Question 9.5: A national park in Namibia determines the weight (in kg) of ......

(a) The point estimate of μ is \bar{x} which is

\hat{\mu }= \bar{x} =\frac{1}{n} \sum\limits_{i=1}^{n}{x_{i}} =\frac{1}{24}\left(450+ \ldots+790 \right) = 667.92.

The variance of σ² can be estimated unbiasedly using s²:

\hat{\sigma }^{2}=s^{2}=\frac{1}{n-1} \sum\limits_{i=1}^{n}{\left(x_{i}-\bar{x}\right)^{2} }

=\frac{1}{23} \left(\left(450 − 667.92\right)^{2}+ \ldots+ \left(790 − 667.92\right)^{2}\right) \approx 18, 035.

(b) The variance is unknown and needs to be estimated. We thus need the t-distribution to construct the confidence interval. We can determine t_{23;0.975} ≈ 2.07 using qt(0.975,23) or Table C.2 (though the latter is not detailed enough), α = 0.05, \bar{x}= 667.97 and \hat{\sigma }^{2} = 18, 035. This yields

I_{l} (X) = \bar{x} − t_{n−1;{1−α}/{2}}· \frac{s}{\sqrt{n} }= 667.92 − t_{23;0.975} · \frac{\sqrt{18, 035}}{\sqrt{24} } \approx 611.17,

I_{u} (X) = \bar{x} + t_{n−1;{1−α}/{2}}· \frac{s}{\sqrt{n} } 667.92 − t_{23;0.975} · \frac{\sqrt{18, 035}}{\sqrt{24} } \approx 724.66.

The confidence interval for μ is thus [611.17; 724.66].

(c) We can reproduce these results in R as follows:

eland <- c(450,730,700,600,620,,790)
t.test(eland)$conf.int