Edexcel S3 — Question 4 11 marks

Exam BoardEdexcel
ModuleS3 (Statistics 3)
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChi-squared test of independence
TypeStandard 2×3 contingency table
DifficultyStandard +0.3 This is a standard chi-squared test of independence with a 2×3 contingency table. Students must state hypotheses, calculate expected frequencies, compute the test statistic, and compare to critical values—all routine procedures for S3. The calculation is straightforward with no conceptual complications, making it slightly easier than average.
Spec5.06a Chi-squared: contingency tables

4. A group of 40 males and 40 females were asked which of three "Reality TV" shows they liked most - Watched, Stranded or One-2-Win. The results were as follows:
\cline { 2 - 4 } \multicolumn{1}{c|}{}WatchedStrandedOne-2-Win
Males21613
Females151015
Stating your hypotheses clearly, test at the \(10 \%\) level whether or not there is a significant difference in the preferences of males and females.

Expected freq. males/watched \(= \frac{36 \times 40}{80} = 18\)
Males/stranded \(= \frac{16 \times 40}{80} = 8\)
AnswerMarks Guidance
Giving expected freqs: 18, 8, 14 and 18, 8, 14M1 A2
\(H_0\): no difference in preference of males and females; \(H_1\): difference in preference of males and femalesA1
OE (O-E)
2118 3
68 -2
1314 -1
1518 -3
108 2
1514 1
\(\therefore \sum \frac{(O-E)^2}{E} = 2.143\)M1 A2
\(\nu = 2, \chi^2_{\text{crit}}(10\%) = 4.605\)M1 A1
\(2.143 < 4.605 \therefore\) not significant. There is no evidence of a difference in preference of males and femalesA1 Total: 11 marks
Expected freq. males/watched $= \frac{36 \times 40}{80} = 18$

Males/stranded $= \frac{16 \times 40}{80} = 8$

Giving expected freqs: 18, 8, 14 and 18, 8, 14 | M1 A2 |

$H_0$: no difference in preference of males and females; $H_1$: difference in preference of males and females | A1 |

| O | E | (O-E) | $\frac{(O-E)^2}{E}$ |
|---|---|-------|-------------------|
| 21 | 18 | 3 | 0.5 |
| 6 | 8 | -2 | 0.5 |
| 13 | 14 | -1 | 0.0714 |
| 15 | 18 | -3 | 0.5 |
| 10 | 8 | 2 | 0.5 |
| 15 | 14 | 1 | 0.0714 |

$\therefore \sum \frac{(O-E)^2}{E} = 2.143$ | M1 A2 |

$\nu = 2, \chi^2_{\text{crit}}(10\%) = 4.605$ | M1 A1 |

$2.143 < 4.605 \therefore$ not significant. There is no evidence of a difference in preference of males and females | A1 | Total: 11 marks
4. A group of 40 males and 40 females were asked which of three "Reality TV" shows they liked most - Watched, Stranded or One-2-Win. The results were as follows:

\begin{center}
\begin{tabular}{ | c | c | c | c | }
\cline { 2 - 4 }
\multicolumn{1}{c|}{} & Watched & Stranded & One-2-Win \\
\hline
Males & 21 & 6 & 13 \\
\hline
Females & 15 & 10 & 15 \\
\hline
\end{tabular}
\end{center}

Stating your hypotheses clearly, test at the $10 \%$ level whether or not there is a significant difference in the preferences of males and females.\\

\hfill \mbox{\textit{Edexcel S3  Q4 [11]}}