Question

**Assume that both Plain and Peanut M&M candies are normally distributed.

1. Suppose that M&M claims that their Plain M&Ms have an equal proportion of red and brown.

a. Test the claim that the proportion of red M&Ms is greater than the proportion of brown M&Ms.

b. Test the claim that the proportion of red M&Ms is less than the proportion of brown M&Ms.

Total number of M&M's: 665

Total number of red M&M's: 76

Total number of brown M&M's: 66

Answer 1

Clearly , it is proportion test .

a.

let's assume that P1 is the proportion of red M&Ms and P2 is the proportion of brown M&Ms .

Hypothesis :

Null hypothesis :

Alternative hypothesis :

Here,

sample size = n = 665

Let ,

Sample proportion of Red M&Ms =

Where ,

  number of red M&Ms in the sample .

also

Sample proportion of Brown M&Ms =

Now we use r-software for testing the hypothesis :

Take level of Significance = 5% = 0.05

> x=c(76,66)
> n=c(665,665)
> prop.test(x,n,conf.level=0.95,alternative="greater")

   2-sample test for equality of proportions with continuity correction

data: x out of n
X-squared = 0.6386, df = 1, p-value = 0.2121
alternative hypothesis: greater
95 percent confidence interval:
-0.01431477 1.00000000
sample estimates:
prop 1 prop 2
0.11428571 0.09924812

Here ,

Decision rule :

if p-value is less than level of significance then we reject null hypothesis otherwise accept null hypothesis .

Here ,

p-value = 0.2121

level of significance = 0.05

since 0.2121>0.05

we can say that , we cannot reject

Therefore , is accepted .

Therefore , Proportion of red M&Ms and brown M&Ms is same .

b.

let's assume that P1 is the proportion of red M&Ms and P2 is the proportion of brown M&Ms .

Hypothesis :

Null hypothesis :

Alternative hypothesis :

Here,

sample size = n = 665

Let ,

Sample proportion of Red M&Ms =

Where ,

  number of red M&Ms in the sample .

also

Sample proportion of Brown M&Ms =

Now we use r-software for testing the hypothesis :

Take level of Significance = 5% = 0.05

> x=c(76,66)
> n=c(665,665)
> prop.test(x,n,conf.level=0.95,alternative="less")

   2-sample test for equality of proportions with continuity correction

data: x out of n
X-squared = 0.6386, df = 1, p-value = 0.7879
alternative hypothesis: less
95 percent confidence interval:
-1.00000000 0.04438996
sample estimates:
prop 1 prop 2
0.11428571 0.09924812

Here ,

Decision rule :

if p-value is less than level of significance then we reject null hypothesis otherwise accept null hypothesis .

Here ,

p-value = 0.7879

level of significance = 0.05

since 0.7879>0.05

we can say that , we cannot reject

Therefore , is accepted .

Therefore , Proportion of red M&Ms and brown M&Ms is same .