To test the claim that the characteristics occur with the same frequency, we can use the Chi-square goodness-of-fit test.
Here is the step-by-step solution:
1. State the Observed Frequencies:
The observed frequencies for the characteristics are:
[tex]\[
\text{Observed } O = \{28, 30, 45, 48, 39, 39\}
\][/tex]
2. Calculate the Expected Frequency:
Since we hypothesize that all characteristics occur with the same frequency, we first need to find the average (expected) frequency. The total number of observed frequencies is summed up:
[tex]\[
\text{Total} = 28 + 30 + 45 + 48 + 39 + 39 = 229
\][/tex]
There are 6 categories, so the expected frequency for each category is:
[tex]\[
E = \frac{\text{Total}}{6} = \frac{229}{6} \approx 38.167
\][/tex]
3. List the Expected Frequencies:
[tex]\[
\text{Expected } E = \{38.167, 38.167, 38.167, 38.167, 38.167, 38.167\}
\][/tex]
4. 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]
Plugging the observed and expected frequencies into the formula:
[tex]\[
\chi^2 = \sum \frac{(28 - 38.167)^2}{38.167} + \frac{(30 - 38.167)^2}{38.167} + \frac{(45 - 38.167)^2}{38.167} + \frac{(48 - 38.167)^2}{38.167} + \frac{(39 - 38.167)^2}{38.167} + \frac{(39 - 38.167)^2}{38.167}
\][/tex]
[tex]\[
\chi^2 = \frac{(28 - 38.167)^2}{38.167} + \frac{(30 - 38.167)^2}{38.167} + \frac{(45 - 38.167)^2}{38.167} + \frac{(48 - 38.167)^2}{38.167} + \frac{(39 - 38.167)^2}{38.167} + \frac{(39 - 38.167)^2}{38.167}
\][/tex]
[tex]\[
\chi^2 \approx 8.249
\][/tex]
5. Conclusion:
The calculated value of the test statistic is approximately [tex]\(8.249\)[/tex]. Therefore, the correct answer is:
[tex]\[
\boxed{\chi^2 = 8.249}
\][/tex]
