What is the standard error of the sample mean?

\frac{\sigma}{\sqrt n} (or \frac{s}{\sqrt n} if \sigma is estimated)

Formula for a symmetric confidence interval for \mu with known \sigma?

\bar x\pm z\frac{\sigma}{\sqrt n}

What do you do if the question gives the variance, not the standard deviation?

Take the square root first to get \sigma, then divide by \sqrt n.

When can s replace \sigma with the z interval?

When the sample is large, so s is a good estimate of \sigma.

Formula for the unbiased estimate of variance from sums?

s^2=\frac{1}{n-1}\left(\sum x^2-\frac{(\sum x)^2}{n}\right)

A claimed mean lies outside a 95\% interval. What do you conclude?

There is evidence at the 5\% level that the true mean is different from the claim; say in which direction, in context.

How does a higher confidence level change the interval?

It needs a larger z, so the interval is wider.

How does a larger sample change the interval?

The standard error \frac{\sigma}{\sqrt n} falls, so the interval is narrower.

When finding a sample size for a given width, how do you round?

Round up to the next whole number.

Confidence intervals for a meanAQA A-Level Further Maths: Flashcards

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What is the distribution of $\bar X$ for a normal population?

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What is the distribution of $\bar X$ for a normal population?: $\bar X\sim N\left(\mu,\ \frac{\sigma^2}{n}\right)$
What is the standard error of the sample mean?: $\frac{\sigma}{\sqrt n}$ (or $\frac{s}{\sqrt n}$ if $\sigma$ is estimated)
Formula for a symmetric confidence interval for $\mu$ with known $\sigma$ ?: $\bar x\pm z\frac{\sigma}{\sqrt n}$
$z$ value for a $90\%$ interval?: $1.645$
$z$ value for a $95\%$ interval?: $1.96$
$z$ value for a $99\%$ interval?: $2.576$
What do you do if the question gives the variance, not the standard deviation?: Take the square root first to get $\sigma$ , then divide by $\sqrt n$ .
When can $s$ replace $\sigma$ with the $z$ interval?: When the sample is large, so $s$ is a good estimate of $\sigma$ .
Formula for the unbiased estimate of variance from sums?: $s^2=\frac{1}{n-1}\left(\sum x^2-\frac{(\sum x)^2}{n}\right)$
A claimed mean lies outside a $95\%$ interval. What do you conclude?: There is evidence at the $5\%$ level that the true mean is different from the claim; say in which direction, in context.
How does a higher confidence level change the interval?: It needs a larger $z$ , so the interval is wider.
How does a larger sample change the interval?: The standard error $\frac{\sigma}{\sqrt n}$ falls, so the interval is narrower.
When finding a sample size for a given width, how do you round?: Round up to the next whole number.

Exam questions on Confidence intervals for a mean

The masses of packets of porridge oats are normally distributed with standard deviation $12$ g. A random sample of $36$ packets has mean mass $504$ g.
Find a $99\%$ confidence interval for the population mean mass.2 marks
The time taken by a train to complete a journey is normally distributed with variance $9$ minutes $^2$ . A random sample of $16$ journeys has mean time $42.5$ minutes.
The rail company claims that the mean journey time is $45$ minutes. Use your interval from (b) to comment on this claim.2 marks
The lengths of trout on a fish farm are normally distributed with standard deviation $2.4$ cm. A random sample of $64$ trout has mean length $31.2$ cm.
Construct a $95\%$ confidence interval for the mean length of trout on the farm.3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).