Question

During summer vacation, a babysitter gives some kids snack size bags of M&M's. She wonders if the candy color distribution in each snack bag matches the distribution of colors reported online that says 24% blue, 13% brown, 16% green, 20% orange, 13% red, and 14% yellow. She has the kids count what they observe in their snack bags. She does a test and the P-value= 0.024. She used a 5% level of significance.

(Do Not Perform The Tests! Just answer the following)

a) Write the null and alternative hypothesis

b) Is it right, left, or two tailed?

c) Is it z, t, or x square or f distribution?

d) What calculator function is used?

e) What decision would you make based on the P-value that resulted when I did the test? Use a level of significance of 5%.

Answer 1

a) Write the null and alternative hypothesis

null hypothesis:

Ho:p1=0.24 p2=0.13 p3=0.16 p4=0.20 p5=0.13 p6=0.14

alternative hypothesis

atleast one proportion differs

b) Is it right, left, or two tailed?

Right tail test

c) Is it z, t, or x square or f distribution?

chi sq distribution

d) What calculator function is used?

chisq.test in R studio

P-value= 0.024.

p<0.05

Reject Ho.

Accept Ha.

There is no sufficient statistical evidence at 5% level of significance to conclude that

candy color distribution in each snack bag matches does match the distribution of colors reported online .

That is

candy color distribution in each snack bag matches does not match the distribution of colors reported online .