6.4.3 Confidence Intervals - GCEALevelMaths9709

In this topic we will learn how to:

Determine and interpret a confidence interval for a population mean in cases where the population is normally distributed with known variance or where a large sample is used.
Determine, from a large sample, an approximate confidence interval for a population proportion.

A confidence interval represents the range values between which we believe the true value of the population parameter lies to a certain confidence level. The confidence level represents how likely the parameter is to lie within this range. For example, When you calculate a $95\%$ confidence interval, the idea is that if $100$ confidence intervals were to be set up, $95$ of them would contain the true value of the unknown population parameter, and $5$ would not contain the unknown population parameter.

$\textbf{\textcolor{gray}{Confidence Interval for normal distribution or}}\\ \textbf{\textcolor{gray}{ large sample}}$ To calculate the confidence interval when the population, $X$ , is normally distributed or when we have a large sample, we use the formula,
$\overline{x} \pm z\left(\sqrt{\frac{\sigma^{2}}{n}}\right)$ Where $\overline{x}$ is the sample mean, $z$ is a $z$ -value from the normal distribution table, $\sigma^{2}$ is variance, $n$ is the sample size.
To calculate the interval width we use the formula,
$2 \times z\left(\sqrt{\frac{\sigma^{2}}{n}}\right)$ $\textbf{\textcolor{gray}{Confidence Interval for a population propotion}}$
To calculate the confidence interval for a population proportion we use the formula,
$p_{s} \pm z\left(\sqrt{\frac{p_{s} \times (1 - p_{s})}{n}}\right)$
Where $p_{s}$ is the sample proportion, $z$ is a $z$ -value from the normal distribution table, $n$ is the sample size.
To calculate the interval width we use the formula,
$2 \times z\left(\sqrt{\frac{p_{s} \times (1 - p_{s})}{n}}\right)$ Note: In some literature $1- p_{s}$ may be denoted as $q_{s}$ .

Let’s look at some past paper questions on confidence intervals.

1. A javelin thrower noted the lengths of $50$ of her throws. The sample mean was $72.3$ m and an unbiased estimate of the population variance was $64.3$ m $^{2}$ . Find a $92\%$ confidence interval for the population mean length of throws by this athlete. (9709/62/M/J/22 number 1)

Let’s identify all the information we have been given by the question,
$\overline{x} = 72.3\ \ \ \ s^{2} = 64.3\ \ \ \ n = 50$ The formula for calculating the confidence interval of a large sample is,
$\overline{x} \pm z\left(\sqrt{\frac{\sigma^{2}}{n}}\right)$ To be able to find the confidence interval we first need to find $z$ , to find $z$ use the formula,
$z = \phi^{-1}(0.5 + 0.005\alpha)$ Where $\alpha$ is the confidence level.

Let’s evaluate $z$ ,
$z = \phi^{-1}(0.5 + 0.005\alpha)$ $z = \phi^{-1}(0.5 + 0.005(92))$ $z = \phi^{-1}(0.96)$ $z = 1.751$ Note: Use the standard normal distribution table to evaluate $\phi^{-1}(0.96)$ . $\phi^{-1}$ means we are using the table in reverse.

Now let’s find the confidence interval,
$\overline{x} \pm z\left(\sqrt{\frac{\sigma^{2}}{n}}\right)$ $72.3 \pm 1.751\left(\sqrt{\frac{64.3}{50}}\right)$ $(70.3, 74.3)$ Note: Since the sample is large $s^{2}$ can be used in place of $\sigma^{2}$ and remember that the confidence interval is a range of values.

Therefore, the $92\%$ confidence interval for the population mean length of throws by this athlete is,
$(70.3, 74.3)$ Note: This means that our confidence interval is from $70.3$ to $74.3$ .

2. Sumita has a six-sided die with faces marked $1,\ ,2,\ 3,\ 4,\ 5,\ 6$ . The probability that the die shows a six on any throw is $p$ . Sumita throws the die $500$ times and finds that it shows a six $70$ times. (9709/63/M/J/20 number 5)

(a) Calculate an approximate $99\%$ confidence interval for $p$ .

Let’s identify all the information we are given by the question,
$p_{s} = \frac{70}{500}\ \ \ \ n = 500$ The formula for calculating the confidence interval of a population proportion is,
$p_{s} \pm z\left(\sqrt{\frac{p_{s} \times (1 - p_{s})}{n}}\right)$ To be able to find the confidence interval we first need to find $z$ , to find $z$ use the formula,
$z = \phi^{-1}(0.5 + 0.005\alpha)$ Where $\alpha$ is the confidence level.

Let’s evaluate $z$ ,
$z = \phi^{-1}(0.5 + 0.005\alpha)$ $z = \phi^{-1}(0.5 + 0.005(99))$ $z = \phi^{-1}(0.995)$ $z = 2.576$ Now let’s find the confidence interval,
$p_{s} \pm z\left(\sqrt{\frac{p_{s} \times (1 - p_{s})}{n}}\right)$ $\frac{70}{500} \pm 2.576\left(\sqrt{\frac{\frac{70}{500} \times \left(1 - \frac{70}{500}\right)}{500}}\right)$ $(0.100, 0.180)$ Therefore, the $99\%$ confidence interval $p$ is,
$(0.100, 0.180)$ (b) Sumita believes that the die is fair. Use your answer to part $(a)$ to comment on her belief.

If the die is fair then the true population proportion is $\frac{1}{6}$ i.e the odds of rolling a six. If $\frac{1}{6}$ lies within the calculated confidence interval above then her claim is supported.

Therefore, the final answer is,
$\frac{1}{6} \approx 0.1666...\textmd{ lies within our calculated CI,}\\ \textmd{ so her belief is justified.}$ 3. Lengths of a certain species of lizard are known to be normally distributed with standard deviation $3.2$ cm. A naturalist measures the lengths of a random sample of $100$ lizards of this species and obtains an $\alpha\%$ confidence interval for the population mean. He finds that the total width of this interval is $1.25$ cm. Find $\alpha$ . (9709/62/F/M/20 number 2)

Let’s identify all the information we have been given in the question,
$\sigma = 3.2\ \ \ \ n = 100\ \ \ \ \textmd{Interval Width} = 1.25$ Let’s use the formula for interval width to help us evaluate $\alpha$ ,
$2 \times z\left(\sqrt{\frac{\sigma^{2}}{n}}\right)$ $2 \times z\left(\sqrt{\frac{(3.2)^{2}}{100}}\right) = 1.25$ $z = \frac{1.25}{2 \times \sqrt{\frac{(3.2)^{2}}{100}}}$ $z = 1.953$ To get $\alpha$ use the formula,
$\alpha = (2\phi(z) - 1) \times 100$ Let’s substitute into the formula, $\alpha = (2\phi(z) - 1) \times 100$ $\alpha = (2\textcolor{#2192ff}{\phi(1.953)} - 1) \times 100$ $\alpha = (2(\textcolor{#2192ff}{0.9746}) - 1) \times 100$ $\alpha = 0.9492 \times 100$ $\alpha = 94.92$ $\alpha = 94.9$ Therefore, the final answer is,
$\alpha = 94.9$