Question

Q.N. 7) According to the manufacturer of M& Ms, 12% of the peanut M&Ms in a bag should be brown, 15% yellow, 12% red, 23% blue, 23% orange, and 15% green. A student randomly selected a bag of peanut M&Ms. He counted the number of M&Ms that were each color and obtained the results shown in the table below. Test whether peanut M&Ms follow the distribution stated by M&M/Mars at the α = 0.05 level of significance.

Color Frequency

Brown 53

Yellow 66

Red 38

Blue 96

Orange 88

Green 59

Use R to solve and show R code

Answer 1

We know that given frequencies are known as observed frequencies.

Hence, total of frequency = N = 400.

Given that, probabilities of the color peanut.

Now we obtain expected frequency = (probabilities*observed frequency)

Now the hypothesis is,

H₀ : Peanut M&Ms follow the distribution stated by M&M/Mars

vs H₁ : peanut M&Ms does not follow the distribution stated by M&M/Mars

If O_i = observed frequency and e_i = expected frequency then the test statistic for testing above hypothesis is

Hence R code for the above problem is

R-code:

o=c(53,66,38,96,88,59) ## o = Observed Frequency

N=sum(o) ## N = Total Frequency

pr=c(0.12,0.15,0.12,0.23,0.23,0.15) ## pr = Given probabilities

e=N*pr ## e = Expected Frequency

chisq.test(x=o,p=pr) ## chisquare test for goodness of fit

## Now manual calculations

chi_test=sum(((o-e)^2)/e) ## chisquare test statistic

pvalue=pchisq(chi_test,df=5,lower.tail=FALSE) ## pvalue for test

print(c(chi_test,pvalue))

R code with output:

> o=c(53,66,38,96,88,59) ## o = Observed Frequency
> N=sum(o) ## N = Total Frequency
> pr=c(0.12,0.15,0.12,0.23,0.23,0.15) ## pr = Given probabilities
> e=N*pr ## e = Expected Frequency
> chisq.test(x=o,p=pr) ## chisquare test for goodness of fit

# Output : Chi-squared test for given probabilities

data: o
X-squared = 3.5687, df = 5, p-value = 0.613

> ## Now manual calculations
>
> chi_test=sum(((o-e)^2)/e) ## chisquare test statistic
> pvalue=pchisq(chi_test,df=5,lower.tail=FALSE) ## pvalue for test
> print(c(chi_test,pvalue))

# Output:
[1] 3.568659 0.613025

From the above output, we get

Chi-square test statistic = 3.5687

p-value = 0.613

Let level of significance = 0.05

Since p-value = 0.613 > level of significance = 0.05, hence we cannot reject the null hypothesis H₀ at 5% level of significance and conclude that peanut M&Ms follow the distribution stated by M&M/Mars at 5% level of significance.