Confidence intervals · Mathematics 6th Year HL

The sample mean as a random variable

Practice★☆☆

A social media analyst wants to study the daily time that Irish teenagers aged between 15 and 18 spend on TikTok*. To do so, they randomly select 400 users in this age…

A social media analyst wants to study the daily time that Irish teenagers aged between 15 and 18 spend on TikTok. To do so, they randomly select 400 users in this age group across the country and record their connection times.

Clearly identify the following elements of the statistical study:

a) The population under study.

b) The sample selected.

c) The sample size.

d) The statistic that would be calculated to summarise the information from the sample.

Solve

Practice★☆☆

It is known that the waiting time in the queue of a popular theme park ride follows a normal distribution with a mean of $\mu = 45$ minutes and a standard deviation of…

It is known that the waiting time in the queue of a popular theme park ride follows a normal distribution with a mean of $\mu = 45$ minutes and a standard deviation of $\sigma = 12$ minutes.

If we choose a random sample of $n = 36$ visitors who have been on the ride:

What distribution does the sample mean $\bar{X}$ follow? State its mean and standard deviation (standard error).
Calculate the probability that the mean waiting time of this sample is greater than $48$ minutes.

Solve

Confidence interval for a mean

Formula for the interval for the mean (σ known)

Random sampling with independent observations is assumed. The interval is exact under a normal population and approximate when the central limit theorem is appropriate. Without normality, a size such as 30 does not by itself guarantee the approximation; if σ is unknown, it is not simply replaced by a sample standard deviation in a small sample.

$IC_{1-\alpha} = \left(\bar{x} - z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}},\ \bar{x} + z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\right)$. $\bar{x}$ is the sample mean, $\sigma$ the population standard deviation (known), $n$ the sample size and $z_{\alpha/2}$ the critical value of $N(0,1)$ for the chosen confidence level: $1.645$ (90%), $1.96$ (95%) or $2.575$ (99%). The maximum permissible error is $E = z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$, half the width of the interval.

Why? When the normal variable $\bar{X}$ is standardised, the central area $1-\alpha$ lies at most $z_{\alpha/2}$ standard deviations from $\mu$; when you solve for $\mu$, that same distance defines the interval symmetrically around the observed sample mean.

For another confidence level, find $z_{\alpha/2}$ in the normal tables: the table gives $P(Z

Practice★☆☆

In statistical inference, the critical value $z_{\alpha/2}$ is essential for determining the width of a confidence interval. It is the value of the standard normal…

In statistical inference, the critical value $z_{\alpha/2}$ is essential for determining the width of a confidence interval. It is the value of the standard normal distribution $N(0, 1)$ that leaves an area $\alpha/2$ to its right, where $1-\alpha$ is the confidence level (in the figure, the central area is $1-\alpha$ and each tail is $\alpha/2$).

Using the standard normal distribution table $N(0, 1)$, find the critical values $z_{\alpha/2}$ corresponding to the following confidence levels:

a) A confidence level of $90\,\%$.

b) A confidence level of $95\,\%$.

c) A confidence level of $99\,\%$.

Solve

Practice★☆☆

A courier company is studying the weight of packages it delivers in a city. The population is assumed to follow a normal distribution with a standard deviation of…

A courier company is studying the weight of packages it delivers in a city. The population is assumed to follow a normal distribution with a standard deviation of $\sigma = 0.8$ kg. A random sample of $n = 64$ packages is taken, obtaining a sample mean of $\bar{x} = 4.2$ kg.

Calculate a confidence interval for the mean weight of all packages delivered by the company, using a confidence level of $95\%$.

Solve

Practice★★☆

For a study on digital reading habits among young people, a confidence interval is to be constructed. Calculate, showing your working, the critical values $z_{\alpha/2}$…

For a study on digital reading habits among young people, a confidence interval is to be constructed. Calculate, showing your working, the critical values $z_{\alpha/2}$ corresponding to the following confidence levels:

a) $94 \%$
b) $98 \%$

Illustrate graphically the position of the critical value in a standard normal distribution for the $94 \%$ case.

Solve

Confidence interval for a proportion

Formula for the interval for a proportion

This Wald interval is approximate: it requires suitable sampling and enough successes and failures (for example, at least 10 of each). If the sample is small or the proportion is close to 0 or 1, it may perform poorly; this is not fixed simply by trimming the ends to [0,1].

$IC_{1-\alpha} = \left(\hat{p} - z_{\alpha/2}\sqrt{\dfrac{\hat{p}\hat{q}}{n}},\ \hat{p} + z_{\alpha/2}\sqrt{\dfrac{\hat{p}\hat{q}}{n}}\right)$, with $\hat{p}$ the proportion observed in the sample, $\hat{q}=1-\hat{p}$, and $n$ the sample size. The structure is identical to that for the mean: point estimate $\pm$ margin of error.

