A-Level Hypothesis testing: worked solution

Question

Large packets are normal with population variance σ²=10.0² g² (from part (d)). A sample of 15 totalled 30075 g (mean 2005 g); testing H0: μ=2000 against H1: μ≠2000 gave p=0.0529. A further Large packet of mass h g is now added, and the 16-packet sample leads to rejecting the claim that the mean is 2 kg at the 4% level. (e) Taking σ=10.0, find the range of h to the nearest gram.

The alternative hypothesis is H1: μ ≠ 2000, so this is a two-tailed test. At the 4% level that puts 2% in each tail, so the critical values are ±z where Φ(z) = 1 − 0.02 = 0.98, giving z = ±2.0537. With σ known and n = 16, the test statistic is Z = (x̄ − 2000)/(σ/√n) = (x̄ − 2000)/(10.0/√16) = (x̄ − 2000)/2.5. The claim μ = 2000 is rejected exactly when |Z| ≥ 2.0537, i.e. when |x̄ − 2000| ≥ 2.0537 × 2.5 = 5.1344. Now express the new sample mean. The 15 packets totalled 30075 g, and adding the new packet of mass h makes the total 30075 + h over 16 packets: x̄ = (30075 + h)/16. So x̄ − 2000 = (30075 + h)/16 − 2000 = (30075 + h − 32000)/16 = (h − 1925)/16. The rejection condition becomes |h − 1925|/16 ≥ 5.1344, so |h − 1925| ≥ 16 × 5.1344 = 82.150. Split the modulus into its two branches: h − 1925 ≥ 82.150 gives h ≥ 2007.15; h − 1925 ≤ −82.150 gives h ≤ 1842.85. Taking whole grams, the smallest integer satisfying h ≥ 2007.15 is 2008, and the largest satisfying h ≤ 1842.85 is 1842. So, to the nearest gram, h ≤ 1842 or h ≥ 2008.