# Question 10 (Hypothesis Testing):

## Part (unlabelled - main hypothesis test):

$H_0: \mu = \frac{35}{6}$ | B1 | AO 1.1 |

$H_1: \mu > \frac{35}{6}$ where $\mu$ = pop mean no. of 2's | B1 | AO 2.5 | **B1B0** one error e.g. undefined $\mu$ or two-tail |

$N(\frac{35}{6}, \frac{175}{36})$ or $N(5.833, 4.861)$ soi | M1 | AO 3.3 | Allow incorrect variance |

$P(X \geq 10) = 1 - P(X < 9.5)$ or $P(X \geq 11) = 1 - P(X < 10.5)$ | M1 | AO 1.1a | $\geq 1$ of these probabilities attempted | $P(X < 9.5)$ or $P(X < 10.5)$ |

$P(X \geq 10) = 0.048$ | A0 | AO 2.1 | BC | $P(X < 9.5) = 0.952$ |

$P(X \geq 11) = 0.017$ (0.04 lies between these hence) | A1 | AO 3.4 | BC | $P(X < 10.5) = 0.983$ (0.96 between these) |

Rejection region is $X \geq 11$; Special case using N~Bin Method B | A1 | AO 2.2a | dep $\geq$ one of above probs seen & correct | rej'n region is $X \geq 11$ |

**Method B:**

$H_0: \mu = \frac{35}{6}$ | B1 | AO 1.1 |

$H_1: \mu > \frac{35}{6}$ where $\mu$ = pop mean no. of 2's | B1 | AO 2.5 | **B1B0** one error e.g. undefined $\mu$ or two-tail |

$N(\frac{35}{6}, \frac{175}{36})$ or $N(5.833, 4.861)$ soi | M1 | AO 3.3 | Allow incorrect variance |

$P(X > a) = 0.04$ soi | M1 | AO 1.1a | $z = \Phi^{-1}(0.96)$ $(= 1.751)$ |

$\frac{35}{6} + 1.751 \times \sqrt{\frac{175}{36}}$ | A1 | AO 2.1 | dep $\Phi^{-1}(0.96)$ attempt. May be implied **BC** |

$= 9.69$ or $9.7$ | A0 | AO 3.4 |

Rejection region is $X \geq 11$ | A1 | AO 2.2a | [7] |

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## Part (i):

Only 784 trees and $810 > 784$ | E1 [1] | AO 2.4 | or other similar |

## Part (ii):

e.g. Each no. not independent of previous no. Each no. is related to the next | E1 [1] | AO 2.3 | Allow 2nd digit of each no. is 1st of next; Consecutive nos share two digits; or similar correct; Digits are re-used; Ignore all else |

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## Part (iii):

$H_0: \mu = 4.2$ | B1 | AO 1.1 | Allow other letters except $X$ or $\bar{X}$ |

$H_1: \mu < 4.2$ where $\mu$ is mean height of trees (in the wood) | B1 | AO 2.5 | One error e.g. undefined $\mu$ or 2-tail: B0B1 |

$\bar{X} \sim N(4.2, \frac{0.8^2}{50})$ and $\bar{X} < 4.0$ or $\bar{X} \leq 4.0$ | M1 | AO 3.3 | Stated or implied; Allow $\bar{X} > 4.0$ or $\bar{X} = 4.0$ | $\Phi^{-1}(0.98) = 2.054$; $4.2 - 2.054 \times \frac{0.8}{\sqrt{50}} = 3.968$ |

$P(\bar{X} < 4.0) = 0.038549...$ or $0.039$ | A1 | AO 3.4 | **BC** Allow 0.038; NB 0.038... implies M1A1 | comp their 3.968 with 4.0 |

Compare $0.02$ | A1 | AO 1.1 | dep $P(\bar{X} < 4.0)$ attempted |

Do not reject $H_0$ | M1 | AO 2.2b | Allow Accept $H_0$; dep $P(\bar{X} < 4.0)$ attempted | Can be implied by conclusion |

There is insufficient evidence that mean height of these trees in the wood is less than 4.2m. | A1f | AO 3.5a | In context, not definite; e.g. "Mean height not less than 4.2m": A0 | [7] |

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