Question 9.6: We are interested in the heights of the players of the two b......

• Let us start with the confidence interval for the “Brose Baskets Bamberg”. Using t _{15;0.975} = 2.1314 (qt(0.975,15) or Table C.2) and α = 0.05, we can determine the confidence interval as follows:

I_{l} (Ba) = \bar{x} − t_{n−1;{1−α}/{2}}· \frac{s}{\sqrt{n} }= 199.06− t_{15;0.975} · \frac{7.047}{\sqrt{16} } = 195.305,

I_{u} (Ba) = \bar{x} + t_{n−1;{1−α}/{2}}· \frac{s}{\sqrt{n} }= 199.06 + t_{15;0.975} · \frac{7.047}{\sqrt{16} } = 202.815.

Thus, we get [195.305; 202.815].

• For the “Bayer Giants Leverkusen”, we use t_{13;0.975} = 2.1604 to get

I_{l} (L) = \bar{x} − t_{n−1;{1−α}/{2}}· \frac{s}{\sqrt{n} }= 196 − t_{13;0.975} · \frac{9.782}{\sqrt{14} } = 190.352,

I_{u} (L) = \bar{x} + t_{n−1;{1−α}/{2}}· \frac{s}{\sqrt{n} }= 196 + t_{13;0.975} · \frac{9.782}{\sqrt{14} } = 201.648.

This leads to a confidence interval of [190.352; 201.648].

• For “Werder Bremen”, we need to use the quantile t_{22,0.975} = 2.0739 which yields a confidence interval of

I_{l} (Br) = \bar{x} − t_{n−1;{1−α}/{2}}· \frac{s}{\sqrt{n} }= 187.52− t_{22;0.975} · \frac{5.239}{\sqrt{23} } = 185.255,

I_{u} (Br) = \bar{x} + t_{n−1;{1−α}/{2}}· \frac{s}{\sqrt{n} }= 187.25+ t_{22;0.975} · \frac{5.239}{\sqrt{23} } = 189.786.

The interval is therefore [185.255; 189.786].

• The mean heights of the basketball teams are obviously larger than the mean height of the football team. The two confidence intervals of the basketball teams overlap, whereas the intervals of the football team with the two basketball teams do not overlap. It is evident that this indicates that the height of football players is substantially less than the height of basketball players. In Chap. 10, we will learn that confidence intervals can be used to test hypotheses about mean differences.