The sample proportion $\hat{p}$ is itself a random variable: for a large sample, $\hat{p} \approx N\left(p, \sqrt{\dfrac{p(1-p)}{n}}\right)$. Its standard deviation is the standard error of the proportion.

Why? A sample proportion is the mean of individual Bernoulli variables with variance $p(1-p)$; by the central limit theorem, the standard deviation of their mean is $\sqrt{p(1-p)/n}$, which is approximated using the sample data.

Practice★☆☆

In a secondary school, it is suspected that $15\%$ of the students are left-handed. A random sample of $n = 100$ students is selected to check this. 1. What distribution…

In a secondary school, it is suspected that $15\%$ of the students are left-handed. A random sample of $n = 100$ students is selected to check this.

What distribution does the sample proportion $\hat{p}$ of left-handed students follow? State its mean and standard deviation.
Calculate the probability that the proportion of left-handed students in the selected sample is less than $18\%$.

Solve

Practice★☆☆

We wish to estimate the proportion of households in a city that separate organic waste for recycling. To do so, a random sample of $n = 400$ households is surveyed, of…

We wish to estimate the proportion of households in a city that separate organic waste for recycling. To do so, a random sample of $n = 400$ households is surveyed, of which $160$ state that they separate this waste.

Determine the confidence interval for the proportion of households that recycle organic waste across the whole city, using a confidence level of $90\%$.

Solve

Practice★★☆

In a survey carried out on $400$ students at a secondary school about the use of streaming platforms, $260$ said that they prefer watching TV series rather than films.…

In a survey carried out on $400$ students at a secondary school about the use of streaming platforms, $260$ said that they prefer watching TV series rather than films.

a) Find the point estimate of the proportion of students who prefer series.

b) Obtain a $97\,\%$ confidence interval for the true proportion of students who prefer series at that school.

Solve

Sample size and margin of error

For a proportion, the margin of error is at most $\dfrac{1}{\sqrt{n}}$ (the radius of the 95% interval in the worst case), as in the Leaving Cert specification.

Practice★☆☆

A LED light bulb factory wishes to estimate the mean lifespan of its products. It is known from previous studies that the standard deviation of the lifespan is…

A LED light bulb factory wishes to estimate the mean lifespan of its products. It is known from previous studies that the standard deviation of the lifespan is $\sigma = 500$ hours.

If the quality control team wants to make an estimation with a confidence level of $95\%$, what minimum sample size $n$ should they choose so that the maximum error incurred is less than $100$ hours?

Solve

Practice★★☆

An estimate is to be made of the proportion of citizens who would be willing to use public transport if it were free. In a prior study the expected proportion is…

An estimate is to be made of the proportion of citizens who would be willing to use public transport if it were free. In a prior study the expected proportion is unknown, so maximum variance ($p = q = 0.5$) is assumed.

Determine the minimum sample size required to estimate this proportion with a maximum margin of error of $4 \%$ and a confidence level of $95.44 \%$ (which corresponds to a critical value of $z_{\alpha/2} = 2$).

Solve

Practice★★★

A mobile phone company wants to estimate the proportion of teenagers who use "dark mode" on their devices at night. In an initial sample of $100$ users, $60\%$ confirmed…

A mobile phone company wants to estimate the proportion of teenagers who use "dark mode" on their devices at night. In an initial sample of $100$ users, $60\%$ confirmed using it.

a) Calculate the $98\%$ confidence interval for the population proportion of users of this mode.

b) Suppose that no prior information about the proportion is available (maximum uncertainty scenario). Determine the sample size needed for the maximum margin of error to be $0.02$ with a confidence level of $95\%$.

c) If the proportion obtained in the initial sample ($0.6$) is used to calculate the sample size under the same conditions as in part b) (margin of error $0.02$ and confidence level $95\%$), how many fewer users would need to be surveyed compared to the maximum uncertainty scenario?

Solve

Relationship between confidence, error and sample size

The margin of error, the confidence level and the sample size are tied together: you cannot improve one without affecting another. Raising the confidence (more certainty) increases $z_{\alpha/2}$ and, all else being equal, widens the interval; the only way to gain precision without losing confidence is to increase $n$.

Practice★★☆

Two secondary schools, A and B, measure the mean time that their 2nd year Baccalaureate students spend studying mathematics per week. In both schools, it is known that…

Two secondary schools, A and B, measure the mean time that their 2nd year Baccalaureate students spend studying mathematics per week. In both schools, it is known that the population standard deviation is $1.5$ hours.

A sample of $100$ students is taken in each school. In school A, the confidence interval obtained for the mean has a width of $0.588$ hours. In school B, the confidence interval obtained for the mean is $(6.2; 7.2)$.

a) Determine the confidence level used in school A.
b) Determine the confidence level used in school B.
c) Which of the two schools has made a more precise estimate? Justify your answer.

