Statistical inference — Question 239

Original QCE Vault practice · 8 marks

Q239 · Practice questionTechnology-activeComplex unfamiliar8 marks

QUESTION 239 (8 marks)

A coral nursery uses

N^{\prime}=kN(1-N/200)

with

N(0)=50

and constant

k>0

, where time is in days. A target is

N(10)\ge150

. An independent calibration sample of 100 estimates of k has mean 0.24 and sample standard deviation 0.08. Use

\overline x\pm1.96s/\sqrt n

for an approximate mean confidence interval. Treat the calibration estimates as independent observations from a population whose mean is the true k; assume normal-interval conditions are satisfied. The diagram shows the boundary profile that reaches the target exactly at day 10.

Solve the logistic equation and find the least k meeting the target.

[3 marks]

Find the approximate 95 percent interval for mean k. Does every k in this interval meet the model target?

[2 marks]

Assuming the same future observed mean and standard deviation, find the least n for which the lower interval endpoint is at least the threshold from (a).

[3 marks]

Question linkSyllabus coverage

WORKED SOLUTION

Practice marking scheme

8 marks

ANSWER

(a)

N(t)=200/(1+3e^{-kt})

;

k\ge\ln9/10

. (b)

(0.22432,0.25568)

day

^{-1}

; yes under the model. (c)

n=60

Worked solution

(a) Separate variables:

\frac{200}{N(200-N)}\,dN=k\,dt

. Since

\frac{200}{N(200-N)}=\frac1N+\frac1{200-N}

, integration gives

\ln\!\left(\frac{N}{200-N}\right)=kt+C_0

. At N(0)=50,

C_0=-\ln3

. Exponentiating and rearranging yields

N(t)=\frac{200}{1+3e^{-kt}}

. At t=10, the target requires

1+3e^{-10k}\le4/3

, hence

e^{-10k}\le1/9

k\ge\ln9/10\approx0.2197224577

. N(10) increases with k, so the boundary is the least qualifying value.

(b) The margin is

1.96(0.08)/10=0.01568

, so the interval is (0.22432,0.25568). Even its lower endpoint exceeds

\ln9/10

, so every k in it meets the specified model target. This is an interval-based assessment under the assumptions, not certainty about the fixed true k.

0.24-0.1568/\sqrt n\ge\ln9/10

. The positive gap is

0.24-\ln9/10

, so

n\ge[0.1568/(0.24-\ln9/10)]^2\approx59.79

. The least integer is 60; using the unrounded threshold confirms that 59 fails and 60 passes. The design is conditional on the assumed future summaries.

Equivalent mathematically justified methods accepted.

Exact answers unless a decimal accuracy is specified.

Part a: Solve the separable equation.

[1 mark]

Part a: Apply the initial value.

[1 mark]

Part a: Solve the target inequality with the correct direction.

[1 mark]

Part b: Calculate both interval endpoints.

[1 mark]

Part b: Compare the lower endpoint with the model threshold.

[1 mark]

Part c: Form the lower-endpoint inequality.

[1 mark]

Part c: Solve using the unrounded threshold.

[1 mark]

Part c: Round up and justify minimality.

[1 mark]

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

View the QCAA syllabus

How many marks did you earn?

Compare your working with the guide above.

Statistical inference — Question 239

QUESTION 239 (8 marks)

Practice marking scheme

Related questions