Statistic Toewijzing 1

Section 1 A

Table 1

  | H0 | H1 | LB | UB | Conclusion |

Question apm>0.75? | pm≤ 0.75 | pm>0.75 | 0.7875 | 0.8725 | Reject H0. Since 0.75 lies outside the 95% confidence interval. Therefore with 95% we conclude that the proportion of the male heads of household among all heads is larger than 0.75. |

Section 1 B

Table 2:

  | H0 | H1 | Type of test | RR | Val | Conclusion |

Question bμ1- μ2>0? | μ1- μ2≤0 | μ1- μ2>0 | Unequal variance approach:T=X1- X2-h(S1n1+S2n2) | T ≥1.6499 | 3.6078 | Reject H0 since val does belong to the rejection region. |

2) We used a test for two independent samples, because the two samples are not related. Their data is not connected in any way.

3) We used unequal-variance approach, because the two variances are different.

S21 = 132.7906 and S22=118.8654

i. Test: H0: σ12σ22=1 against H1: σ12σ22≠1

ii. Test statistics: T=S12S22 =132.7906 / 118.8654 = 1.1172

iii. Reject H0 ↔ T≥t0.05,298=1.6499

iv. Val= 3.6078

v. Conclusion: Reject H0 since val does belong to the rejection region.

Section 1 C

BMI of adult heads of households |

| N | Minimum | Maximum | Mean | Std. Deviation |

BMI | 300 | 16,19 | 35,24 | 24,3877 | 3,21841 |

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Table 3:

| H0 | H1 | Type Test | RR | Val | Conclusion |

Question cµ(BMI)<25? | µ(BMI)≥25 | µ(BMI)<25 | t-distribution:T=X-S√n

| T ≤ -1.6499 | T= -3.2954 | Reject H0 since val belongs to the rejection region |

Question dσ2 (BMI male)≠11.2? | σ2 (BMI male)=11.2 | σ2 (BMI male) ≠11.2 | Chi-square distribution:W=(n-1)S2σ2 | W≤206,2736 or W≥293,5128 | W=210,49 | Do not reject H0 since val does not belong to the rejection region |

Section 1 D

BMI male heads of household |

| N | Minimum | Maximum | Mean | Std. Deviation | Variance |

BMI male heads | 249 | 17,75 | 35,24 | 24,6444 | 3,08325 | 9,506 |

2) We assumed that the sample is normally distributed.

Figure 1:...