Consider a multinomial experiment with n = 360 and k = 3 The null hypothesis to

Question

Consider a multinomial experiment with n = 360 and k = 3 The null hypothesis to be tested is H_0: p_1 = 0.60, P_2 = 0.25, and P_3 = 0.15. The observed frequencies resulting from the experiment are Choose the appropriate alternative hypothesis. All population proportions differ from their hypothesized values. At least one of the population proportions differs from its hypothesized value. Calculate the value of the test statistic (Round intermediate calculations to 4 decimal places and your final answer to 2 decimal places.) Test statistic Approximate the p-value 0.050

Explanation / Answer

Here we have to test the hypothesis that,

H0 : p1 = 0.60, p2 = 0.25 and p3 = 0.15

H1 : At least one of the population proportions differ from its hypothesized value.

Assume alpha = level of significance = 1% = 0.01

The test statistic is,

X2 = (O-E)2 / E

where O is observed frequency and

E is expected frequency.

Here observed frequencies are 230, 80 and 50.

And the expected frequency we have to calculate from hypothesized proportions and n.

Given that n = 360

Expected frequency = n*p

For p1 = 0.6,

E = 360*0.6 = 216

For p2 = 0.25,

E = 360*0.25 = 90

For p3 = 0.15

E = 360*0.15 = 54

Therefore expected frequencies are 216, 90 and 54.

The complete table of observed and expected frequencies are :

The test statistic X2 = 2.3148

Assume alpha = 1% = 0.01

P-value we can find by using EXCEL.

syntax is :

=CHIDIST(x,deg_freedom)

where x is test statistic value.

deg_freedom = k-1 = 3-1 = 2

P-value = 0.3143

P-value > alpha

Accept H0 at 1% level of significance.

P-value > 0.100

Conclusion : There is sufficient evidence to say that p1 = 0.60, p2 = 0.12 and p3 = 0.15.

O E (O-E)^2/E 230 216 0.9074074 80 90 1.1111111 50 54 0.2962963 total 2.3148148

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