Question

Hi! I would like to receive help with the following question for Probabilities and Statistics, and I thank you in advance for your kind help!

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Mars Company claims that 10% of M&M candies in a random bag are green.

PART I: Suppose we choose randomly several samples of bags containing 50 M&Ms, and record the % of greens in each bag.

a) If we plot a histogram showing the proportions of green candies in the different bags, what shape would you expect? Can you approximate the histogram by a Normal model? Why or why not?

b) Where should the center of the histogram be? What should the standard deviation of the proportions be?\

PART II: Suppose we work with bags containing 200 M&Ms, and we compute the % of greens in each bag.

a) Why is it appropriate now to use a Normal model to describe the distribution of the proportion of greens?

b) Use the 68-95-99.7 Rule to describe his this proportion might vary from bag to bag.

c) How would this model change if the bags contained even more M&Ms?

PART III: In a really large bag of M&Ms, we found 12% of 500 M&Ms were green. Does this refute Mars’ claim of 10%? Explain.

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Thank you for your help! I would really appreciate it!

Answer 1

a)

here, p=0.10

n=50

X be number of candies that are green

X~binomial(n=50, p =0.10 )

if p=0.5, binomial distrbution is symmetrical.

since, p=0.10<0.50, so, shape of histogram is skewed to right.

here, np=5<10 ,  

so, we cannot approximate the histogram by a Normal model

b)

center of histogram,p=0.10

std dev = √p(1-p)/n = √(0.10*0.90/50)=0.0424

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part II

a)

now, np=200*0.10 = 20>10 ,

and n(1-p) = 200*0.90 =180>10

since, now both np and n(1-p)>10  

so, we can approximate the histogram by a Normal model

b)

center of histogram=0.10

Standard Error ,    SE = √( p(1-p)/n ) =    0.0212

according to empirical rule,

68% of observation of data lie witin 1 std dev away from mean,µ±1σ = (0.0789,0.1212)

95% of observation of data lie witin 2 std dev away from mean,µ±2σ =(0.0576,0.1424)

99.7 of observation of data lie witin 3 std dev away from mean, µ±3σ =(0.0364 , 0.1636)

c)

if the bags contained even more M&Ms , then model will approach more and more normal distibution , because std error will decrease,as sample size increase.

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part III

Ho :   p =    0.1      
H1 :   p ╪   0.1      
              
Level of Significance,   α =    0.05      
Sample Size,   n =    500      
              
Sample Proportion ,    p̂ =    0.12      
              
Standard Error ,    SE = √( p(1-p)/n ) =    0.0134   
              
Z Test Statistic =    Z = ( p̂-p)/SE =    1.4907   
              
  
p-Value   =   0.136037128 [from z table]
Conclusion:     p value>α ,do not reject null hypothesis

hence, there is no sufficient evidence to reject men's claim at α=0.05