Mars inc. claims that they produce m& ms with the following distributions:
brown 30% red 20% yellow 20%
orange 10% green 10% blue 10%
how many m& ms must be sampled to construct the 97% confidence interval for the proportion of red m& ms in that bag if we want a margin of error of ± .15?
a) 33
b) 36
c) 34
d) 26
e) 25
f) none of the above

C) 34

Step-by-step explanation:

1) Some definitions

By definition the margin of error (ME) is the error that tell to us how many percentage points your results will differ from the real population value, on this case our parameter of interest is  pr = proportion of red M&M's

ME = Critical value x Standard error of the sample =0.15.

The proportion of red M&M's follows a normal distribution, and our critical value would be from the normal standard distribution on this case

2) Calculate the critical value

a) Compute alpha (α): α = 1 - (confidence level / 100)  = 1- 0.97 = 0.03

b) Calculate the critical probability (p*): p* = 1 - α/2  = 1 - (0.03/2) = 0.985

c) Find the z-score using the cumulative probability obtained at step b)

On this case P(Z<z) = 0.985 , the value of z = 2.17 using the normal standard table

3) Calculate n from the formula of ME

The margin of error for a proportion is given by this formula

ME = z sqrt{{pr(1-pr)/n}}

Squaring both sides :

(ME/z) ^2  = (pr(1-pr))/n  

And solving for n we got

n = (pr(1-pr))/(ME/z)^2 = (0.2x0.8)/ (0.15/2.17)^2 = 33.488

We need to round up the sample in order to ensure that the confidence level of 97% is meeted, and on this case the answer would be 34.