Question

The hypothesis being tested is:

H0: Plain M&Ms follow the stated distribution

Ha: Plain M&Ms does not follow the stated distribution

	observed	expected	O - E	(O - E)² / E
Red	63	89.440	-26.440	7.816
Orange	155	137.600	17.400	2.200
Yellow	90	89.440	0.560	0.004
Green	101	137.600	-36.600	9.735
Blue	181	137.600	43.400	13.689
Brown	98	96.320	1.680	0.029
	688	688.000	0.000	33.473
	33.47	chi-square
	5	df
	3.03E-06	p-value

The p-value is 0.0000.

Since the p-value (0.0000) is less than the significance level (0.05), we can reject the null hypothesis.

Therefore, we can conclude that Plain M&Ms does not follow the stated distribution.

A different significance level would not have changed the conclusion.

3. Use the distribution below to test that Halloween Skittles follow the stated distribution. Discuss your choice of significance level. Would a different significance level have changed your conclusion?
Skittles states the following distribution:
Red = 18%, Orange = 22%, Green = 17%, Blue = 20%, Purple = 22%, Other (Zombie color) = 1%

Answer 1

observed frequencey, O	expected proportion	expected frequency,E	(O-E)		(O-E)²/E
63	0.180	123.84	-60.84		29.889
155	0.220	151.36	3.64		0.088
90	0.170	116.96	-26.96		6.214
101	0.200	137.60	-36.60		9.735
181	0.220	151.36	29.64		5.804
98	0.010	6.880	91.12		1206.810

chi square test statistic,X² = Σ(O-E)²/E =   1258.541              
                  
level of significance, α=   0.01              
Degree of freedom=k-1=   6   -   1   =   5
                  
P value =   0.0000   [ excel function: =chisq.dist.rt(test-stat,df) ]          
Decision: P value < α, Reject Ho                  

Therefore, we can conclude that Plain M&Ms does not follow the stated distribution.

A different significance level would not have changed the conclusion.