Solve

Practice★★☆

An "eco-friendly" clothing brand claims that $30\,\%$ of its customers are under $20$ years old. To check this, a random sample of $500$ customers is taken and it is…

An "eco-friendly" clothing brand claims that $30\,\%$ of its customers are under $20$ years old. To check this, a random sample of $500$ customers is taken and it is observed that $135$ of them are under $20$ years old.

a) Construct a $99\,\%$ confidence interval for the proportion of customers under $20$ years old based on the sample data.

b) In view of the result, can we consider the brand's claim to be consistent with the data obtained at a $99\,\%$ confidence level? Explain why.

Solve

Practice★★★

The reaction time of pilots using a state-of-the-art flight simulator follows a normal distribution with unknown mean and standard deviation of $0.05$ seconds. On taking…

The reaction time of pilots using a state-of-the-art flight simulator follows a normal distribution with unknown mean and standard deviation of $0.05$ seconds. On taking a sample of $64$ pilots, a mean of $0.42$ seconds is observed.

a) Obtain the confidence interval for the population mean at a confidence level of $97\%$.

b) The simulator manufacturer claims that the mean reaction time is $0.40$ seconds. Is this claim compatible with the interval obtained in the previous part? Justify your answer based on the relationship between confidence intervals and population parameters.

c) If it is decided to double the sample size and reduce the confidence level to $90\%$, how will this affect the width of the interval? Calculate the factor by which the width is reduced or increased.

Solve

Lab: intervals, simulation and decision-making

A fictitious survey by a county council randomly selects 400 households from a large population; 160 say they separate organic waste. Assume independent observations. Design the record sheet beforehand: population, sampling frame, selection, non-response, variable and confidence. A voluntary selection through a social network is not validated by calculating a narrow interval.

from math import sqrt, ceil

from statistics import NormalDist

from random import Random

normal = NormalDist()

n, successes, confidence = 400, 160, 0.90

p = successes/n

z = normal.inv_cdf((1+confidence)/2)

E = z*sqrt(p*(1-p)/n)

print('CI for proportion:', p-E, p+E)

print('n for error 0.04 at 95%:', ceil(normal.inv_cdf(.975)**2*.25/.04**2))

# Simulation of a CI for a mean with known sigma:

rng = Random(2026)

mu, sigma, size, repetitions = 50, 10, 25, 1000

margin = normal.inv_cdf(.975)*sigma/sqrt(size)

covered = 0

for _ in range(repetitions):

    mean = sum(rng.gauss(mu,sigma) for _ in range(size))/size

    covered += mean-margin <= mu <= mean+margin

print('Fraction of intervals that cover mu:', covered/repetitions)

Intervals change from sample to sample; the model parameter stays fixed:

Change $n$ and the confidence level, keeping the same observed proportion, and compare widths. Distinguish standard error from margin. For quantitative data, also calculate the mean and the sample standard deviation (with divisor $n-1$) using a spreadsheet: the latter estimates $\sigma$, but it does not automatically turn a normal interval into an exact one if $\sigma$ is unknown. Hand in the data, the procedure, the checks and the limits of the decision.

Let's practice! 10 medium exercises in this section

★★☆ For a study on digital reading habits among young people, a confidence interval is to be…★★☆ An esports team wants to know the mean daily training time of its players. It is known…

Summary

Under normality and independence, $\bar{X} \sim N\left(\mu, \frac{\sigma}{\sqrt{n}}\right)$: the sample mean varies less the larger $n$ is.
Under the stated conditions, CI for the mean: $\bar{x} \pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$. CI for a proportion: $\hat{p} \pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}\hat{q}}{n}}$.
The confidence level refers to the procedure, not to whether $\mu$ (or $p$) is or is not in that particular interval.
Sample size: $n = \left(\frac{z_{\alpha/2}\sigma}{E}\right)^2$ for the mean, $n=\frac{z_{\alpha/2}^2 pq}{E^2}$ for a proportion; with no information on $p$, use $p=q=0.5$. Back to the start: the opinion poll of about 1,000 people has a margin of error of roughly $1/\sqrt{1000} \approx 3\%$.
Halving the error multiplies $n$ by $4$. More confidence with the same $n$ widens the interval; more precision without losing confidence requires a larger sample.

Let's practice! 10 medium exercises in this section

★★☆ For a study on digital reading habits among young people, a confidence interval is to be…★★☆ An esports team wants to know the mean daily training time of its players. It is known…★★☆ A technology company is testing the battery life of a new smartphone model. It is assumed…★★☆ In a survey carried out on $400$ students at a secondary school about the use of…

Let's practice! 10 hard exercises in this section

★★★ A well-known content creator is analysing how long his followers stay connected during…★★★ A mobile phone company wants to estimate the proportion of teenagers who use "dark mode"…★★★ The reaction time of pilots using a state-of-the-art flight simulator follows a normal…★★★ A study on sustainable habits reveals that $15\%$ of young people in a city travel…