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QCAA Mathematical Methods — Interval estimates for proportions Questions & Answers

Q20

2020

QCAA

Paper 2

5 marks

Q20

5 marks

Assuming the approximate normality of sample proportions ( $\hat{p}_1$ and $\hat{p}_2$ ) and based on two independent samples, the approximate confidence interval for the difference of two proportions is given by

$\left( \hat{p}_1 - \hat{p}_2 - z \sqrt{\frac{\hat{p}_1 (1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2 (1 - \hat{p}_2)}{n_2}} , \hat{p}_1 - \hat{p}_2 + z \sqrt{\frac{\hat{p}_1 (1 - \hat{p}_1)}{n_1} + \frac{\hat{p}_2 (1 - \hat{p}_2)}{n_2}} \right)$

If the approximate confidence interval for the difference between two proportions does not contain 0, this provides evidence that the two proportions are not equal.

The data in the table shows the observed frequencies of two drink preferences for independent samples of people who live in Town A and Town B.

Town	Tea	Coffee	Total
A	111	105	216
B	150	107	257

Using the approximate 99% confidence interval for the difference of two proportions, determine if there is evidence to conclude that drink preference is associated with the town where the person lives.

Reveal Answer

$p_1 =$ proportion of Town A who prefer to drink tea
$p_2 =$ proportion of Town B who prefer to drink tea

The sample proportions are:
$\hat{p}_1 = \frac{111}{216}$
$\hat{p}_2 = \frac{150}{257}$

Using the 99% confidence interval for the difference of two proportions
$\left(\frac{111}{216} - \frac{150}{257} - 2.576\sqrt{\frac{\frac{111}{216}(1-\frac{111}{216})}{216} + \frac{\frac{150}{257}(1-\frac{150}{257})}{257}}, \frac{111}{216} - \frac{150}{257} + 2.576\sqrt{\frac{\frac{111}{216}(1-\frac{111}{216})}{216} + \frac{\frac{150}{257}(1-\frac{150}{257})}{257}}\right)$

$= (-0.188, 0.048)$

This interval contains zero; therefore, there is no evidence in the data to say that the two proportions are different, i.e. preference to drink tea does not depend on where the person lives.

Marking Criteria

Descriptor	Marks
correctly determines the sample proportions	1
establishes confidence interval for the difference of two proportions	1
determines 99% confidence interval	1
interprets 99% confidence interval to determine equality of proportions	1
shows logical organisation communicating key steps	1

Q6

2023

VCAA

Paper 1

4 marks

Q6

Let $\hat{P}$ be the random variable that represents the sample proportion of households in a given suburb that have solar panels installed.

From a sample of randomly selected households in a given suburb, an approximate 95% confidence interval for the proportion $p$ of households having solar panels installed was determined to be $(0.04, 0.16)$ .

Q6b

Use $z = 2$ to approximate the 95% confidence interval.

Q6a

1 mark

Find the value of $\hat{p}$ that was used to obtain this approximate 95% confidence interval.

Reveal Answer

$(0.04 + 0.16) \div 2 = 0.1$

$\hat{p} = 0.1$

Marking Criteria

Descriptor	Marks
Calculates the correct value of $\hat{p}$	1

Q6b

2 marks

Find the size of the sample from which this 95% confidence interval was obtained.

Reveal Answer

$0.06 = 2 \times \sqrt{\frac{0.1 \times 0.9}{n}}$

$0.03 = \sqrt{\frac{0.1 \times 0.9}{n}}$

$(0.03)^2 = \frac{0.1 \times 0.9}{n}$

$(0.03)^2 = \frac{9}{100n}$

$n = 100$

Marking Criteria

Descriptor	Marks
Sets up a correct equation involving $n$ using the margin of error or confidence interval bounds	1
Calculates the correct sample size, $n = 100$	1

Q6c

1 mark

A larger sample of households is selected, with a sample size four times the original sample.

The sample proportion of households having solar panels installed is found to be the same.

By what factor will the increased sample size affect the width of the confidence interval?

Reveal Answer

Confidence interval width is halved (reduced or decreased by a factor of 2; altered by a factor of $\frac{1}{2}$ ).

