Statistical inference — Question 257

Original QCE Vault practice · 12 marks

Q257 · Practice questionTechnology-activeVery complex unfamiliar12 marks

QUESTION 257 (12 marks)

Three sensors measure an unknown mean

\mu

, but their population means are

\mu+\beta

\mu-\beta

and

\mu+2\beta

, where

\beta

is an unknown fixed calibration offset. Each sensor supplies

n

independent normal readings; the three samples are independent. Population standard deviations are 3, 2 and 1 respectively. Let their sample means be

\overline X_1,\overline X_2,\overline X_3

. Consider

M=a\overline X_1+b\overline X_2+c\overline X_3

. You may use linearity of expectation,

\operatorname{Var}(M)=(9a^2+4b^2+c^2)/n

, and that

M

is normal. Use 1.96 for a 95% mean interval.

Derive the conditions on

a,b,c

that make

M

unbiased for

\mu

for every possible

\mu,\beta

. Express

a,b

in terms of

c

[3 marks]

Find the unbiased estimator with smallest variance, allowing real weights, and prove it is the unique optimum. Determine its variance.

[4 marks]

With

n=120

, observed sample means are 12.42, 12.08 and 12.58. Calculate the optimal estimate and a 95% confidence interval for

\mu

, to three decimal places. State whether 12 is contained in the unrounded interval.

[3 marks]

Find the smallest equal sample size per sensor that makes the total 95% interval width at most 0.50, using the optimal weights.

[2 marks]

Question linkSyllabus coverage

WORKED SOLUTION

Practice marking scheme

12 marks

ANSWER

(a)

a+b+c=1

a-b+2c=0

;

a=(1-3c)/2

b=(1+c)/2

. (b)

(a,b,c)=(10,56,23)/89

;

\operatorname{Var}(M)=157/(89n)

. (c) Estimate

12.247

; interval

(12.010,12.485)

; 12 is excluded. (d)

n=109

readings per sensor.

Worked solution

(a)

E(M)=(a+b+c)\mu+(a-b+2c)\beta

. For this to equal

\mu

for all values, the coefficients must be 1 and 0 respectively. Solving gives

a=(1-3c)/2

and

b=(1+c)/2

(b) Substitution gives

n\operatorname{Var}(M)=(13-46c+89c^2)/4=\frac{89}{4}(c-23/89)^2+157/89

. The positive squared coefficient proves a unique global minimum at

c=23/89

. The constraints then give

a=10/89

and

b=56/89

, and the minimum variance is

157/(89n)

(c)

m=(10(12.42)+56(12.08)+23(12.58))/89=12.2474157\ldots

. Its standard error is

\sqrt{157/(89(120))}

. The interval

m\pm1.96\sqrt{157/10680}

(12.0097753\ldots,12.4850561\ldots)

. Its unrounded lower endpoint is above 12, so 12 is not contained in the interval.

(d) Total width is

2(1.96)\sqrt{157/(89n)}

. Requiring it to be at most 0.50 gives

n\ge(1.96/0.25)^2(157/89)=108.4280809\ldots

. The smallest integer is 109; 108 gives a width above the target.

Equivalent mathematically justified methods accepted.

Exact answers unless a decimal accuracy is specified.

Part a: Form the estimator expectation.

[1 mark]

Part a: Identify both unbiasedness constraints.

[1 mark]

Part a: Solve for

a,b

in terms of

c

[1 mark]

Part b: Substitute the constraints into the variance.

[1 mark]

Part b: Complete the square or differentiate to obtain the optimal

c

[1 mark]

Part b: Calculate the other weights and prove uniqueness.

[1 mark]

Part b: Obtain the minimum variance.

[1 mark]

Part c: Calculate the optimally weighted estimate.

[1 mark]

Part c: Use the correct standard error to obtain the interval.

[1 mark]

Part c: Use the unrounded endpoints to decide containment.

[1 mark]

Part d: Form the correct width inequality.

[1 mark]

Part d: Round the sample-size bound upwards to the minimum integer.

[1 mark]

Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.

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Statistical inference — Question 257

QUESTION 257 (12 marks)

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