Question

Mar’s, In. claims that its M&M candies are distributed with the color percentages of 20% brown, 20% yellow, 30% red, 10% orange, 10 % green, and 10% blue. A classroom exercise involving a random sample of 460 M&M’s resulted in the observed frequencies were: 90 brown, 94 yellow, 99 red, 64 orange, 51 green, and 62 blue. Test the claim that the color distribution is as claimed by Mars, Inc (α= 0.05). What can you conclude if the test statistic is greater than the critical value?

Answer 1

Null hypothesis is rejected. There is no or little evidence to support the claim made by Mars.In

Explanation:

Solution:-

- A claim is made by Mars.In that the M&M candies in a packet are distributed by color percentage given below.

- A random sample of N = 460 M&M's was taken and the frequencies of different colors were observed as given below.

- We are to test a claim made by the Mars.In regarding the color distribution of M&Ms at significance level α = 0.05.

- Compute the expected frequency of distribution for color as per Mars.In claim. Use the following formula for expected outcome:

Expected = N*pi

Where, pi : The percentages for each color.

- The table for expected and observed frequencies for each color is tabulated below.

Colors Percentage Expected Observed

Brown 20% 460*0.2 = 92 90

Yellow 20% 460*0.2 = 92 94

Red 30% 460*0.3 = 138 99

Orange 10% 460*0.1 = 46 64

Green 10% 460*0.1 = 46 51

Blue 10% 460*0.1 = 46 62

- To test the claim for color distribution of M&Ms using X^2 - test. We will state the Null and Alternate hypothesis as follows:

Null Hypothesis: The distribution is colors is as its claimed by the company

Alternate Hypothesis: The distribution is colors is not its claimed by the company

- We will first determine the X^2 statistics value using the following relation:

`X^2-test = Sum [ ((Oi- Ei)^2)/(Ei) ]\\\\X^2-test = ((90- 92)^2)/(92) + ((94- 92)^2)/(92) + ((99- 138)^2)/(138) + ((64- 46)^2)/(46) + ((51- 46)^2)/(46) + ((62- 46)^2)/(46) \\\\X^2-test = (4)/(92) + (4)/(92) + (1521)/(138) + (324)/(46) + (25)/(46) + (256)/(46) \\\\X^2-test = 0.04347 + 0.04347 + 11.0217 + 7.04348 + 0.54348 + 5.56522\\\\X^2-test = 24.2608`

- The rejection region is defined by the significance level ( α ) = 0.05 and degree of freedom. The bound ( critical value ), X^2-critical is determined using the look-up table:

degree of freedom = number of observation category - 1 = 6 - 1 = 5

P ( X < X^2 - critical ) = 0.025

X^2 - critical = 12.8

- All the test values of X^2 > X^2-critical lie in the rejection region.

24.2608 > 12.8

X^-test > X^2-critical

Hence, Null hypothesis is rejected.

Conclusion: As per Chi-square test the claim made by the Mars.In has little or no evidence to be true; hence, the color distribution of M&M is not what is claimed.