Question

Problem 8 Problem Statement Gazelle has developed 7 new items for her designer auto parts business, and she wants to do a quick marketing test. She recruits 200 subjects, and she presents the 7 items to each customer. She then records the number of items that they decide to purchase, ending up with this data: Sales Count Customers 15 59 46 26 13 3 200 6 Total Thus, there were 5 customers who purchased no items, 15 customers who purchased 1 item, 33 customers who purchased 2 items, and so on. Gazelle wonders if this data could be modelled using a binomial distribution, and using the observed data Gazelle estimates the probability of success as p- 0.48137 Part (a) Assuming a binomial model with a probability of success p- 0.48137 and a total of 200 ulate the expected counts for each customers, calc category Part (b) Calculate the chi-squared statistic using the expected counts from part (a) Part (c) How many degrees of freedom does this chi-squared statistic have? Part (d) Conduct a hypothesis test that a binomial distribution with prob- ability of success p 0.48137 provides a good fit to this data.

Answer 1

x	p	Oi	Ei	(Oi-Ei)^2/Ei
0	0.010093	5	2.018522	4.403822883
1	0.065573	15	13.11453	0.271071914
2	0.182585	33	36.51703	0.338733251
3	0.282446	59	56.48923	0.111596055
4	0.262154	46	52.43087	0.78877262
5	0.145992	26	29.19844	0.350362015
6	0.045168	13	9.033578	1.741558483
7	0.005989	3	1.197797	2.711592251
	1	200	200	10.71750947
			p-value	0.151424581

formulas in Excel

x	p	Oi	Ei	(Oi-Ei)^2/Ei
0	=BINOM.DIST(A2,7,0.48137,0)	5	=$C$13*B2	=(C2-D2)^2/D2
=1+A2	=BINOM.DIST(A3,7,0.48137,0)	15	=$C$13*B3	=(C3-D3)^2/D3
=1+A3	=BINOM.DIST(A4,7,0.48137,0)	33	=$C$13*B4	=(C4-D4)^2/D4
=1+A4	=BINOM.DIST(A5,7,0.48137,0)	59	=$C$13*B5	=(C5-D5)^2/D5
=1+A5	=BINOM.DIST(A6,7,0.48137,0)	46	=$C$13*B6	=(C6-D6)^2/D6
=1+A6	=BINOM.DIST(A7,7,0.48137,0)	26	=$C$13*B7	=(C7-D7)^2/D7
=1+A7	=BINOM.DIST(A8,7,0.48137,0)	13	=$C$13*B8	=(C8-D8)^2/D8
=1+A8	=BINOM.DIST(A9,7,0.48137,0)	3	=$C$13*B9	=(C9-D9)^2/D9
	=SUM(B2:B10)	=SUM(C2:C10)	=SUM(D2:D10)	=SUM(E2:E10)
			p-value	=CHISQ.DIST.RT(E13,7)

a)
Expected count are in column Ei
b)
TS = 10.7175
c)
df = r-1 = 7
d)
p-value = 0.1514
p-value > 0.05
hence we fail to reject the null hypothesis

we conclude that there is not sufficient evidence that data does not follow binomial distribution with p = 0.48137