Standard +0.3 This is a standard chi-squared test of independence with a 2×3 contingency table. Students must calculate expected frequencies using row/column totals, compute the test statistic, compare to critical values, and interpret at different significance levels. While it requires multiple steps and careful arithmetic, it follows a completely routine procedure taught explicitly in S3 with no novel problem-solving or insight required. Slightly easier than average due to its formulaic nature.
A random sample of 200 people were asked which hot drink they preferred from tea, coffee and hot chocolate. The results are given below.
\cline { 3 - 6 }
\multicolumn{2}{|c|}{}
\multirow{2}{*}{Total}
\cline { 3 - 5 }
\multicolumn{2}{|c|}{}
Tea
Coffee
Hot Chocolate
\multirow{2}{*}{Gender}
Males
57
26
11
94
\cline { 2 - 6 }
Females
42
47
17
106
Total
99
73
28
200
Test, at the \(5 \%\) significance level, whether or not there is an association between type of drink preferred and gender. State your hypotheses and show your working clearly. You should state your expected frequencies to 2 decimal places.
State what difference using a \(0.5 \%\) significance level would make to your conclusion. Give a reason for your answer.
Conclude there is an association between type of drink preferred and gender (or they are not independent)
A1
Based on their \(X^2\) and their \(\chi^2\) critical value. Must mention "drink" and "gender"/"sex". Contradictory statements score A0. Full accuracy: \(X^2 = 8.911619...\), p-value 0.0116–0.0117
Conclude there is no association between type of drink preferred and gender (or they are independent); or the conclusion would change (if correct \(H_0\) was rejected in part (a))
B1
Any one of these
# Question 5:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0$: There is no association between type of drink and gender; $H_1$: There is an association between type of drink and gender | B1 | Must mention "drink" and "gender"/"sex". Use of "relationship"/"correlation"/"connection" is B0 |
| Expected frequencies: Male: Tea 46.53, Coffee 34.31, Hot Choc 13.16; Female: Tea 52.47, Coffee 38.69, Hot Choc 14.84 | M1, A1 | $\frac{(\text{Row Total})(\text{Column Total})}{\text{Grand Total}}$; all expected frequencies correct |
| At least 2 correct terms for $\frac{(O-E)^2}{E}$ or $\frac{O^2}{E}$: 2.3559, 2.0127, 0.3545, 2.0892, 1.7849, 0.3144 | dM1 | Dependent on 1st M1; accept 2 s.f. accuracy |
| All 6 terms correct to 2 d.p. or better | A1 | Allow truncation |
| $X^2 = \sum\frac{(O-E)^2}{E}$ or $\sum\frac{O^2}{E} - 200\ ;= 8.9116...$ | dM1; A1 | 8.9 or awrt (8.88–8.91). If 8.9 seen without expected frequencies: special case M0A0M1A1M1A1 |
| $\nu = (2-1)(3-1) = 2$ | B1 | Implied by correct critical value of 5.991 |
| $\chi^2_2(0.05) = 5.991 \Rightarrow$ CR: $X^2 \geqslant 5.991$ | B1ft | |
| Conclude there is an association between type of drink preferred and gender (or they are not independent) | A1 | Based on **their** $X^2$ and **their** $\chi^2$ critical value. Must mention "drink" and "gender"/"sex". Contradictory statements score A0. Full accuracy: $X^2 = 8.911619...$, p-value 0.0116–0.0117 |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\chi^2_2(0.005) = 10.597 \Rightarrow$ CR: $X^2 \geqslant 10.597$ | B1 | Critical value of 10.597 |
| Conclude there is no association between type of drink preferred and gender (or they are independent); **or** the conclusion would change (if correct $H_0$ was rejected in part (a)) | B1 | Any one of these |
\begin{enumerate}
\item A random sample of 200 people were asked which hot drink they preferred from tea, coffee and hot chocolate. The results are given below.
\end{enumerate}
\begin{center}
\begin{tabular}{ | l | l | c | c | c | c | }
\cline { 3 - 6 }
\multicolumn{2}{|c|}{} & \multicolumn{3}{|c|}{\begin{tabular}{ c }
Type of drink \\
preferred \\
\end{tabular}} & \multirow{2}{*}{Total} \\
\cline { 3 - 5 }
\multicolumn{2}{|c|}{} & Tea & Coffee & Hot Chocolate & \\
\hline
\multirow{2}{*}{Gender} & Males & 57 & 26 & 11 & 94 \\
\cline { 2 - 6 }
& Females & 42 & 47 & 17 & 106 \\
\hline
\multicolumn{2}{|l|}{Total} & 99 & 73 & 28 & 200 \\
\hline
\end{tabular}
\end{center}
(a) Test, at the $5 \%$ significance level, whether or not there is an association between type of drink preferred and gender. State your hypotheses and show your working clearly. You should state your expected frequencies to 2 decimal places.\\
(b) State what difference using a $0.5 \%$ significance level would make to your conclusion. Give a reason for your answer.\\
\hfill \mbox{\textit{Edexcel S3 2014 Q5 [12]}}