Answer to Question #154648 in Statistics and Probability for chi 1

Question

Consider a multinomial experiment involving

n = 150 trials and k = 4 cells. The observed frequencies

resulting from the experiment are shown

in the accompanying table, and the null hypothesis

to be tested is as follows:

H0: p1 = .3, p2 = .3, p3 = .2, p4 = .2

Cell 1 2 3 4

Frequency 38 50 38 24

Test the hypotheses, using  = .05.

Answer 1

Given a multinomial experiment involving n=150 trials and k=4 cells. For this the data is as follows

Cell 1 2 3 4

Frequency 38 50 38 24

The given null hypothesis is

H₀: p₁ = 0.3, p₂ = 0.3, p₃ = 0.2, p₄ = 0.2

Alternative hypothesis is

H₁: At least one p₁ is not equil to its specific value

The test statistic for the Chi-Square goodness-of-fit (χ²) is given by,

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

e_i = expected frequencies

f_i = observed frequencies

Given level of significance α=0.05

Using MINITAB, the following are the steps to get the χ²-statistic for the Account receivable data.

1) Click Stat, Table, and Chi-square Goodness-of-fit Test (One Variable)….

2) Type the observed values into the Observe Counts and specify the variable name.

3) Click Proportions specified by historical counts and Input constants type the values of the proportions under the null hypothesis (0.3, 0.3, 0.2, 0.2)

4) Click OK.

We get the following output

From the above output,

The test statistic χ²= 4.97778

The P-value is 0.173

Here we observe that, the P-value is greater than the level of the significance 0.05, so we fail to reject the null hypothesis. Therefore, there is enough evidence to support the claim that the given proportions are equal to the specified values.