14 | 15 | 16 | 15 | |
Brown | 2 | 2 | 3 | 3 |
Yellow | 5 | 0 | 2 | 2 |
Red | 2 | 2 | 3 | 2 |
Orange | 2 | 1 | 1 | 0 |
Green | 1 | 5 | 2 | 2 |
Blue | 2 | 5 | 5 | 6 |
Using the data from your M&M’s, calculate the following confidence intervals with a 95% confidence,
a.) The proportion of Brown M&M’s in a bag.
b.) The proportion of Orange M&M’s in a bag.
c.) The proportion of Primary Colored (red, yellow, blue combined) M&M’s in a bag.
z score for 95% confidence interval is z = 1.96 (using z distribution table)
(A) Number of brown M&M's = 10
total number of M&M's = 14+15+16+15 = 60
proportion of brown M&M's = 10/60 = 0.167
Confidence interval =
this implies
(B) Number of brown M&M's = 4
total number of M&M's = 14+15+16+15 = 60
proportion of brown M&M's = 4/60 = 0.067
Confidence interval =
this implies
(C) Number of yellow, red and blue M&M's = 36
total number of M&M's = 14+15+16+15 = 60
proportion of yellow, red and blue M&M's = 36/60 = 0.6
Confidence interval =
this implies
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