Question

3. A multinomial experiment with k-4 cells and n=201 produced the data shown in the one-way table to the right. Complete parts a through e. CELL 1 CELL 2 CELL 3 CELL 4 60 Do these data provide sufficient evidence to conclude that the multinomial probabilities differ? Test using alpha=0.10. What is the null and alternative hypothesis? II. Calculate the test statistic. (6) III. Calculate the p-value. IV. What is your conclusion?

Answer 1

I.

Null hypothesis: There is no significant difference between the observed proportion and the expected proportion i.e. p1=0.25, p2 = 0.25, p3 = 0.25, and p4 = 0.25

Alternative hypothesis: Some of the proportion differ significantly.

II.

The Chi-Square statistic is obtained using the formula,

(Observed-Expected) Expected

where the observed and expected frequencies are,

	Observed	Expected
	60	201*0.25=50.25
	55	201*0.25=50.25
	44	201*0.25=50.25
	42	201*0.25=50.25
Sum	201	201

Now,

Observed, $O_i$	Expected,	(O; - E;) Ei
60	50.25	(60 - 50.2552 - = 1.892 50,25
55	50.25	0.449
44	50.25	0.777
42	50.25	1.354
Sum		4.473

X2 = 4.473

III.

The P-value for the chi-square = 4.473 is obtained from the chi-square distribution table for degree of freedom = k - 1 = 4 - 1 = 3

P-value = 0.2147

IV.

P-value = 0.0000 < 0.10

Since the p-value for the observed chi-square statistic is less than 0.10 at a 10% significance level. the null hypothesis is not rejected. Hence the multinominal experiment follows the same probability.