Solution assignment week 1
a.
∗
∗
∗
(𝑛 − 1) = 𝑡%.%#'
(33) < 𝑡%.%#'
(30) = 2.042, a conservative
With the right-tail quantile 𝑡!/#
∗
(𝑛 − 1) ∙
approximation of the 95% confidence interval is given by 𝑥̅()*)+,,-./011,+)+2 ± 𝑡!/#
3!"#"$%%&'()**%$"$+
√+
≈ 12.18 ± 2.042 ∙
5'.#'%
√67
= (6.839,17.521).
b.
(i) (1 point)
𝐻% : 𝜇()*)+,,-. = 𝜇,11,+)+2 ,
𝐻8 : 𝜇()*)+,,-. > 𝜇,11,+)+2
(ii) (2 points)
𝑥̅()*)+,,-. − 𝑥̅,11,+)+2
12.18
𝑡=𝑠
=
= 4.657
()*)+,,-./011,+)+2
15.250A
@
√34
√𝑛
(iii) (1 point)
The degrees of freedom are 𝑛 − 1 = 34 − 1 = 33.
(iv) (1 point)
∗ (40)
∗ (𝑛
∗ (33)
∗ (30)
One sided alternative: critical value 1.684 = 𝑡%.%'
< 𝑡%.%'
− 1) = 𝑡%.%'
< 𝑡%.%'
=
1.697
(v) (1 point)
∗ (30)
∗ (33),
Since 𝑡 = 4.657 > 1.697 = 𝑡%.%'
> 𝑡%.%'
𝐻% is rejected. Hence the average ball possession
of Feyenoord is significantly higher than that of their opponent.
c.
Two possible argumentations:
No. The 90% (two-sided) confidence interval should be used for this one-sided test to get the correct
critical value in the lower tail.
Yes. Even the 95% confidence interval does not contain the hypothesized mean difference of 0, which
corresponds to rejection of the one-sided test at a significance level of 2.5%. Hence the test at a
significance level of 5% would be rejected as well.
d.
The boxplot of the differences on the right shows two outliers, hence the t test is not appropriate.