Porisma

Worked solution

Use raw and shifted summaries for unbiased estimates

Find unbiased estimates of the population mean and variance from each sample.

(a) ∑x=15508\sum x=15508, ∑x2=2413891\sum x^2=2413891, n=100n=100.

(b) ∑(t−36.9)=−12\sum(t-36.9)=-12, ∑(t−36.9)2=79.21\sum(t-36.9)^2=79.21, n=40n=40.

CJC · Discussion · Question 2

Answer

(a) Mean 155.08155.08, variance 90.090.0; (b) mean 36.636.6, variance 1.941.94.

(a) Raw totals

xˉ=15508/100=155.08\bar x=15508/100=155.08. The unbiased variance is s2=199(2413891−155082100)=24751275≈90.0s^2=\dfrac1{99}\left(2413891-\dfrac{15508^2}{100}\right)=\dfrac{24751}{275}\approx90.0, to 3 s.f.

How did this part go?

(b) Shifted totals

The mean is tˉ=36.9+(−12)/40=36.6\bar t=36.9+(-12)/40=36.6. Subtracting a constant does not change variance, so s2=139(79.21−(−12)240)=75613900≈1.94s^2=\dfrac1{39}\left(79.21-\dfrac{(-12)^2}{40}\right)=\dfrac{7561}{3900}\approx1.94, to 3 s.f. Use the same centring correction on the shifted totals.

How did this part go?

Common mistakes

Σ(t−36.9)² is not already the sum of squared deviations from the sample mean. Subtract the squared shifted total divided by n.

Solution notes

No marks are printed for this question; none are invented.

Read the hint