gaokao 2017 Q19

gaokao · China · national-II-arts 12 marks Chi-squared test of independence
(12 points)
To compare the yields of an old and a new breeding method, a survey was conducted on 100 aquaculture farms, recording the yield (in kg) of a certain aquatic product. The frequency distribution histogram is given.
(1) Let $A$ denote the event ``the yield using the old breeding method is less than 50 kg''. Estimate the probability of $A$.
(2) Complete the contingency table below, and use the chi-squared test to determine whether we can be 99\% confident that yield is related to breeding method.
Yield $< 50 \text{ kg}$Yield $\geq 50 \text{ kg}$
Old breeding method
New breeding method

$P(K^2 \geq k)$0.0500.0100.001
$k$3.8416.63510.828

$$K^2 = \frac{n(ad - bc)^2}{(a+b)(c+d)(a+c)(b+d)}$$
(12 points)

To compare the yields of an old and a new breeding method, a survey was conducted on 100 aquaculture farms, recording the yield (in kg) of a certain aquatic product. The frequency distribution histogram is given.

(1) Let $A$ denote the event ``the yield using the old breeding method is less than 50 kg''. Estimate the probability of $A$.

(2) Complete the contingency table below, and use the chi-squared test to determine whether we can be 99\% confident that yield is related to breeding method.

\begin{center}
\begin{tabular}{ | l | l | l | }
\hline
 & Yield $< 50 \text{ kg}$ & Yield $\geq 50 \text{ kg}$ \\
\hline
Old breeding method & & \\
\hline
New breeding method & & \\
\hline
\end{tabular}
\end{center}

\begin{center}
\begin{tabular}{ c | l l r }
$P(K^2 \geq k)$ & 0.050 & 0.010 & 0.001 \\
\hline
$k$ & 3.841 & 6.635 & 10.828 \\
\hline
\end{tabular}
\end{center}

$$K^2 = \frac{n(ad - bc)^2}{(a+b)(c+d)(a+c)(b+d)}$$