AI HL · AHL 8
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The t-interval for a mean
AHL 8.11
Menu > Statistics > Confidence Intervals > t Interval
Screenshot and video to come
Menu > Statistics > Confidence Intervals > t Interval
Screenshot and video to come
Screenshot and video to come
The population is normal with μ = 50 and σ = 8, so we know where μ is. Each press takes more random samples and builds an interval for the mean from each. Orange intervals miss μ.
The picture shows the latest 60 intervals, the newest at the top. The count is of all of them. Changing the level or the kind of interval keeps the same samples.
Press 10 more to start.
A random sample of eight batteries was tested. Their lifetimes, in hours, were 31.4, 28.9, 35.2, 30.1, 33.7, 27.6, 32.8, 29.5. The lifetimes are normally distributed. Find a 90% confidence interval for the mean lifetime of this make of battery.
Have a go first. Then open the working one line at a time.
The population variance is not given: t-interval
State which interval you use and why. Unknown variance means t, whatever the sample size.
From the calculator: t Interval, Data, the list, C Level = 0.90
Choose Data as the input method, then the list, a frequency of 1 and C = 0.90.
\(29.4 < \mu < 32.9\) hours
Give 3 significant figures and say it in context.
We are 90% confident the mean lifetime is between these values
Not 'there is a 90% chance that \(\mu\) is here': \(\mu\) is fixed.
The masses of 18 randomly chosen pumpkins from a farm have mean 62.4 kg and \(s_{n-1}=5.1\) kg. Assume the masses are normally distributed. Find a 95% confidence interval for the mean mass of the farm's pumpkins.
Have a go first. Then open the working one line at a time.
Variance unknown: t-interval
Unknown variance means t, and the t value is bigger than 1.96 for a small sample.
From the calculator: t Interval, Stats, \(\bar x=62.4\), \(s_{n-1}=5.1\), \(n=18\), C Level 0.95
Use Stats input. You must enter n as well.
\(59.9 < \mu < 64.9\) kg
The interval is centred on \(\bar x\).
The reaction times (seconds) of seven students are 0.62, 0.55, 0.71, 0.58, 0.66, 0.6, 0.64. Assume they are normally distributed. Find a 95% confidence interval for the mean reaction time: find \(\bar x\), \(s_{n-1}\), the degrees of freedom, the critical t value and the margin of error.
Have a go first. Then open the working one line at a time.
From the calculator: \(\bar x=0.6229\), \(s_{n-1}=0.05314\)
Use the sample standard deviation that divides by \(n-1\).
\(\text{df}=7-1=6\); inverse t for an upper tail of 0.025: \(t=2.447\)
For 95% confidence, 2.5% is in each tail. Use the calculator's inverse t.
\(t\times\dfrac{s_{n-1}}{\sqrt n}=2.447\times\dfrac{0.05314}{\sqrt7}=0.0491\)
Standard error times the critical value.
\(\bar x\pm 0.0491\): \(0.574 < \mu < 0.672\)
For 90%, the critical value falls to \(t=1.943\) and the interval narrows to \(0.584 < \mu < 0.662\).
A machine fills bags with a true mean mass of 250 g. A student takes 20 random samples of 10 bags and finds a 95% confidence interval for the mean from each. The diagram shows the 20 intervals (the dots are the sample means) and the true mean (the vertical line). How many of the intervals do not contain 250 g, and what does the 95% mean?
Have a go first. Then open the working one line at a time.
Look for intervals that do not cross the line: 1
Count the intervals that stop before the line or start after it.
\(\frac{19}{20}=95\%\) contain the true mean
About 95% is what we expect. It will not be exactly 95% every time.
If we took many samples and built an interval from each, about 95% of those intervals would contain the true mean
The 95% describes the method, not one interval. Any one interval either contains the mean or it does not.