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2015 A-Level · Paper 2 · Question 8

2015 H2 Maths Paper 2, Question 8

Hypothesis testing 7 marks one-tailed test of a mean
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Solution

7 marks

Let XX be the mass, in kg, of a pineapple from the stall, and let n=8n = 8 be the sample size.

Finding the unbiased estimates

Summarising the sample: ∑x=7.06\sum x = 7.06 and ∑x2=6.2696\sum x^{2} = 6.2696.

x‾=∑xn=7.068=0.8825\overline{x} = \frac{\sum x}{n} = \frac{7.06}{8} = 0.8825
s2=1n−1(∑x2−(∑x)2n)=17(6.2696−7.0628)=0.00559 (3 s.f.)s^{2} = \frac{1}{n - 1}\left( \sum x^{2} - \frac{(\sum x)^{2}}{n} \right) = \frac{1}{7}\left( 6.2696 - \frac{{7.06}^{2}}{8} \right) = 0.00559\ \text{(3 s.f.)}

∴\therefore the unbiased estimate of the population mean is x‾=0.8825 kg\overline{x} = 0.8825\ kg, and the unbiased estimate of the population variance is s2=0.00559 kg2s^{2} = 0.00559\ kg^{2} (3 s.f.).

Setting up the test

Let μ\mu be the population mean mass, in kg, of pineapples sold at the stall. The stall owner’s claim is μ≥0.9\mu \geq 0.9, so we test whether the sample gives evidence against this at the 10%10\% level of significance.

H0: μ=0.9H_{0}:\ \mu = 0.9
H1: μ<0.9H_{1}:\ \mu < 0.9

where the test is carried out at the 10%10\% significance level.

Note: since the population variance is unknown and is estimated from a small sample (n=8<30n = 8 < 30), we need the population of pineapple masses to be normally distributed (as given) to use a tt-test.

Approach

the alternative hypothesis is one-tailed (“doubt the claim that the mean is at least 0.90.9” means checking whether the mean is actually less than 0.90.9), so this is a lower-tailed test.

Under H0H_{0}, the test statistic is

T=X‾−0.9S/n∼t(n−1),T = \frac{\overline{X} - 0.9}{S/\sqrt{n}} \sim t(n - 1),

where X‾\overline{X} is the sample mean and S2S^{2} is the unbiased estimate of the population variance.

t=0.8825−0.90.00559/8t = \frac{0.8825 - 0.9}{\sqrt{0.00559/8}}
=−0.662 (3 s.f., by GC).= - 0.662\ \text{(3 s.f., by GC).}

By GC, the corresponding one-tailed pp-value is

p-value=P(T≤−0.662)=0.265 (3 s.f.).p\text{-value} = P(T \leq - 0.662) = 0.265\ \text{(3 s.f.).}
To check your answer: the critical value for a one-tailed test at the 10%10\% level with 77 degrees of freedom is t=−1.415t = - 1.415; since −0.662>−1.415- 0.662 > - 1.415, the test statistic does not fall in the critical region, consistent with p=0.265>0.10p = 0.265 > 0.10. ✓

Since p=0.265>0.10p = 0.265 > 0.10, we do not reject H0H_{0}.

∴\therefore there is insufficient evidence, at the 10%10\% level of significance, to conclude that the mean mass of pineapples is less than 0.9 kg0.9\ kg, so there is no evidence to doubt the stall owner’s claim.

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