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

Answer:

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

Step-by-step 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:

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

- 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.