Marking Criteria

Descriptor	Marks
States the correct factor by which the width is affected (e.g., $\frac{1}{2}$ , halved, or decreased by a factor of 2)	1

Q15

2024

QCAA

Paper 1

4 marks

Q15

A survey was conducted to understand whether people support a new policy.
Using a z-score of 2, the approximate confidence interval for the population proportion of people who support the policy was calculated as $\left(\frac{3}{10}, \frac{7}{10}\right)$ .

Q15a

1 mark

Determine the margin of error.

Reveal Answer

The confidence interval corresponds to $(\hat{p}-E, \hat{p}+E)$ , where E is the margin of error about $\hat{p}$ .
$\hat{p} + E = \frac{7}{10}$
$\hat{p} - E = \frac{3}{10}$
subtracting:
$\therefore 2E = \frac{7}{10} - \frac{3}{10} = \frac{4}{10}$
$\therefore E = \frac{2}{10} = \frac{1}{5}$

Marking Criteria

Descriptor	Marks
Correctly determines the margin of error	1

Q15b

3 marks

Determine the number of people surveyed.

Reveal Answer

$\hat{p} + E = \frac{7}{10}$
$\hat{p} + \frac{2}{10} = \frac{7}{10}$
$\hat{p} = \frac{5}{10} = \frac{1}{2}$
Upper CI value = $\hat{p} + z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$
$\frac{7}{10} = \frac{1}{2} + 2\sqrt{\frac{\frac{1}{2}\left(1-\frac{1}{2}\right)}{n}}$
$\frac{1}{5} = 2\sqrt{\frac{\frac{1}{2}\left(1-\frac{1}{2}\right)}{n}}$
$\frac{1}{5} = 2\sqrt{\frac{\frac{1}{4}}{n}}$
$\frac{1}{5} = 2\frac{\sqrt{\frac{1}{4}}}{\sqrt{n}}$
$\frac{1}{5} = 2\frac{\frac{1}{2}}{\sqrt{n}}$
$\frac{1}{5} = \frac{1}{\sqrt{n}}$
$\frac{1}{10} = \sqrt{\frac{1}{4n}}$
$\frac{1}{100} = \frac{1}{4n}$
$100 = 4n$
$n = 25$
25 people were surveyed.

Marking Criteria

Descriptor	Marks
Determines the value of the sample proportion $\hat{p}$	1
Substitutes $\hat{p}$ and z-score into the confidence interval formula	1
Determines the number of people surveyed	1

Q10

2022

VCAA

Paper 2

1 mark

Q10

1 mark

An organisation randomly surveyed 1000 Australian adults and found that 55% of those surveyed were happy with their level of physical activity.

An approximate 95% confidence interval for the percentage of Australian adults who were happy with their level of physical activity is closest to

A

$(4.1, 6.9)$

B

$(50.9, 59.1)$

C

$(52.4, 57.6)$

D

$(51.9, 58.1)$

E

$(45.2, 64.8)$

Reveal Answer

A

$(4.1, 6.9)$

This interval is incorrect because it does not center around the sample proportion of 55%.

B

$(50.9, 59.1)$

This interval is too wide. It corresponds to an approximate 99% confidence interval, which uses a z-score of 2.576 instead of 1.96.

C

$(52.4, 57.6)$

This interval is too narrow. It corresponds to an approximate 90% confidence interval, which uses a z-score of 1.645 instead of 1.96.

D

$(51.9, 58.1)$

Correct Answer

This is correct. The 95% confidence interval is calculated using $\hat{p} \pm 1.96\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ , which gives $0.55 \pm 1.96\sqrt{\frac{0.55(0.45)}{1000}} \approx 0.55 \pm 0.031$ , or $(51.9\%, 58.1\%)$ .

E

$(45.2, 64.8)$

This interval is much too wide. It represents the 95% confidence interval if the sample size was $n=100$ instead of $n=1000$ .

Q3

2021

VCAA

Paper 2

1 mark

Q3

1 mark

A box contains many coloured glass beads.

