Porisma

Worked solution

Invert a two-sided test without imposing integer means

A random sample of 50 packets of ‘Nilo Bar’ snacks is weighed and the mass, xx grams, is recorded. The results are summarised by

∑(x−150)=−250,∑(x−150)2=3000.\sum(x-150)=-250,\qquad\sum(x-150)^2=3000.

The population mean mass is μ\mu grams. A test is carried out at the 1% significance level with the following hypotheses:

H0:μ=μ0,H1:μ≠μ0.H_0:\mu=\mu_0,\qquad H_1:\mu\ne\mu_0.

Given that H0H_0 is rejected in favour of H1H_1, find the set of possible values of μ0\mu_0. marks

[2009/IJC/II/10(b)]

CJC · Review problems · Question 2

Answer

μ0≤142.8230269\mu_0\le 142.8230269 or μ0≥147.1769731\mu_0\ge 147.1769731 (exact boundaries in working). Source integer-cutoff final set is too restrictive for real-valued means.

Estimate the sampling variance

xˉ=150−250/50=145\bar x=150-250/50=145. The unbiased variance is s2=3000−(−250)2/5049=175049.\begin{aligned}&s^2\\ & =\frac{3000-(-250)^2/50}{49}\\ & =\frac{1750}{49}.\end{aligned} Under H0H_0, Xˉ≈N(μ0,1750/2450)\bar X\approx N(\mu_0,1750/2450) by the large-sample normal approximation.

How did this part go?

Use both rejection tails

The two-sided 1% test rejects if ∣145−μ01750/2450∣≥Φ−1(0.995).\left|\frac{145-\mu_0}{\sqrt{1750/2450}}\right|\ge\Phi^{-1}(0.995). The critical magnitude is 2.5758293042.575829304, giving μ0≤142.8230269\mu_0\le 142.8230269 or μ0≥147.1769731\mu_0\ge 147.1769731. Keep the exact expression 145±Φ−1(0.995)1750/2450145\pm\Phi^{-1}(0.995)\sqrt{1750/2450} for boundary decisions.

How did this part go?

Compare the printed final set

The source working gives boundaries 142.823 and 147.1769, but its final set says μ0≤142\mu_0\le 142 or μ0≥148\mu_0\ge 148 while declaring μ0∈R\mu_0\in\mathbb R. That final set discards valid real values between 142 and 142.823 or 147.177 and 148. It would follow only with an additional integer restriction, which the question does not state. The real-valued rejection set above is the intended statistical calculation; the source final answer is retained here as a qualified discrepancy.

How did this part go?

Compare your sketch

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Snack masses: 1% two-sided rejection: standard normal, two-tail rejection region, critical magnitude 2.57583, total shaded probability 0.01 (0.005 each tail).
Snack masses: 1% two-sided rejection: standard normal, two-tail rejection region, critical magnitude 2.57583, total shaded probability 0.01 (0.005 each tail).
Snack masses: 1% two-sided rejection: standard normal, two-tail rejection region, critical magnitude 2.57583, total shaded probability 0.01 (0.005 each tail).

Common mistakes

Population means need not be integers even when observations are discrete; here packet masses are not restricted to integers either.

Solution notes

Source page 4 reverses an inequality sign in an intermediate line. Its final integer cutoffs conflict with its own real-valued domain. Correct inequalities and exact critical-value boundaries are shown, while both printed numeric stages are retained.

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