Answer to Question #226075 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

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} Cell & 1 & 2 & 3 & 4 \\ \hline Frequency & 76 & 100 & 76 & 48 \\ \hdashline \end{array}$$

The null hypothesis to be tested is as follows:

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

We seek a test statistic that will detect a lack of fit of the observed cell counts to our hypothesized (null) expected cell counts based on the hypothesized cell probabilities.

These expected values are:

$$E(n_1)=np_1=300(0.3)=90$$

$$E(n_2)=np_2=300(0.3)=90$$

$$E(n_3)=np_3=300(0.2)=60$$

$$E(n_4)=np_4=300(0.2)=60$$

$$\chi^2=\displaystyle\sum_{all\ cells}\dfrac{(observed-expected)^2}{expected}=\displaystyle\sum_{i}\dfrac{(n_i-E(n_i))^2}{E(n_i)}$$

$$\chi^2=\dfrac{(76-90)^2}{90}+\dfrac{(100-90)^2}{90}+\dfrac{(76-60)^2}{60}+\dfrac{(48-60)^2}{60}$$

$$\chi^2=\dfrac{196}{90}+\dfrac{100}{90}+\dfrac{256}{60}+\dfrac{144}{60}$$

$$\chi^2=\dfrac{448}{45}$$

$$\chi^2\approx9.9556$$

5. $9.9556$