DHS 2025 H2 Math Prelim Paper 2 — Hypothesis Testing (worked solution)

Hypothesis Testing: Multi-part hypothesis test — DHS 2025 H2 Math Prelim Paper 2

Dunman High School (DHS)2025 PrelimPaper 2●●● Challenging12 marks

What this question tests

Multi-part hypothesis test.

Question

Mr Dough's factory produces packets of flour for sale in local supermarkets packed in 2 sizes — Regular-sized packets and Large-sized packets. Automated machines weigh and pack the flour. A recent major power outage resulted in all his machines being needed to be recalibrated upon restarting.

Regular-sized packets are supposed to have a mean mass of 1 kg. Mr Dough suspects that his machines now pack Regular-sized packets that are underweight. He took a sample of his Regular-sized packets and measured their masses, \(x\) g (correct to nearest 5 g). The results are recorded as follows:

Mass, \(x\) (g)	No. of Regular-sized packets
980	5
985	6
990	7
995	7
1000	9
1005	6
1010	5
1015	3
1020	2

Determine the unbiased estimate of the population variance of the mass of Regular-sized packets.
Explain whether Mr Dough should conduct a one-tail or a two-tail test.
Carry out the test at 5% level of significance and give your conclusion in context. Explain whether it is necessary to assume that the mass of Regular-sized packets of flour follow a normal distribution.

Large-sized packets of flour are known to have a normal distribution with a mean mass of 2 kg. Mr Dough also suspects that his machines are now not packing the Large-sized packets according to the required mass correctly. A sample of 15 Large-sized packets of flour gave a total mass of 30075 g.

Assuming Large-sized packets of flour have a population variance of \(\sigma^2\,\mathrm{g}^2\), an appropriate test was carried out which gave a \(p\)-value of 0.0529.

Showing all necessary workings, prove that \(\sigma = 10.0\) correct to 3 significant figures.
Another packet of Large-sized packet of flour with mass \(h\) g is then added to the sample of 15. The combined sample of Large-sized packets resulted in a rejection of the claim that the mean mass of a Large-sized packet of flour is 2 kg at 4% level of significance.
Using \(\sigma = 10.0\), find the possible range of values of \(h\), giving your answer to the nearest gram.

Show full worked solution▾

(a)

Using GC, unbiased estimate of the population variance:

\(s^2 = 123\) g\(^2\) (3 s.f.)

The unbiased estimate of population variance is \(s^2\), not \(\sigma^2\). We can only get an estimate from sample data.

(b)

A one-tail (lower-tail) test, as \(H_1: \mu < 1000\).

(c)

Let \(\mu\) be the population mean mass (g) of a Regular-sized packet of flour.

\(H_0: \mu = 1000\) \(H_1: \mu < 1000\)

Perform 1-tail test at 5% level of significance.

Under \(H_0\), by Central Limit Theorem (since \(n = 50\) is large): \[\bar{X} \sim \mathrm{N}\!\left(1000,\,\frac{122.69}{50}\right) \text{ approximately}\]

\(\bar{x} = 997.4\) (exact), \(n = 50\).

\(p\)-value \(= 0.0485\) (3 s.f.).

Since \(p\)-value \(= 0.0485 < 0.05\), we reject \(H_0\).

There is sufficient evidence at 5% level of significance to conclude that the mean mass of a Regular-sized packet of flour is less than 1000 g.

CLT can only approximate the distribution of \(\bar{X} = \dfrac{X_1 + X_2 + \cdots + X_n}{n}\) (sample mean) for large \(n\); it cannot be used to approximate the distribution of \(X\) (mass of individual packet).

(d)

Let \(Y\) g be the mass of a Large-sized packet, \(\mu_Y\) be the population mean.

\(H_0: \mu_Y = 2000\) \(H_1: \mu_Y \neq 2000\)

Under \(H_0\): \(\bar{Y} \sim \mathrm{N}\!\left(2000,\,\dfrac{\sigma^2}{15}\right)\).

\(\bar{y} = \dfrac{30075}{15} = 2005\).

Since \(2005 > 2000\), \(p\)-value \(= 2\mathrm{P}(\bar{Y} \ge 2005) = 0.0529\).

\[\mathrm{P}(\bar{Y} \ge 2005) = 0.0529/2\] \[\frac{2005 - 2000}{\sigma/\sqrt{15}} = z_{\text{cal}}\] \[\frac{5\sqrt{15}}{\sigma} = 1.935736\ldots\]

(e)

\(\text{H}_0\): \(\mu_Y = 2000\)

\(\text{H}_1\): \(\mu_Y \neq 2000\)

Perform 2-tail test at 4% level of significance

Under \(\text{H}_0\), \(\bar{Y}_{\text{new}} \sim \mathrm{N}\!\left(2000,\ \dfrac{10.0^2}{16}\right)\)

\[ \bar{y}_{\text{new}} = \frac{\sum y + h}{16} = \frac{30075 + h}{16} \]

Figure for this question

Since \(\text{H}_0\) is rejected: \(\bar{y}_{\text{new}}\) lies in critical region

The critical/rejection region should not use strict inequality – it must include the “equal sign”, i.e. “\(\le\), \(\ge\)”. Some students carelessly used sample size \(n = 15\) instead of 16.

Answer: (a) \(s^2 = 123\) g\(^2\) (3 s.f.) (b) One-tail (lower-tail) test. (c) Reject \(H_0\). Sufficient evidence that mean mass \(< 1000\) g at 5\% level. (d) \(\sigma = 10.0\) (3 s.f.) (shown) (e)

p{0.45\linewidth} l p{0.45\linewidth}@{}} \(\bar{y}_{\text{new}} \leq 1994.8656\)	or	\(\bar{y}_{\text{new}} \geq 2005.1344\) [6pt] \(\dfrac{30075+h}{16} \leq 1994.8656\)	or	\(\dfrac{30075+h}{16} \geq 2005.1344\) [10pt] \(h \leq 1842.85\)	or	\(h \geq 2007.15\) [6pt] \(\therefore h \leq 1842\) (nearest g)	or	\(h \geq 2008\) (nearest g)

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