A recent national poll showed the reading habits of adults during the last 12 months: $$\begin{array} { l r }
3 \text { or more books } & 5 \
2 \text { boxlss } & 27 \
1 \text { book } & 38 \
\text { No hooks } & 30
\end{array}$$ A poll in a small Midwestern town shows the following percentages:
$$\begin{array} { l r }
3 \text { or more broks } & 1 \
2 \text { books } & 22 \
1 \text { boxk } & 39 \
\text { No hooks } & 38
\end{array}$$ The observed chi-squared statistic is ______.

Question

A recent national poll showed the reading habits of adults during the last 12 months: $$\begin{array} { l r } 
3 \text { or more books } & 5 \
2 \text { boxlss } & 27 \
1 \text { book } & 38 \
\text { No hooks } & 30
\end{array}$$ A poll in a small Midwestern town shows the following percentages:
$$\begin{array} { l r } 
3 \text { or more broks } & 1 \
2 \text { books } & 22 \
1 \text { boxk } & 39 \
\text { No hooks } & 38
\end{array}$$ The observed chi-squared statistic is ______.

Quizplus · Accepted Answer

The chi-squared statistic is calculated using the formula $\chi^2 = \sum \frac{(O - E)^2}{E}$, where $O$ is the observed frequency and $E$ is the expected frequency. The expected frequencies are based on the national poll percentages applied to the total number of respondents in the small Midwestern town. Assuming the town's poll had 100 respondents for simplicity, the expected frequencies (E) for each category would be calculated based on the national percentages, and the observed frequencies (O) are given by the town's poll. The calculation for each category would look something like this:- For "3 or more books": $E = 100 \times 0.05 = 5$, $O = 1$, so $\frac{(1 - 5)^2}{5} = \frac{16}{5}$.- For "2 books": $E = 100 \times 0.27 = 27$, $O = 22$, so $\frac{(22 - 27)^2}{27} = \frac{25}{27}$.- For "1 book": $E = 100 \times 0.38 = 38$, $O = 39$, so $\frac{(39 - 38)^2}{38} = \frac{1}{38}$.- For "No books": $E = 100 \times 0.30 = 30$, $O = 38$, so $\frac{(38 - 30)^2}{30} = \frac{64}{30}$.Adding these values together gives the total chi-squared statistic, which matches the value in option D, 6.28.