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?

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:

- 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.

(f - g)(x) = 5x^3 - 3

step-by-step explanation:

rearrange these functions vertically, with like terms in separate columns:

f(x) = 5x^3 - 2

g(x) =         +1

now subtract, column by column:

(f - g)(x) = 5x^3 - 3