SCSA Mathematics Methods Interval estimates for proportions

Exclusion bias: Athletes with unexplained respiratory symptoms who do not seek medical intervention, or who do not identify themselves as athletes to their doctor, will not be represented in the sample.

The sample proportion
$\hat{p} = \frac{25}{71}$
Hence the 95% confidence interval is given by
$95\% \text{ CI} = \left( \frac{25}{71} - 1.96\sqrt{\frac{\frac{25}{71}\left(1-\frac{25}{71}\right)}{71}}, \ \frac{25}{71} + 1.96\sqrt{\frac{\frac{25}{71}\left(1-\frac{25}{71}\right)}{71}} \right)$
$= (0.2410, 0.4632)$

The margin of error is given by
$E = \frac{0.4632 - 0.2410}{2} = 0.1111$

The margin of error would decrease if the sample size is increased.

The margin of error would increase if the confidence level is increased.

Since the sample proportion is less than 0.5 (0.5 is the sample proportion yielding the largest margin of error), a decrease in sample proportion would move the margin of error away from being maximum. Hence the margin of error would decrease.

We require
$0.04 \ge 1.96\sqrt{\frac{0.5(1-0.5)}{n}}$
$\Rightarrow \left(\frac{0.04}{1.96}\right)^2 \ge \frac{0.25}{n}$
$\Rightarrow n \ge 0.25\left(\frac{1.96}{0.04}\right)^2$
$= 600.25$
Since $n$ must be an integer the minimum value of the sample size is $n = 601$.

Yes. The claimed American proportion is outside the confidence interval for Australian athletes, so it is possible to conclude that the proportions are different at the 95% level of confidence.

Analytical procedure:

$\sigma=\sqrt{\dfrac{p(1-p)}{n}}$ (formula book)

$0.04=\sqrt{\dfrac{p(1-p)}{120}}$

$0.0016=\dfrac{p(1-p)}{120}$

$0.192=p(1-p)$

$0.192=p-p^2$

$p^2-p+0.192=0$

Use GDC equation solver or graph $y=0.04$ and the
square root function to find intersections:

$p=0.25921$

$=26\%$

or

$p=0.7408$

$=74\%$

Both population proportions (26% and 74%) are less than
80%.

Therefore, the bus route proposal would not be discussed
at a council meeting.

Using GDC to determine confidence interval associated with $n = 200, \hat{p} = 0.25, z = 1.96$

$(0.19, 0.31)$

Combining results
$n = 450, \hat{p} = \frac{11}{45}$

Using GDC
$(0.2047, 0.2842)$

By combining the results, the sample size is increased and the confidence interval width is reduced.
The new sample statistic provides a better estimate for the population parameter.

Using approximation to the normal distribution
Mean $= 0.24$
Standard deviation $= \sqrt{\frac{0.24 \times 0.76}{200}} = 0.0302$
Using GDC
$P(\hat{p} > 0.30) = 0.0235$

$p_1 =$ 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.

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

$\hat{p} = 0.1$

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

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

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

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

$n = 100$

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

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:
$\therefore 2E = \frac{7}{10} - \frac{3}{10} = \frac{4}{10}$
$\therefore E = \frac{2}{10} = \frac{1}{5}$

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

$n > \left( \frac{1.96\sqrt{0.5 \times 0.5}}{0.01} \right)^2 = 9604$

$\varepsilon = 2.5758 \times \sqrt{\frac{0.5 \times 0.5}{500}} = 0.058$
that is, within 5.8%

1. The sample is at a fixed time, so only people around at that time will be sampled.
2. The location is fixed, so:
   (i) only people at that location will be sampled or
   (ii) not everyone from the suburb will pass by that area, so this is not a random sample of the residents.

$\hat{p} = \frac{102}{121} = 0.84$

Using normal distribution
$\mu = 0.9$
$\sigma = \sqrt{\frac{0.9 \times 0.1}{121}}$

Using GDC
$P(\hat{p} < 0.85)$
$= 0.0334$

There is 3.34% chance of shipping less than 85% of orders when the population parameter is 0.90.
Observing a sample proportion of 0.843 (or even lower) would have occurred by chance less than 3.34% of the time if the retailer's claim is true. Therefore, we suspect that the retailer's claim is dubious.

Answers could include:

• the sample only includes books from the newest printing press, which may perform differently to the other three presses
• the sample was gathered over a single narrow time period, so books printed later in the week are not included (this may involve different operators and the performance of the presses might change during their use over the week).

The sample could be randomly selected:

• from books printed across the entire duration of the print run
• from all printing presses.

The distribution of $\hat{p}$ can be approximated as

$\hat{p} \sim \text{N}(0.05, 0.0002375)$

Hence,

$\text{P}(\hat{p} < 0.04) = 0.2582$

A 95% confidence interval is given by

$95\% \text{ CI} = (0.1 - 0.024, 0.1 + 0.024) = (0.0760, 0.1240)$

The proportion of books with errors claimed by the publisher was $\frac{10}{200} = 0.05$. This proportion is not within the confidence interval and so there is sufficient evidence to conclude that the claim of the publisher is incorrect at the 95% confidence level.

The margin of error could be decreased by

• increasing the size of the sample
• decreasing the confidence level.

For the worst-case scenario set $\hat{p} = 0.5$. Solving for a margin of error equal to 0.02 gives

$0.02 = 1.96\sqrt{\frac{0.5(1 - 0.5)}{n}}$

$\Rightarrow n = 2401$

A sample size of at least 2401 would ensure the margin of error is at most 0.02.

$\hat{p} = \frac{17}{50}$
$= 0.34$

Using the formula:
Variance $= \frac{\hat{p}(1-\hat{p})}{n}$ to estimate variance
$= \frac{0.34(1-0.34)}{50}$
$= 0.004488$
Standard deviation $= 0.066993$

Using GDC InvN
Area $= 0.95$
Std dev $= \sqrt{0.004488}$
$\mu = 0.34$

CI $= (0.20869, 0.47130)$

From Q11b) we are 95% confident that the proportion of people using public transport is approximately between 21% and 47%. 50% is outside of this range.
The claim is not supported.

The sample proportion has distribution
$\hat{p} \sim N\left(0.35, \frac{0.35(1 - 0.35)}{100}\right)$
$\hat{p} \sim N(0.35, 0.002275)$
Hence
$P(\hat{p} \le 0.28) = 0.0711$

• Single location: only sampling people who visit a particular store. People who live a long way from the store are less likely to shop there than people who live locally, and so are less likely to be included/represented in the sample
• Single time: only sampling between 9am and 10am on Tuesday. People who have 9am–5pm work commitments are less likely to shop at this time than people who do not work these hours and so are less likely to be included/represented in the sample.

$\hat{p} = \frac{0.3488 + 0.2520}{2}$
$= 0.3004$
Hence the number of people who purchase free-range eggs is
$n = 243 \times 0.3004$
$= 73$

The margin of error is
$E = \frac{0.3488 - 0.2520}{2}$
$= 0.0484$
Hence we have
$0.0484 = k\sqrt{\frac{0.3004(1 - 0.3004)}{243}}$
$\Rightarrow k = 1.6458$
Then
$P(-1.6458 \le Z \le 1.6458) = 0.9002$
Hence the confidence level was 90%.