Tangent Tuition
Specialist Mathematics · Past QCAA questions All 24 questions
2023 · Paper 2 · Technology-active

Question 19

[7 marks] Technology-active
Unit 4 · Topic 5Statistical inference Tool Kit 9.5The inversion drill Tool Kit 9.4Constructing a confidence interval
The question

The height of Year 9 students at a school is assumed to be normally distributed with a population mean height of $\mu$ cm.

A teacher at the school measured the height of all the students in her Year 9 class. This data was used to calculate an approximate 95% confidence interval for $\mu$ of $(163.7,\ 166.9)$ cm.

The teacher repeated the procedure using data from another Year 9 class. Although this class had the same number of students, its data produced an approximate 95% confidence interval for $\mu$ of $(167.8,\ 172.4)$ cm.

Using the same data, the teacher recalculated the approximate confidence intervals for $\mu$ for each class using a confidence level of $x\%$. She observed that the upper bound of the confidence interval from her Year 9 class now equalled the lower bound of the confidence interval from the other Year 9 class.

Determine the value of $x$. Give your answer rounded to one decimal place.

Watch the situation first

Stretch both intervals until they just touch

Watch the two 95% intervals grow as the confidence level goes up. You will see that the means stay put while both half-widths stretch by the same factor.

Each class gives you a 95% confidence interval The centre is x̄ and the half-width is 1.96 × s√n At 95% there is a 0.9 cm gap between them Raising the confidence level stretches both intervals They just touch when z ≈ 2.413 That z-score gives a confidence level of x ≈ 98.4 160 162 164 166 168 170 172 174 176 cm teacher’s class other class 163.7 166.9 167.8 172.4 x̄₁ = 165.3 x̄₂ = 170.1 0.9 cm gap 167.27 163.33 172.93 teacher’s class: (163.7, 166.9) other class: (167.8, 172.4) same number of students in each class half-widths: 1.6 and 2.3 s₁√n = 1.61.96 ≈ 0.816 s₂√n = 2.31.96 ≈ 1.173 a higher confidence level needs a bigger z which makes both intervals wider 165.3 + 0.816z = 170.1 − 1.173z both intervals use the same z 1.989z = 4.8, so z ≈ 2.413 up from 1.96 at 95% x ≈ 98.4 P(−2.413 < Z < 2.413) ≈ 0.984

The other class has the wider interval, so it stretches further towards the touching point.

Now for the mathematics

Change the confidence level until the intervals meet

Each interval is $\bar{x}\pm z\frac{s}{\sqrt{n}}$. Move the slider to change the confidence level, which changes $z$ for both classes at once.

teacher’s class other class
z-score
for both classes
Teacher’s upper bound
165.3 + 0.816z
Other class lower bound
170.1 − 1.173z
Gap between them

QCAA marking guide · 7 marks

Work through the solution one mark at a time

Try each step yourself before you reveal it. The first four marks come from working backwards from the two 95% intervals.

Step 11 mark

Both of the given intervals are 95%, so you need the z-score that leaves 95% in the middle of the standard normal curve.

$$P(-z<Z<z)=0.95\quad\Rightarrow\quad z=1.96$$

You can also use $z\approx2$.

Marker: correctly determines the $z$-score associated with a 95% CI

Step 21 mark

A confidence interval is centred on the sample mean, so each mean is the midpoint of its interval.

$$\bar{x}_{1}=\frac{163.7+166.9}{2}=165.3\qquad\bar{x}_{2}=\frac{167.8+172.4}{2}=170.1$$

You keep follow-through marks from here if you slip later.

Marker: correctly determines the sample means for both classes

Step 31 mark

Let $s_{1}$ be the sample standard deviation for the teacher’s class, with sample size $n$. The upper bound is the mean plus the margin.

$$165.3+1.96\frac{s_{1}}{\sqrt{n}}=166.9\quad\Rightarrow\quad\frac{s_{1}}{\sqrt{n}}=\frac{1.6}{1.96}\approx0.816$$

You never need $s_{1}$ or $n$ on their own. You only need them combined as $\frac{s_{1}}{\sqrt{n}}$.

Marker: determines a relationship between the sample standard deviation and the sample size for the teacher’s class

Step 41 mark

Do the same for the other class, using its own standard deviation $s_{2}$ and the same $n$.

$$170.1+1.96\frac{s_{2}}{\sqrt{n}}=172.4\quad\Rightarrow\quad\frac{s_{2}}{\sqrt{n}}=\frac{2.3}{1.96}\approx1.173$$

The two classes have the same $n$, but you get a bigger value here because this class has a bigger spread of heights.

Marker: determines a relationship between the sample standard deviation and the sample size for the other class

Step 5 · insight mark ★1 mark

Let $z_{x}$ be the z-score for an $x\%$ interval. Both new intervals use the same $z_{x}$, and the upper bound of the teacher’s class equals the lower bound of the other class.

$$\bar{x}_{1}+z_{x}\frac{s_{1}}{\sqrt{n}}=\bar{x}_{2}-z_{x}\frac{s_{2}}{\sqrt{n}}$$

$$165.3+0.816z_{x}=170.1-1.173z_{x}$$

You get a plus on the left because it is an upper bound, and a minus on the right because it is a lower bound.

Marker: determines an equation in terms of the $z$-score associated with the new CIs using the data from the two classes

Step 61 mark

Collect the $z_{x}$ terms on one side and solve.

$$1.989z_{x}=4.8\quad\Rightarrow\quad z_{x}\approx2.413$$

If you rounded early and got $z_{x}\approx2.4$, you can still earn this mark.

Marker: determines the $z$-score associated with the new CI calculations

Step 71 mark

Now you turn the z-score back into a confidence level. It is the area between $-z_{x}$ and $z_{x}$.

$$x=100\times P(-2.413<Z<2.413)\approx98.4$$

$x\approx98.4$

Use normal cdf on your calculator with a lower bound of $-2.413$ and an upper bound of $2.413$. You can also give $98.3$ if you rounded early, and it is fine to write the % symbol.

Marker: determines the confidence level for the new CI calculations, rounded to one decimal place

Putting it all together

Each phrase of the question gave you something

“an approximate 95% confidence interval”
You know the original intervals use $z=1.96$.
“$(163.7,\ 166.9)$ … $(167.8,\ 172.4)$”
You get each sample mean from the midpoint and each margin from the half-width.
“the same number of students”
You can use one $n$ for both classes, so you only need $\frac{s_{1}}{\sqrt{n}}$ and $\frac{s_{2}}{\sqrt{n}}$.
“Using the same data … a confidence level of $x\%$”
The means and $\frac{s}{\sqrt{n}}$ values stay the same, and only $z$ changes.
“the upper bound … equalled the lower bound”
You set $\bar{x}_{1}+z_{x}\frac{s_{1}}{\sqrt{n}}$ equal to $\bar{x}_{2}-z_{x}\frac{s_{2}}{\sqrt{n}}$.
“rounded to one decimal place”
You give $x=98.4$.
What makes this complex unfamiliar

You are given two finished intervals but none of the sample means, standard deviations or the sample size. You have to work backwards from each interval to its mean and margin, then forwards again with a new z-score.

The detail I would highlight is that one confidence level means one z-score for both classes. The QCAA often ask you to reverse a confidence interval, so you need to be comfortable pulling $\bar{x}$ and $\frac{s}{\sqrt{n}}$ out of the two bounds.

Only mark this done when you could do it without help. Reading the solution does not count.

Back to the index

Question wording and marking-guide steps are from the 2023 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2023, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.