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Worked solution

Find significance levels that reject a wage claim

In year 2008, a survey was done by the Ministry of Manpower (MOM) to determine the amount of daily wages, denoted as XX, that part-time workers were receiving. Past records show that on average, the wages that each worker received was $30.

A year later, MOM wants to find out if there is a change in the mean wage. Data from 90 workers are collected and their wages are summarized as shown below:

xˉ=31,∑(x−31)2=1947.\bar x=31,\qquad\sum(x-31)^2=1947.

Calculate the unbiased estimate of the variance of XX . mark

MOM decides to do a test on the collected data. Find the range of values of α\alpha if MOM were to conclude that there is a change in the mean wage at α%\alpha\% level of significance. marks

[2009/NJC/II/7(part of)]

CJC · Review problems · Question 3

Answer

s2=1947/89s^2=1947/89. Reject for 100p≤α<100100 p\le\alpha<100, where p≈0.042529119p\approx 0.042529119; lower endpoint≈4.25291%.

Unbiased variance

Since 31 is the sample mean, the sum is already centered: s2=1947/89=21.87640449s^2=1947/89=21.87640449 dollars squared.

How did this part go?

Two-sided test and calculator method

Let μ\mu be the population mean daily wage. Test H0:μ=30H_0:\mu=30 against H1:μ≠30H_1:\mu\ne 30. For n 90, under H0H_0, Xˉ≈N(30,(1947/89)/90)\bar X\approx N(30,(1947/89)/90). Thus z=31−30(1947/89)/90=2.028305098.\begin{aligned}&z\\ & =\frac{31-30}{\sqrt{(1947/89)/90}}\\ & =2.028305098.\end{aligned} The two-sided p-value is 2Φ(−∣z∣)=0.0425291192\Phi(-|z|)=0.042529119. The source Z-Test Stats screen uses null 30, estimated standard deviation 4.677221878, sample mean 31, n 90 and the not-equal alternative; its p≈.0425289811 agrees to displayed precision.

How did this part go?

Invert the significance comparison

Rejection requires p≤α/100p\le\alpha/100, so α≥100p≈4.2529119\alpha\ge 100 p\approx 4.2529119, with α<100\alpha<100 for a nondegenerate significance percentage. The source writes 4.26≤α<1004.26\le\alpha<100, an upward-rounded sufficient threshold, but the exact lower endpoint is 100p100 p, not 4.26. Use the unrounded p-value when classifying boundary cases.

How did this part go?

Common mistakes

The p-value is not the probability that the null hypothesis is true. Rounding a threshold can change the stated set of possible significance levels.

Solution notes

Source p=.042528 and screen p≈.0425289811 differ insignificantly from independent standard-normal evaluation .042529119. Its final 4.26 threshold is conservative rounding; exact endpoint retained separately.

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