Consider a situation in which we expect one-third of the observed values to be in each of three categories. We can use an ?²goodness-of-fit test to test whether the frequencies of offspring are as expected. If the numbers of values in each category are 13, 18, and 29, and using the table of critical values shown, what is the conclusion of our test?
?
$\begin{array}{c}\chi^{2} \text { critical values }\\\text { for } d f=2\\\begin{array}{ll}\text { Sig. level } & \text { Value } \\\hline 0.05 & 5.99 \\0.025 & 7.38 \\0.01 & 9.21\end{array}\end{array}$

Question

Consider a situation in which we expect one-third of the observed values to be in each of three categories. We can use an ?²goodness-of-fit test to test whether the frequencies of offspring are as expected. If the numbers of values in each category are 13, 18, and 29, and using the table of critical values shown, what is the conclusion of our test?
?

\begin{array}{c}\chi^{2} \text { critical values }\\\text { for } d f=2\\\begin{array}{ll}\text { Sig. level } & \text { Value } \\\hline 0.05 & 5.99 \\0.025 & 7.38 \\0.01 & 9.21\end{array}\end{array}

Quizplus · Accepted Answer

To determine the conclusion of the test, we first calculate the expected frequencies and then the chi-square statistic. With 60 total observations (13+18+29) and expecting one-third in each category, the expected frequency for each category is 20. The chi-square statistic is calculated as $\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$, where $O_i$ and $E_i$ are the observed and expected frequencies, respectively. Substituting the given values, we get $\chi^2 = \frac{(13-20)^2}{20} + \frac{(18-20)^2}{20} + \frac{(29-20)^2}{20} = \frac{49}{20} + \frac{4}{20} + \frac{81}{20} = 6.7$. This value exceeds the critical value of 5.99 at the 0.05 significance level, indicating we reject the null hypothesis and conclude there is good evidence to suggest the frequencies differ from those expected.

Consider a Situation in Which We Expect One-Third of the Observed