How many QCAA Mathematical Methods questions cover Interval estimates for proportions?

AusGrader has 66 QCAA Mathematical Methods questions on Interval estimates for proportions, all with instant AI grading and detailed marking feedback.

QCAA (QCE) Mathematical Methods — Interval estimates for proportions Questions & Answers | AusGrader

Q3

2024

QCAA

Paper 1

1 mark

Q3

1 mark

A sample of size $n$ can be used to obtain a sample proportion $\hat{p}$ .
An approximate margin of error for the population proportion can be obtained using the formula

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

If the level of confidence is increased from 95% to 99%, then

A

the associated z-value would decrease, so $E$ would increase.

B

the associated z-value would increase, so $E$ would increase.

C

the associated z-value would decrease, so $E$ would decrease.

D

the associated z-value would increase, so $E$ would decrease.

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

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

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.

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

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.

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.

$\hat{p} + E = \frac{7}{10}$
$\hat{p} + \frac{2}{10} = \frac{7}{10}$
$\hat{p} = \frac{5}{10} = \frac{1}{2}$
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

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.

$p_1 =$ 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

QCAA Mathematical Methods Interval estimates for proportions

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