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, 5 lbs of Almonds and 4 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 chi-squared test (also chi-square or χ2 test) is a statistical hypothesis test that is valid to perform when the test statistic is chi-squared distributed under the null hypothesis, specifically Pearson's chi-squared test and variants thereof. Pearson's chi-squared test is used to determine whether there is a statistically significant difference between the expected frequencies and the observed frequencies in one or more categories of a contingency table.

The data is given as follow:

The experimental weight of cashews = 40 % of 20lbs

= 8

The experimental weight of Brazil Nuts = 15 % of 20lbs

= 3

The experimental weight of Almonds = 20 % of 20lbs

= 4

The experimental weight of peanuts = 25 % of 20lbs

= 5

The observed value is :

Cashews = 5
Brazil Nuts= 6
Almonds= 5
Peanuts= 4

So, [tex]{\displaystyle \chi }[/tex] =[tex]\frac{(O-E)^{2}}{E}[/tex]

[tex]{\displaystyle \chi }[/tex]= [tex]\frac{(5-8)^{2}}{8} + \frac{(6-3)^{2}}{3} + \frac{(5-4)^{2}}{4} +\frac{(4-5)^{2}}{5}[/tex]

= 4.575

pvalue = p([tex]{\displaystyle \chi }^{2}[/tex] > 4.575) = [tex]{\displaystyle \chi }^{2}[/tex] cdf(4.575, 99999, 3)

= 0.208 >α

Hence, Fail to reject.

Learn ore about chi square test here:

https://brainly.com/question/16865619?referrer=searchResults

#SPJ2