A-Level Hypothesis testing: worked solution

Question

The manager now tests whether machine B's mean differs from 1.5 kg. Machine B's masses are Normal with variance 0.00198 kg². A random sample of size n has mean 1.489 kg. At the 5% level, find the values of n for which there is enough evidence that the mean differs from 1.5 kg.

Step 1 - state the hypotheses. "Differs from" gives a TWO-tailed test. H0: mu = 1.5 H1: mu is not equal to 1.5 Test at the 5% significance level. Step 2 - state the test statistic. The population is normal and its variance is KNOWN (0.00198 kg²), so use a z-test rather than a t-test. Under H0, Z = (xbar - 1.5) / sqrt(0.00198/n) ~ N(0, 1) Step 3 - state the critical region. For a two-tailed test at 5%, the 5% is split into 2.5% in each tail. H0 is rejected, i.e. there is sufficient evidence, when |z| > 1.96 (more precisely 1.95996) Step 4 - substitute the sample values, xbar = 1.489. z = (1.489 - 1.5) / sqrt(0.00198/n) Numerator: 1.489 - 1.5 = -0.011 Denominator: sqrt(0.00198/n) = sqrt(0.00198)/sqrt(n) = 0.0444972/sqrt(n) z = -0.011 sqrt(n) / 0.0444972 Taking the modulus: |z| = 0.011 sqrt(n) / 0.0444972 = 0.24721 sqrt(n) Step 5 - impose the rejection condition and solve. 0.24721 sqrt(n) > 1.95996 sqrt(n) > 1.95996 / 0.24721 = 7.9284 Square both sides (both are positive): n > 7.9284² = 62.86 Step 6 - interpret. n > 62.9 (3 s.f.), and n must be a whole number, so the smallest sample size giving sufficient evidence that the mean differs from 1.5 kg is n = 63, and any n >= 63 works.