SCSA Mathematics Methods Interval estimates for proportions

The sample proportion of heads is given by

$\hat{p} = \frac{30}{50} = 0.6$

Hence, the 90% confidence interval is

$0.6 - 1.645\sqrt{\frac{0.6 \times 0.4}{50}} \le p \le 0.6 + 1.645\sqrt{\frac{0.6 \times 0.4}{50}}$

$0.4860 \le p \le 0.7140$

$X \sim \text{Bin}(20, 0.9)$

The expected value of $X$ is given by

$E(X) = 20 \times 0.9$

$= 18$

The variance of $X$ is given by

$Var(X) = 20 \times 0.9 \times 0.1$

$= 1.8$

If three confidence intervals did not contain the true proportion, then 17 did contain the true proportion.

$\text{P}(X = 17) = 0.1901$

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.

We are given that $X \sim \text{N}(50, 10^2)$. Hence

$\text{P}(X \ge 64) = 0.0808$

and so, we expect 8.08% of Australian 15-year-old students to obtain a score of at least 64.

Solving for $\text{P}(X \ge x) = 0.01$ yields $x = 73.26\dots$. Hence the minimum score required for a student to be in the top 1% is 73.27 (or 74 if reporting only integer scores).

$\text{P}(43 \le X \le 57) = 0.5161$

The sample proportion $\hat{p}$ of students classified as 'average' has a mean of

$\mu = p = 0.5161$

and standard deviation

$\sigma = \sqrt{\frac{0.5161(1 - 0.5161)}{50}} = 0.0707$

Hence, we use the approximate distribution $\hat{p} \sim \text{N}(0.5161, 0.0707^2)$. The probability that $\hat{p}$ is no more than 0.46 given by

$\text{P}(\hat{p} < 0.46) = 0.2137$

Let $Z \sim \text{N}(0, 1)$. Given that

$\text{P}(Y \le 82) = \text{P}(Z \le -1.2) = 0.1151$

and

$\text{P}(Y \ge 130) = \text{P}(Z \ge 2) = 0.0228$

It follows that

$-1.2 = \frac{82 - \mu}{\sigma}$ and $2 = \frac{130 - \mu}{\sigma}$

where $\mu$ and $\sigma$ are the mean and standard deviation of $Y$ respectively. Solving these two equations yields

$\mu = 100$

$\sigma = 15$

Given that $Y = aX + b$ it follows from the means and standard deviations of $X$ and $Y$ that

$100 = 50a + b$

$15 = 10a$

Hence $a = 1.5$ and $b = 25$.