A random sample of 48 beads is selected and it is found that the proportion of blue-coloured beads in this sample is 0.125

Based on this sample, a 95% confidence interval for the proportion of blue-coloured glass beads is

A

(0.0314, 0.2186)

B

(0.0465, 0.2035)

C

(0.0018, 0.2482)

D

(0.0896, 0.1604)

E

(0.0264, 0.2136)

Reveal Answer

A

(0.0314, 0.2186)

Correct Answer

This is correct. The 95% confidence interval is calculated using the formula $\hat{p} \pm z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ , where $\hat{p} = 0.125$ , $n = 48$ , and the critical value $z \approx 1.96$ .

B

(0.0465, 0.2035)

This is incorrect. This interval uses a z-score of approximately 1.645, which corresponds to a 90% confidence interval rather than the requested 95%.

C

(0.0018, 0.2482)

This is incorrect. This interval uses a z-score of approximately 2.58, which calculates a 99% confidence interval instead of a 95% confidence interval.

D

(0.0896, 0.1604)

This is incorrect. This interval is much too narrow, likely resulting from an error in calculating the standard error or omitting the z-score multiplier.

E

(0.0264, 0.2136)

This is incorrect. This interval does not match the correct calculation for a 95% confidence interval and likely stems from an arithmetic error.

Q8

2023

QCAA

Paper 1

1 mark

Q8

1 mark

A sample of size $n$ was used to estimate a population proportion. An approximate margin of error of 3% was calculated using $z = 1.96$ . Given the sample proportion was 0.6, determine $n$ .

A

$n = \frac{\left( \frac{0.03}{1.96} \right)^2}{0.24}$

B

$n = \frac{0.24}{\left( \frac{0.03}{1.96} \right)^2}$

C

$n = \frac{\left( \frac{0.03}{1.96} \right)^2}{2.4}$

D

$n = \frac{2.4}{\left( \frac{0.03}{1.96} \right)^2}$

Reveal Answer

A

$n = \frac{\left( \frac{0.03}{1.96} \right)^2}{0.24}$

This option incorrectly inverts the formula. The sample size $n$ is calculated by dividing the variance estimate $\hat{p}(1-\hat{p})$ by the squared term $\left(E/z\right)^2$ , not the other way around.

B

$n = \frac{0.24}{\left( \frac{0.03}{1.96} \right)^2}$

Correct Answer

Rearranging the margin of error formula $E = z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ to solve for $n$ gives $n = \frac{\hat{p}(1-\hat{p})}{(E/z)^2}$ . With $\hat{p}=0.6$ , the numerator is $0.6(0.4) = 0.24$ .

C

$n = \frac{\left( \frac{0.03}{1.96} \right)^2}{2.4}$

This option incorrectly inverts the fraction and uses the wrong value for the product $\hat{p}(1-\hat{p})$ . The calculation $0.6 \times 0.4$ equals $0.24$ , not $2.4$ .

D

$n = \frac{2.4}{\left( \frac{0.03}{1.96} \right)^2}$

While the structure of the formula is correct, the numerator is calculated incorrectly. The value of $\hat{p}(1-\hat{p})$ is $0.6 \times 0.4 = 0.24$ , not $2.4$ .

Q19

2021

QCAA

Paper 1

4 marks

Q19

4 marks

A firm aims to have 95% confidence in estimating the proportion of office workers who respond to an email in less than an hour to within $\pm 0.05$ .

A survey has never been undertaken before, so no past data is available.

The firm believes that if the proportion is 0.5, then this will result in the largest variability in the sample proportion.

Based on this, determine the sample size needed using the approximate value of $z = 2$ for the 95% confidence interval.

Justify the choice of 0.5 for the proportion.

Reveal Answer

interval margin $= z \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}}$

$0.05 = 2 \sqrt{\frac{0.5(1-0.5)}{n}}$

rearranging
$n = \frac{2^2 \times 0.5(0.5)}{(0.05)^2}$
$n = 400$

The largest sample size will result when $\hat{p}(1 - \hat{p})$ is maximised in the numerator, therefore generating largest $n$ value.

