* Construct a 95% Confidence Interval for the proportion of blue M&Ms’ candies.
Solution:
The confidence interval is given as p±z1-α2 p1-pn where, p represents the proportion of success=proportion of blue M&Ms’ candies=0.213944 n represents sample size=9194
Z1-α2for 95% confidence level=1.96 Therefore, the required Confidence Interval is given as p±z1-α2 p1-pn
=0.213944±1.960.2139441-0.2139449194
=0.213944±0.008383
=(0.205561, 0.222327)

* Construct a 95% Confidence Interval for the proportion of orange M&Ms’ candies.

Solution:
The confidence interval is given as p±z1-α2 p1-pn where, p represents the proportion of success=proportion of orange M&Ms’ candies=0.214596 n represents sample size=9194
Z1-α2for 95% confidence level=1.96 Therefore, the required Confidence Interval is given as p±z1-α2 p1-pn
=0.214596±1.960.2145961-0.2145969194
=0.214596±0.008392
=(0.206204, 0.222988)

* Construct a 95% Confidence Interval for the proportion of green M&Ms’ candies.

Solution:
The confidence interval is given as p±z1-α2 p1-pn where, p represents the proportion of success=proportion of green M&Ms’ candies=0.183163 n represents sample size=9194
Z1-α2for 95% confidence level=1.96 Therefore, the required Confidence Interval is given as p±z1-α2 p1-pn
=0.183163±1.960.1831631-0.1831639194
=0.183163±0.007907
=(0.175256, 0.821070)

* Construct a 95% Confidence Interval for the proportion of yellow M&Ms’ candies.

Solution:
The confidence interval is given as p±z1-α2 p1-pn where, p represents the proportion of success=proportion of yellow M&Ms’ candies=0.130629 n represents sample size=9194
Z1-α2for 95% confidence level=1.96 Therefore, the required Confidence Interval is