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Worked solution

Find the spending rejection boundary without over-rounding

A department store reports that average spending per customer over the past six months is $215 with standard deviation $100. During a week-long sales promotion, it takes a random sample of 100 visiting customers and records their spending. It tests at 5% whether the promotion increased average spending.

(i) State, with a reason, whether the test requires assuming normally distributed customer spending.

(ii) State the null and alternative hypotheses. Given that the null hypothesis is rejected, show that the sample mean is greater than $231.45.

[2007/SAJC/II/11(a)]

CJC · Discussion · Question 3

Answer

No individual normality assumption is required for the source CLT test. H0:μ=215H_0:\mu=215, H1:μ>215H_1:\mu>215; exact normal-approximation rejection condition xˉ≥231.4485362695…\bar x\ge231.4485362695\ldots. The printed strict bound is overstrong.

(i) CLT applies to the sample mean

With independent representative observations and a stable finite variance, the source uses CLT for n=100n=100; individual spending need not itself be normal. The sample mean, not each customer’s spending, is approximately normal. Retaining the historical SD 100 for the promotion population is an additional model assumption.

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(ii) p-value and critical-value routes

Test H0:μ=215H_0:\mu=215 against H1:μ>215H_1:\mu>215. Under H0H_0, Xˉ≈N(215,1002/100)=N(215,100)\bar X\approx N(215,100^2/100)=N(215,100). Rejection with the source convention p≤0.05p\le0.05 means PH0(Xˉ≥xˉ)≤0.05P_{H_0}(\bar X\ge\bar x)\le0.05, or z=(xˉ−215)/10≥Φ−1(0.95)z=(\bar x-215)/10\ge\Phi^{-1}(0.95). Both methods give xˉ≥215+10(1.644853627)=231.4485362695….\begin{aligned}&\bar x\ge215\\ &{}+10(1.644853627)\\ & =231.4485362695\ldots.\end{aligned}

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(ii) Qualify the printed “show” statement

The question asks for xˉ>231.45\bar x>231.45 while its working states xˉ≥231.45\bar x\ge231.45 after rounding. Neither is the exact real-valued rejection boundary. For example, 231.449 exceeds 231.4485362695 and rejects, but is below 231.45. Thus the requested strict bound does not follow as stated. The boundary rounds to 231.45; if the mean itself is restricted to a 0.01 grid, the least rejecting grid value is 231.45, with equality included. Do not infer that grid merely from individual prices being in cents.

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5% upper spending boundary: standard normal curve, right rejection tail area 0.05, critical z 1.644853627.
5% upper spending boundary: standard normal curve, right rejection tail area 0.05, critical z 1.644853627.
5% upper spending boundary: standard normal curve, right rejection tail area 0.05, critical z 1.644853627.

Common mistakes

Keep the full critical boundary before comparing real-valued means. The source’s rounded ≥ statement and the question’s > statement are not interchangeable.

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