Confidence Intervals for a Proportion

An inspector takes a random sample of $32$ tennis balls and determines a confidence interval for the population proportion of grade A balls produced. The confidence interval is $(0.7382, 0.9493)$, correct to four decimal places. Find the level of confidence that the population proportion of grade A balls is within the interval, as a percentage correct to the nearest integer.

1. 1

   The sample proportion is the midpoint of the interval, and the margin of error is half its width.

   $$\hat{p} = \frac{0.7382 + 0.9493}{2} \approx 0.84375, \quad E \approx 0.10555$$
2. 2

   Use $E = z\sqrt{\hat{p}(1-\hat{p})/32}$ and solve for $z$.

   $$z = \frac{E}{\sqrt{\hat{p}(1-\hat{p})/32}} \approx 1.645$$
3. 3

   A $z$ value of about $1.645$ corresponds to a $90\%$ confidence level. The examiner report notes many students computed $\hat{p}$ incorrectly or simply gave $95\%$.

   $$\text{confidence level} = 90\%$$

VCAA 2023 Mathematical Methods Exam 2, Section B Q4g

In one random sample of $50$ pieces of luggage, $10$ are labelled heavy. (i) Use this sample to find an approximate $90\%$ confidence interval for $p$, the population proportion of luggage labelled heavy, correct to three decimal places. (ii) A second random sample of $50$ pieces is selected, and its approximate $90\%$ confidence interval for $p$ is wider than the one in part (i). State the minimum and maximum possible number of pieces of luggage labelled heavy in the second sample.

1. 1

   (i) The sample proportion is $\hat{p} = \frac{10}{50} = 0.2$, and for $90\%$ confidence $z = 1.6449$.

   $$0.2 \pm 1.6449\sqrt{\frac{0.2 \times 0.8}{50}} = (0.107, 0.293)$$
2. 2

   (ii) The width is proportional to $\sqrt{\hat{p}(1-\hat{p})}$, so a wider interval needs $\hat{p}(1-\hat{p}) > 0.16$, i.e. $\hat{p}$ strictly between $0.2$ and $0.8$.

   $$0.2 < \hat{p} < 0.8 \implies 10 < X < 40$$
3. 3

   So the heavy count $X$ ranges from $11$ to $39$. The report notes many found the minimum $11$ but gave a wrong maximum such as $50$.

   $$\text{minimum } X = 11, \quad \text{maximum } X = 39$$

VCAA 2024 Mathematical Methods Exam 2, Section B Q4e

A polling company surveys $n = 200$ randomly chosen voters and finds $50$ support a new policy. Construct an approximate $95\%$ confidence interval for the population proportion $p$, correct to three decimal places.

1. 1

   Find the sample proportion $\hat{p}$ from $50$ out of $200$.

   $$\hat{p} = \frac{50}{200} = 0.25$$
2. 2

   Find the standard error using $\sqrt{\hat{p}(1-\hat{p})/n}$.

   $$\sqrt{\frac{0.25 \times 0.75}{200}} = \sqrt{0.0009375} \approx 0.030619$$
3. 3

   For $95\%$ confidence the $z$ value is $1.96$. The margin of error is $z$ times the standard error.

   $$1.96 \times 0.030619 \approx 0.060$$
4. 4

   The interval is $\hat{p}$ plus or minus the margin of error.

   $$0.25 \pm 0.060$$

In a sample of $n = 100$ households, the proportion owning an electric vehicle is $\hat{p} = 0.4$. Find the margin of error for an approximate $90\%$ confidence interval, then state the interval. Use $z = 1.645$.

1. 1

   Write the standard error with $\hat{p} = 0.4$ and $n = 100$.

   $$\sqrt{\frac{0.4 \times 0.6}{100}} = \sqrt{0.0024} \approx 0.048990$$
2. 2

   The margin of error is $z$ times the standard error, with $z = 1.645$ for $90\%$.

   $$M = 1.645 \times 0.048990 \approx 0.081$$
3. 3

   Build the interval as $\hat{p} \pm M$.

   $$0.4 \pm 0.081$$

A researcher wants a $95\%$ confidence interval with a margin of error no larger than $0.04$. Using the safest case $\hat{p} = 0.5$, find the smallest sample size $n$ needed. Use $z = 1.96$.

1. 1

   Start from the margin of error formula and set it to be at most $0.04$.

   $$1.96 \sqrt{\frac{0.5 \times 0.5}{n}} \le 0.04$$
2. 2

   Rearrange to make $n$ the subject.

   $$n \ge \left(\frac{1.96}{0.04}\right)^2 \times 0.25$$
3. 3

   Evaluate the right hand side.

   $$n \ge 49^2 \times 0.25 = 600.25$$
4. 4

   Sample size must be a whole number, and you must round up so the margin stays under the limit.

   $$n = 601$$