A-Level Hypothesis testing: worked solution

Question

The cafe's premium coffee has caffeine N(mu, 200) mg^2. The owner now claims its mean is below 120 mg. A random sample of n cups has mean 116.6 mg; a 2.5% test supports the claim. Find the possible values of n.

Let X be the caffeine content of one cup of premium coffee, X ~ N(mu, 200), so the population variance sigma² = 200 is KNOWN. For a sample of n cups the sample mean satisfies X-bar ~ N(mu, 200/n). Step 1: State the hypotheses. The owner claims the mean is BELOW 120 mg, so this is a one-tail (lower-tail) test: H0: mu = 120 H1: mu < 120 tested at the 2.5% significance level. Step 2: Write down the test statistic. Because sigma² is known, use the z-statistic Z = (x-bar - mu0)/sqrt(sigma²/n) = (116.6 - 120)/sqrt(200/n), which is N(0, 1) under H0. Step 3: Say what "supports the claim" means. The claim is supported when H0 is rejected, i.e. when the test statistic falls in the lower-tail critical region. For a lower-tail test at 2.5% the critical value is the z with P(Z < z) = 0.025, namely z = -1.95996. So the condition is (116.6 - 120)/sqrt(200/n) < -1.95996. Step 4: Solve the inequality for n. The numerator is 116.6 - 120 = -3.4, and sqrt(200/n) = sqrt(200)/sqrt(n), so the left-hand side is -3.4 sqrt(n)/sqrt(200): -3.4 sqrt(n)/sqrt(200) < -1.95996 Multiply both sides by -1, which REVERSES the inequality sign: 3.4 sqrt(n)/sqrt(200) > 1.95996 Multiply both sides by sqrt(200)/3.4, which is positive, so the direction is unchanged: sqrt(n) > 1.95996 x sqrt(200)/3.4 sqrt(n) > 1.95996 x 14.14214/3.4 sqrt(n) > 27.71806/3.4 sqrt(n) > 8.15237 Both sides are positive, so squaring keeps the direction: n > 66.461 Step 5: Interpret. n is a number of cups, so it must be a positive integer, and the smallest integer above 66.461 is 67. Any larger sample also gives a significant result, so {n in Z : n >= 67}.