096

If the p-value is less than 0.05, we reject the null hypothesis.

Option c is correct.

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