Edexcel S3 2018 June — Question 2 12 marks

Exam BoardEdexcel
ModuleS3 (Statistics 3)
Year2018
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChi-squared goodness of fit
TypeChi-squared goodness of fit: Binomial
DifficultyStandard +0.3 This is a standard chi-squared goodness of fit test for a binomial distribution with straightforward calculations. Part (a) is simple arithmetic (finding sample mean proportion), part (b) requires basic binomial probability calculations, and part (c) follows the standard chi-squared test procedure taught in S3. The question is slightly easier than average because the parameter is estimated from data (reducing degrees of freedom is taught), expected frequencies are partially given, and the procedure is routine with no conceptual challenges.
Spec5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.06b Fit prescribed distribution: chi-squared test

  1. A random sample of 75 packets of seeds is selected from a production line. Each packet contains 12 seeds. The seeds are planted and the number of seeds that germinate from each packet is recorded. The results are as follows.
Number of seeds that
germinate from each packet
6 or
fewer
789101112
Number of packets0351828174
  1. Show that the probability of a randomly selected seed from this sample germinating is 0.82 A gardener suggests that a binomial distribution can be used to model the number of seeds that germinate from a packet of 12 seeds. She uses a binomial distribution with the estimated probability 0.82 of a seed germinating. Some of the calculated expected frequencies are shown in the table below.
    Number of seeds that
    germinate from each packet
    6 or
    fewer
    789101112
    Expected frequency\(s\)2.807.97\(r\)22.0418.266.93
  2. Calculate the value of \(r\) and the value of \(s\), giving your answers correct to 2 decimal places.
  3. Test, at the \(10 \%\) level of significance, whether or not these data suggest that the binomial distribution is a suitable model for the number of seeds that germinate from a packet of 12 seeds. State your hypotheses clearly and show your working.

\begin{enumerate}
  \item A random sample of 75 packets of seeds is selected from a production line. Each packet contains 12 seeds. The seeds are planted and the number of seeds that germinate from each packet is recorded. The results are as follows.
\end{enumerate}

\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | }
\hline
\begin{tabular}{ l }
Number of seeds that \\
germinate from each packet \\
\end{tabular} & \begin{tabular}{ c }
6 or \\
fewer \\
\end{tabular} & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
Number of packets & 0 & 3 & 5 & 18 & 28 & 17 & 4 \\
\hline
\end{tabular}
\end{center}

(a) Show that the probability of a randomly selected seed from this sample germinating is 0.82

A gardener suggests that a binomial distribution can be used to model the number of seeds that germinate from a packet of 12 seeds.

She uses a binomial distribution with the estimated probability 0.82 of a seed germinating. Some of the calculated expected frequencies are shown in the table below.

\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | }
\hline
\begin{tabular}{ l }
Number of seeds that \\
germinate from each packet \\
\end{tabular} & \begin{tabular}{ c }
6 or \\
fewer \\
\end{tabular} & 7 & 8 & 9 & 10 & 11 & 12 \\
\hline
Expected frequency & $s$ & 2.80 & 7.97 & $r$ & 22.04 & 18.26 & 6.93 \\
\hline
\end{tabular}
\end{center}

(b) Calculate the value of $r$ and the value of $s$, giving your answers correct to 2 decimal places.\\
(c) Test, at the $10 \%$ level of significance, whether or not these data suggest that the binomial distribution is a suitable model for the number of seeds that germinate from a packet of 12 seeds. State your hypotheses clearly and show your working.

\hfill \mbox{\textit{Edexcel S3 2018 Q2 [12]}}