Maximum occurs at $\hat{p} = 0.5$

Marking Criteria

Descriptor	Marks
correctly selects the interval margin formula	1
substitutes values into the formula	1
determines sample size n	1
verifies the firm's decision to use $\hat{p} = 0.5$ using mathematical reasoning	1

Q3

2025

QCAA

Paper 2

1 mark

Q3

1 mark

A streaming service surveyed a random sample of 100 customers, and 60 customers said that they prefer to watch movies rather than miniseries.

Based on these results, the 95% confidence interval for the proportion of customers who prefer to watch movies was (0.5, 0.7).

Which statement shows the correct interpretation of this outcome?

A

The streaming service can be 95% confident that the proportion of customers who prefer to watch movies is between 50% and 70%.

B

There is a probability between 50% and 70% that most customers prefer to watch movies 95% of the time.

C

60% of the sampled customers prefer to watch movies between 50% and 70% of the time.

D

60 customers in the random sample prefer to watch movies 95% of the time.

Reveal Answer

A

The streaming service can be 95% confident that the proportion of customers who prefer to watch movies is between 50% and 70%.

Correct Answer

This is the correct interpretation. A 95% confidence interval means we can be 95% confident that the true population proportion lies within the calculated interval, which is 50% to 70%.

B

There is a probability between 50% and 70% that most customers prefer to watch movies 95% of the time.

This is incorrect because a confidence interval estimates the true population proportion, not the probability or frequency of how often customers prefer movies.

C

60% of the sampled customers prefer to watch movies between 50% and 70% of the time.

This is incorrect. The interval (0.5, 0.7) represents the estimated range for the true population proportion, not the percentage of time individuals prefer to watch movies.

D

60 customers in the random sample prefer to watch movies 95% of the time.

This is incorrect because the 95% refers to the confidence level of the interval estimating the population proportion, not the percentage of time customers prefer movies.

Q12

2025

QCAA

Paper 1

4 marks

Q12

4 marks

A teacher conducted a survey about student involvement in school activity groups. They found that of a random sample of 25 students, five students were members of the environment group.

Determine the approximate confidence interval for the proportion of students at the school who are members of the environment group, using $z = 1$ . Express your answer using fractions in simplified form.

Reveal Answer

$n = 25$
$\hat{p} = \frac{5}{25} = \frac{1}{5}$

\begin{align*} \hat{p} - z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} &= \frac{1}{5} - \sqrt{\frac{\frac{1}{5}\left(1-\frac{1}{5}\right)}{25}}\\ &= \frac{1}{5} - \sqrt{\frac{4}{25^2}}\\ &= \frac{1}{5} - \frac{\sqrt{4}}{25}\\ &= \frac{1}{5} - \frac{2}{25}\\ &= \frac{5}{25} - \frac{2}{25}\\ &= \frac{3}{25} \end{align*}

Similarly the upper bound is $\frac{7}{25}$

Confidence interval $\left(\frac{3}{25}, \frac{7}{25}\right)$

Marking Criteria

Descriptor	Marks
correctly determines the sample proportion	1
substitutes the sample size, sample proportion and z-value into at least one end of approximate confidence interval formula	1
determines the approximate margin of error	1
determines the lower and upper end of the approximate confidence interval	1

Q8

2022

QCAA

Paper 1

1 mark

Q8

1 mark

In a survey, 80 respondents exercised daily, while 120 did not. When calculating the approximate 95% confidence interval for the proportion of people who exercise daily, the margin of error is

A

$1.96\sqrt{\frac{0.4(1-0.4)}{200}}$

B

$0.95\sqrt{\frac{0.4(1-0.4)}{200}}$

C

$1.96\sqrt{\frac{0.67(1-0.67)}{120}}$

D

$0.95\sqrt{\frac{0.67(1-0.67)}{120}}$

Reveal Answer

A

$1.96\sqrt{\frac{0.4(1-0.4)}{200}}$

Correct Answer

This is the correct formula for the margin of error. The sample proportion is $\hat{p} = \frac{80}{80+120} = 0.4$ , the total sample size is $n=200$ , and the critical value ( $z^*$ ) for a 95% confidence interval is approximately 1.96.

