A1-pound bag of candy-coated chocolate-covered peanuts contained 224 pieces of candy, each colored brown, orange, green, or yellow. test the null hypothesis that the machine filling these bags treats the four colors of candy equally likely; that is, test h0 : pb = po = pg = py = 1 4 the observed values were 42 brown, 64 orange, 53 green, and 65 yellow candies. you may select the significance level or give an approximate p-value.

The colours are equally likely is the conclusion.

Step-by-step explanation:

Given that a 1-pound bag of candy-coated chocolate-covered peanuts contained 224 pieces of candy, each colored brown, orange, green, or yellow.

(Two tailed chi square test)

We assume H0 to be true and find the expected frequencies.  Then calculate chi square and check the p value

OrangwBrownGreen  Yellow  

Observed64425365224

Expected56565656224

O-E8-14-39  

chisqurae1.1428571433.50.1607142861.4464285716.25

df3    

p value=0.1000

let us assume alpha is 5%

since p value is >5% we accept H0

The colours are equally likely is the conclusion.    

b is the final answer! : )

see explanation

step-by-step explanation:

since the sequence is geometric there is a common ratio r between consecutive terms, that is

r = 4 ÷ - 2 = - 2

r = - 8 ÷ 4 = - 2

r = 16 ÷ 8 = - 2

the next term - 13 must be incorrect as it should be - 32 for the sequence to be geometric.

the first 4 terms are therefore geometric with common ratio r = - 2