Answer:

1. (a) The 99% confidence interval is approximately \((196.35,\ 201.98)\) g, or to 3 significant figures \((196,\ 202)\) g.
2. (b) The greatest possible value is \(k \approx 200.8\) g.

1. (a)

   The sample mean is

   \(\bar{x}=\frac{2390}{12}=199.1667\).

   Since the population variance is not given, use the unbiased sample variance:

   \(s^2=\frac{1}{12-1}\left(476117-\frac{2390^2}{12}\right)=\frac{326}{33}=9.8788\).

   For a 99% confidence interval with \(n=12\), there are \(11\) degrees of freedom, so the required two-tailed critical value is

   \(t_{0.995,11}=3.106\).

   The standard error is

   \(\sqrt{\frac{s^2}{n}}=\sqrt{\frac{9.8788}{12}}=0.9073\).

   Hence the confidence interval is

   \(199.1667\pm 3.106(0.9073)=199.1667\pm 2.818\).

   So the 99% confidence interval is

   \((196.35,\ 201.98)\) g, approximately \((196,\ 202)\) g.

2. (b)

   The test is of \(H_0:\mu=k\) against \(H_1:\mu\lt k\), so it is a lower-tail one-sided \(t\)-test with \(11\) degrees of freedom.

   At the 5% significance level, the lower-tail critical value is

   \(-t_{0.95,11}=-1.796\).

   The test statistic is

   \(T=\frac{\bar{x}-k}{s/\sqrt{n}}=\frac{199.1667-k}{0.9073}\).

   For the test not to support the alternative hypothesis, the statistic must not fall in the rejection region, so

   \(\frac{199.1667-k}{0.9073}\ge -1.796\).

   Solving for \(k\):

   \(199.1667-k\ge -1.796(0.9073)\),

   \(k\le 199.1667+1.796(0.9073)\).

   Therefore

   \(k\le 200.796\ldots\).

   So the greatest possible value of \(k\) is approximately \(200.8\) g.