Answer:

1. (a) Assuming the children’s estimates are independent observations from a normal population with mean \(\theta\), a 95% confidence interval for \(\theta\) is \((82.3^\circ,\,86.5^\circ)\).
2. (b) These assumptions may not be appropriate because the angle is acute and the estimates are close to \(90^\circ\), so the distribution of estimates may be skewed or truncated rather than normal.

1. (a) Let the estimates be treated as a sample from a normal population with mean \(\theta\), and assume the observations are independent. Since the population variance is unknown and the sample size is small, use a \(t\)-interval with \(10-1=9\) degrees of freedom.

   The sample mean is

   \(\bar{x}=\frac{84+85+77+85+84+87+86+88+83+85}{10}=\frac{844}{10}=84.4\).

   Also,

   \(\sum x^2=71314\),

   so the unbiased sample variance is

   \(s^2=\frac{1}{9}\left(71314-\frac{844^2}{10}\right)=\frac{134}{15}=8.933\ldots\).

   Thus

   \(s=\sqrt{8.933\ldots}=2.989\ldots\).

   For a 95% confidence interval with 9 degrees of freedom, \(t_{0.975,9}=2.262\).

   Therefore the confidence interval is

   \(\bar{x}\pm t_{0.975,9}\frac{s}{\sqrt{n}}=84.4\pm 2.262\sqrt{\frac{8.933\ldots}{10}}\).

   The margin of error is

   \(2.262\sqrt{\frac{8.933\ldots}{10}}=2.138\ldots\).

   Hence the 95% confidence interval is

   \((84.4-2.138\ldots,\,84.4+2.138\ldots)=(82.262\ldots,\,86.538\ldots)\).

   So, to 1 decimal place, the interval is \((82.3^\circ,\,86.5^\circ)\).

2. (b) The normal model may not be suitable here because the triangle is acute, so the true angle is below \(90^\circ\), and the estimates are quite close to \(90^\circ\). This means the possible estimates may be limited or skewed near \(90^\circ\), rather than being well modelled by a normal distribution.