Question

She tasted Dr. Pepper 3 times and each time she had a probability of liking Dr. Pepper, which was 0.75. Each taste was independent, and we said that the count of the number of likes followed a Binomial Distribution. I observed Tess drinking the Dr. Pepper 3 times over 100 trials and recorded the number of likes for each trial. I ended up with the frequencies shown below. Run a chi-square analysis that will test the goodness of fit to the binomial distribution. The expected frequencies under the null hypothesis are a binomial (That is, the expected counts should use the probabilities of the number of likes that are calculated using the binomial formula). Therefore, in order to get the expected probabilities, you must calculate, P(X=0), P(X=1), P(X=2), and P(X=3) using a binomial with n=3 and p=.75.

# of Likes	0	1	2	3
Count	5	15	25	55

Answer 1

Null Hypothesis :H_o : The given data fits to Binomial distribution

Alternate hypothesis:H_a : The given data does not fit to Binomial distribution

Test Statistic

O : Observed Count

E: Expected Count

Given, O :

# of Likes	0	1	2	3
Count	5	15	25	55

Binomial pmf for p = 0.75 and n =3

Expected Count = 100 x P(X)

E:

X:# of Likes	0	1	2	3
P(X)	0.0156	0.1406	0.4219	0.4219
E:Count = 100*P(X)	1.56	14.06	42.19	42.19

O	E	O-E	(O-E)2	(O-E)2/E
5	1.56	3.44	11.8336	7.5856
15	14.06	0.94	0.8836	0.0628
25	42.19	-17.19	295.4961	7.0039
55	42.19	12.81	164.0961	3.8895
			Total	18.5419

Test Statistic

Degrees of freedom = 4-1 =3

Assumes level of significance : = 0.05

As p-value : 0.00034 < 0.05 ; Reject the null hypothesis.

There is sufficient evidence to conclude  that the given data does not fit to Binomial distribution