Suppose we want to test the color distribution claim on the M&M’s website that a bag of plain M&M’s is made up of 10% blue, 10% orange, 10% green, 20% red, 20% yellow, and 30% brown. We select a sample of 400 plain M&M’s and found the following: Color Blue Orange Green Red Yellow Brown Frequency 30 48 55 66 70 131 Is there evidence to doubt the color distribution claimed by the website? Use =0.05

Calculated χ² = 13.425

χ² (5,0.025) >14.45 and χ²(5,0.975) <1.24

The given data does not fall in the critical region so we accept H0 and conclude there is no evidence to doubt the color distribution claimed by the website.

Step-by-step explanation:

Color             Blue      Orange     Green    Red   Yellow    Brown

Frequency     30         48              55        66         70         131

Expected      40           40              40        80          80        120

H0:  The bag of plain M&Ms is made up of 10% blue, 10% orange, 10% green, 20% red, 20% yellow, and 30% brown

Ha: The color distribution is not equal to  the distribution stated in the null hypothesis.

Calculate chi square

χ² = (30-40)² /40 + (48-40)²/40 + (55-40)²/40 + (66-80)²/80 + (70-80)²/80 + (131-120)²/120

χ² = 2.5 + 1.6 + 5.625 + 2.45 + 1.25= 13.425

The critical region for χ²  for 5 degrees of freedom with ∝= 0.05 is

χ² (5,0.025) >14.45 and χ²(5,0.975) <1.24

The given data does not fall in the critical region so we accept H0 and conclude there is no evidence to doubt the color distribution claimed by the website.

36/60

36/2=18

60/2=30

18/30

18/3=6

30/3=10

6/10

6/2=3

10/2=5

your final answer would be 3/5 of an

hope this !

step-by-step explanation: