Given the 5 categories in the table below, test the claim that the categories are equally likely to be selected at a [tex]$\alpha =0.05$[/tex] significance level.

\begin{tabular}{|c|c|c|}
\hline
Category & Observed Frequency & Expected Frequency \\
\hline
A & 19 & 15.6 \\
\hline
B & 18 & 15.6 \\
\hline
C & 22 & 15.6 \\
\hline
D & 8 & 15.6 \\
\hline
E & 11 & 15.6 \\
\hline
\end{tabular}

a. Complete the table by calculating the expected frequencies.

b. What is the chi-square test statistic? Round to three decimal places.

[tex]x^2 = \square[/tex]

[tex]\square[/tex]



Answer :

To test the claim that the categories are equally likely to be selected at the [tex]\(\alpha = 0.05\)[/tex] significance level, we can perform a chi-square goodness-of-fit test.

### Step-by-step solution:

a. Complete the table by calculating the expected frequencies:

Given that there are five categories and we are testing that they are equally likely, we calculate the expected frequency for each category. The total observed frequency is the sum of the observed frequencies:

[tex]\[ 19 + 18 + 22 + 8 + 11 = 78 \][/tex]

Since there are 5 categories, the expected frequency for each category becomes:

[tex]\[ \frac{78}{5} = 15.6 \][/tex]

Thus, the expected frequencies for each category are (as filled in your table):

[tex]\[ \text{Expected Frequency (for each category)} = 15.6 \][/tex]

b. Calculate the chi-square test statistic:

The chi-square test statistic is calculated using the formula:

[tex]\[ \chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \][/tex]

where:
- [tex]\(O_i\)[/tex] = observed frequency for category [tex]\(i\)[/tex]
- [tex]\(E_i\)[/tex] = expected frequency for category [tex]\(i\)[/tex]

Let’s calculate this for each category:

1. Category A:
[tex]\[ O_1 = 19, \quad E_1 = 15.6 \][/tex]
[tex]\[ \frac{(19 - 15.6)^2}{15.6} = \frac{(3.4)^2}{15.6} \approx 0.742 \][/tex]

2. Category B:
[tex]\[ O_2 = 18, \quad E_2 = 15.6 \][/tex]
[tex]\[ \frac{(18 - 15.6)^2}{15.6} = \frac{(2.4)^2}{15.6} \approx 0.369 \][/tex]

3. Category C:
[tex]\[ O_3 = 22, \quad E_3 = 15.6 \][/tex]
[tex]\[ \frac{(22 - 15.6)^2}{15.6} = \frac{(6.4)^2}{15.6} \approx 2.631 \][/tex]

4. Category D:
[tex]\[ O_4 = 8, \quad E_4 = 15.6 \][/tex]
[tex]\[ \frac{(8 - 15.6)^2}{15.6} = \frac{(-7.6)^2}{15.6} \approx 3.705 \][/tex]

5. Category E:
[tex]\[ O_5 = 11, \quad E_5 = 15.6 \][/tex]
[tex]\[ \frac{(11 - 15.6)^2}{15.6} = \frac{(-4.6)^2}{15.6} \approx 1.347 \][/tex]

Summing these values gives the chi-square test statistic:

[tex]\[ \chi^2 \approx 0.742 + 0.369 + 2.631 + 3.705 + 1.347 = 8.795 \][/tex]

Thus, the chi-square test statistic is:

[tex]\[ \chi^2 = 8.795 \][/tex]

So, the expected frequencies for each category are 15.6, and the chi-square test statistic, rounded to three decimal places, is [tex]\(8.795\)[/tex].

[tex]\[ \boxed{8.795} \][/tex]