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

Respuesta :

Answer:

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.

good job