Answer to Question #226073 in Statistics and Probability for Carrie

Question

Consider a multinomial experiment involving n D 300 and that the observed frequencies are given

in the following table:

Cell 1 2 3 4

Frequency 76 100 76 48

The investigator wants to test the null hypothesis: H0 : p1 D 0:3; p2 D 0:3; p3 D 0:2; p4 D 0:2:

The test statistic is

1. 302

2. 6:0786

3. 0:9876

4. 1

5. 9:9556

Answer 1

$H_0: p_1= 0.3, \; p_2= 0.3, \; p_3=0.2, \; p_4 = 0.2$

The test used here is a chi-square goodness-of-fit test.

The alternative hypothesis H₁ is defined as at least one is not equal to its specified value.

The expected cell count for each cell is given by $e_i=np_i$ , where n=300 the number of trials and p_i is the probability associated with that cell.

Thus, the expected values are as follows:

Thus, the above-obtained table gives the expected frequencies.

The chi-square goodness-of-fit test statistic is

$χ^2 = \sum^k_{i=1} \frac{(f_i-e_i)^2}{e_i}$

Thus, the value of the test statistic is derived as follows:

Thus, the value of the test statistic is $χ^2 = 9.9556$