B

$0.95\sqrt{\frac{0.4(1-0.4)}{200}}$

This option incorrectly uses the confidence level (0.95) as the multiplier instead of the critical $z$ -score. For a 95% confidence interval, the critical value is 1.96.

C

$1.96\sqrt{\frac{0.67(1-0.67)}{120}}$

This option uses an incorrect sample size and proportion. The denominator must be the total number of respondents ( $n=200$ ), not just those who did not exercise ( $120$ ), and the proportion is calculated relative to that total.

D

$0.95\sqrt{\frac{0.67(1-0.67)}{120}}$

This option uses the wrong multiplier (0.95 instead of 1.96) and incorrect values for the sample size and proportion. The sample size should be the total number of respondents ( $n=200$ ).

Q18

2025

QCAA

Paper 2

5 marks

Q18

5 marks

The results of an employee satisfaction survey of 500 employees at a large company are presented to board members.

The results include a 95% confidence interval for the proportion of satisfied employees. The lower end of the confidence interval is 0.648.

A board member would like to use the survey results to make the claim that the proportion of the satisfied employees in the entire company is larger than 75%.

Evaluate the reasonableness of the claim.

Reveal Answer

Set up the 95% confidence interval:

$\left( \hat{p} - z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \hat{p} + z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \right)$

$\left( 0.648, \hat{p} + z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \right)$

Use a GDC with a 95% central area, $z = -1.960$ and $1.960$

The lower cut-off of the 75% CI is given by:

0.648 = \hat{p} - 1.960\sqrt{\frac{\hat{p}(1-\hat{p})}{500}}

Solving for $\hat{p}$ using a GDC:

\begin{align*} 0.648 &= \hat{p} - 1.960\sqrt{\frac{\hat{p}(1-\hat{p})}{500}}\\ \hat{p} &= 0.6886 \end{align*}

Substituting $\hat{p}$ into the upper CI formula:

\begin{align*} &= \hat{p} + z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\\ &= 0.6886 + 1.960\sqrt{\frac{0.6886(1-0.6886)}{500}}\\ &= 0.729 \end{align*}

The 95% CI is therefore: $(0.648, 0.729)$

The survey suggests we can be 95% confident the population proportion is between 0.648 and 0.729, i.e. between 64.8% and 72.9%.

The claim of 75% for the population proportion is outside of the 95% confidence interval in the survey, so the claim is not reasonable.

Marking Criteria

Descriptor	Marks
correctly determines the z-score for a 95% confidence interval	1
determines an equation involving the lower cut-off of the CI formula	1
determines the sample proportion for the survey results	1
determines the upper end of the 95% confidence interval	1
determines if the claim is reasonable	1

Q3

2021

QCAA

Paper 2

1 mark

Q3

1 mark

A random sample of people were surveyed about the most important factor when deciding where to shop. The results appear in the table.

Factor	Percentage (%)
Price	40
Quality of merchandise	30
Service	15
Shopping environment	15

If the sample size was 1200, the approximate 95% confidence interval for the proportion of people who identified price as the most important factor is

A

(0.395, 0.405)

B

(0.386, 0.414)

C

(0.377, 0.423)

D

(0.372, 0.428)

Reveal Answer

A

(0.395, 0.405)

This interval is too narrow and does not account for the standard error associated with the sample size. It implies a margin of error of only $0.005$ , which is mathematically incorrect for these parameters.

B

(0.386, 0.414)

This interval represents approximately a 68% confidence level. It uses a margin of error of 1 standard error ( $1 \times SE$ ) instead of the $1.96 \times SE$ required for 95% confidence.

C

(0.377, 0.423)

This interval corresponds to a 90% confidence level. It uses a critical value of $Z \approx 1.645$ instead of the $Z \approx 1.96$ required for a 95% confidence interval.

D

(0.372, 0.428)

Correct Answer

Using the formula $\hat{p} \pm 1.96 \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ with $\hat{p}=0.40$ and $n=1200$ , the margin of error is $\approx 0.028$ , resulting in the interval $(0.372, 0.428)$ .

