Question

Imagine a basketball player named Shack. He is getting old, doesn’t run too well and has always been a poor free-throw shooter. He decides to work on the last of these as follows:Five times per day for the next 80 days he will shoot four free throws and count thenumber of successes that he achieves.Thus, Shack will collect n = 400 numerical values, with each value being one of: 0, 1, 2, 3 or 4.Shack wants to use the Goodness of Fit Test to test whether these 400 values behave as if they come from a binomial distribution. Note p is unknown.His O’s are below.Outcome01234Oi251181399325(1)State the null hypothesis and alternative hypothesis. (2)Estimateπ.(3)Fill the following table and carry out a multinomial 2test with =0.05 BY HAND. Remember to state your conclusions.

Outcome 0 1 2 3 4

Oi 25 118 139 93 25

Pi

Ei

Oi-Ei

(Oi-Ei)2

(Oi-Ei)2/Ei

Answer 1

1) The null and alternative hypothesis

H0 : The distribution of success Shack achieves follow Binomial distribution with n=4, p =0.5

Ha : The distribution of success is different from  Binomial distribution

2) As p is unknown we assume , p =0.5

3) Let X be the number of success

X follow Binomial with n=4 , p=05 ( according to null hypothesis )

We know that

Similarly we calculate other probabilities

X	P(X)	Ei	Oi	Oi-Ei	(Oi-Ei)^2	(Oi-Ei)^2/Ei
0	0.0625	25	25	0	0	0
1	0.25	100	118	18	324	3.24
2	0.375	150	139	-11	121	0.806666667
3	0.25	100	93	-7	49	0.49
4	0.0625	25	25	0	0	0
	1	400	400			4.536666667

Test statistic

= 4.54

degrees of freedom = 5-1 =4

P value =0.3378

Since P value > 0.05

The difference is not significant

We fail to reject H0

There is not sufficient evidence to conclude that the distribution of success is different from Binomial distribution.