Q10

2020

QCAA

Paper 1

1 mark

Q10

1 mark

Two types of material (A and B) are being tested for their ability to withstand different temperatures. A random selection of both materials was subjected to extreme temperature changes and then classified according to their condition after they were removed from the testing facility. The results are shown in the table.

	Material
	A	B
Broke completely	25	43
Showed defects	35	38
Remained intact	35	24
Total	95	105

An approximate 95% confidence interval for the probability that material A will break completely or show defects is given by

$\left( c - 1.96\sqrt{\frac{c(1-c)}{n}}, c + 1.96\sqrt{\frac{c(1-c)}{n}} \right)$

The values of $c$ and $n$ are

A

$\frac{60}{95}$ and 95

B

$\frac{60}{200}$ and 95

C

$\frac{140}{200}$ and 95

D

$\frac{60}{200}$ and 200

Reveal Answer

A

$\frac{60}{95}$ and 95

Correct Answer

This is the correct answer. The sample size $n$ is the total number of Material A items tested ( $95$ ). The proportion $c$ is the sum of Material A items that broke completely ( $25$ ) or showed defects ( $35$ ) divided by the sample size, resulting in $c = \frac{25+35}{95} = \frac{60}{95}$ .

B

$\frac{60}{200}$ and 95

This option incorrectly calculates the proportion $c$ by using the grand total of all materials ( $200$ ) in the denominator. Since the question asks specifically about the probability for Material A, the denominator must be the total count of Material A ( $95$ ).

C

$\frac{140}{200}$ and 95

This option uses incorrect values for the proportion. It appears to use the total sample size ( $200$ ) in the denominator and likely aggregates defects across both materials, rather than restricting the calculation to the specific outcomes and sample size of Material A.

D

$\frac{60}{200}$ and 200

This option incorrectly identifies the sample size $n$ as the total for both materials ( $200$ ). Since the inference is about Material A specifically, $n$ should be the total count of Material A ( $95$ ).

Q17

2020

QCAA

Paper 2

4 marks

Q17

4 marks

In a survey of 326 lecturers, 303 said that on at least one occasion a mobile phone had rung in a lecture they were giving.

Determine the sample size required to conduct a follow-up survey that provides 95% confidence that this one-point estimate is correct to within $\pm 0.02$ of the population proportion.

Reveal Answer

$\hat{p} = \frac{303}{326}$

Using confidence interval formula (margin of error)
$0.02 = 1.96 \times \sqrt{\frac{\frac{303}{326} \times \frac{23}{326}}{n}}$
Using GDC

$n = 629.778$

$\therefore$ a sample of 630 lecturers would be required.

Marking Criteria

Descriptor	Marks
Correctly determines the sample proportion	1
Establishes equation in $n$	1
Determines $n$	1
Determines reasonable value of $n$	1

Q8

2025

VCAA

Paper 2

1 mark

Q8

1 mark

A random sample of $n$ Victorian households is taken to estimate the proportion of all Victorian households that have vegetable gardens. The approximate 95% confidence interval calculated using this sample is $(0.248, 0.552)$ , correct to three decimal places.

The number of households, $n$ , in the sample is

A

$10$

B

$28$

C

$40$

D

$49$

Reveal Answer

A

$10$

Incorrect. Substituting $n=10$ into the margin of error formula $1.96\sqrt{0.4(0.6)/n}$ yields approximately $0.304$ , which is much larger than the required margin of error of $0.152$ .

B

$28$

Incorrect. Substituting $n=28$ into the margin of error formula yields approximately $0.181$ , which does not match the interval's actual margin of error of $0.152$ .

C

$40$

Correct Answer

Correct. The sample proportion is the midpoint $\hat{p} = 0.4$ and the margin of error is $0.152$ . Solving the margin of error formula $1.96\sqrt{0.4(0.6)/n} = 0.152$ gives $n \approx 40$ .

D

$49$

Incorrect. Substituting $n=49$ into the margin of error formula yields approximately $0.137$ , which is smaller than the required margin of error of $0.152$ .

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