AI HL · AHL 8
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z-intervals, t-intervals and choosing between them
AHL 8.12
Menu > Statistics > Confidence Intervals > z Interval
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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.
The masses of newborn lambs on a farm are normally distributed with standard deviation 0.62 kg. A random sample of 30 lambs has mean mass 4.18 kg. Find a 95% confidence interval for the mean mass of newborn lambs on the farm.
Have a go first. Then open the working one line at a time.
Population standard deviation known: z-interval
Known variance means z. State it.
From the calculator: z Interval, Stats, \(\sigma=0.62\), \(\bar x=4.18\), \(n=30\), C Level 0.95
The calculator asks for the population \(\sigma\).
\(3.96 < \mu < 4.40\) kg
Check: \(4.18\pm1.96\times\frac{0.62}{\sqrt{30}}\).
A call centre knows the standard deviation of its call lengths is 1.8 minutes. Nine randomly chosen calls lasted 12.4, 15.1, 13.8, 11.9, 14.6, 13.2, 12.7, 14.9, 13.5 minutes. Assume the lengths are normally distributed. Find a 90% confidence interval for the mean call length.
Have a go first. Then open the working one line at a time.
\(\sigma\) known: z-interval
The 1.8 is the population standard deviation.
From the calculator: z Interval, Data, the list, \(\sigma=1.8\), C Level 0.90
Choose Data and enter \(\sigma\) yourself: the calculator does not use the sample's.
\(12.6 < \mu < 14.6\) minutes
The mean of the list is 13.57.
The masses of packets of rice are normally distributed with standard deviation 12 g. A sample of 25 packets has mean 340 g. Find the 90% and the 99% confidence intervals for the mean mass, and say which is wider and why.
Have a go first. Then open the working one line at a time.
\(\sigma\) known: z-interval
If only a sample standard deviation were given, use the t-interval.
\(336 < \mu < 344\), width \(7.90\)
From the calculator: z Interval, C Level 0.90.
\(334 < \mu < 346\), width \(12.4\)
Same data, C Level 0.99.
The 99% interval is wider
To be more sure of catching \(\mu\) you need a wider interval. Higher confidence means a wider interval.
A bottling plant says its bottles hold 500 ml. A random sample of 9 bottles has mean 502.3 ml and standard deviation 3.9 ml (dividing by n). Assume the volumes are normal. (a) Find \(s_{n-1}\). (b) Find a 95% confidence interval for the mean volume. (c) Does the interval support a claim that the mean is more than 500 ml?
Have a go first. Then open the working one line at a time.
\(s_{n-1}=\sqrt{\dfrac{9}{8}}\times3.9=4.137\)
The calculator's Stats input wants \(s_{n-1}\).
From the calculator: t Interval, Stats, \(\bar x=502.3\), \(s_{n-1}=4.137\), \(n=9\), C Level 0.95: \(499 < \mu < 505\)
\(\sigma\) is unknown, so t.
500 lies inside the interval
The interval does not rule out a mean of 500, so it does not support a mean above 500.
The heights of 50 sunflowers are in the table. Assume they are normally distributed. Find an approximate 95% confidence interval for the mean height, and say why it is approximate.
Have a go first. Then open the working one line at a time.
Midpoints: 110, 130, 150, 170, 190; frequencies: 6, 13, 20, 8, 3
Use the midpoint of each class as its value.
From the calculator: 1-Variable Stats on the midpoints with the frequencies: \(\bar x=145.6\), \(s_{n-1}=21.11\)
Put the frequencies in the frequency list.
t Interval, Stats: \(140 < \mu < 152\)
\(\sigma\) unknown, so t.
Every height is replaced by its class midpoint
We do not know the true heights, so \(\bar x\) and \(s_{n-1}\) are only estimates.
Ten students sit a spelling test before and after a revision course. Scores before: 41, 35, 52, 47, 38, 44, 50, 36, 43, 49. Scores after: 44, 40, 51, 52, 45, 43, 56, 41, 47, 53. Assume the differences are normally distributed. Find a 90% confidence interval for the mean improvement and say whether it shows the course helps.
Have a go first. Then open the working one line at a time.
After − before: 3, 5, −1, 5, 7, −1, 6, 5, 4, 4
Work with one list of differences. Each difference is one observation.
From the calculator: t Interval, Data, the list of differences, C Level 0.90: \(2.13 < \mu_d < 5.27\)
Variance unknown, so t. \(\mu_d\) is the mean difference.
The whole interval is above 0
So the mean improvement is likely positive: the course appears to help. If 0 were inside, there would be no evidence of a change.
The waiting times at a clinic are not normally distributed, but their standard deviation is known to be 9.4 minutes. A random sample of 75 patients has mean wait 23.8 minutes. Find a 95% confidence interval for the mean wait. Is it valid?
Have a go first. Then open the working one line at a time.
\(\sigma\) known: z-interval
The 9.4 is the population standard deviation.
From the calculator: z Interval, Stats, \(\sigma=9.4\), \(\bar x=23.8\), \(n=75\), C Level 0.95: \(21.7 < \mu < 25.9\)
Read the interval.
n = 75 is large, so by the central limit theorem \(\bar X\) is approximately normal even though the waits are not
If the sample were small and the population not normal, the z-interval would not be justified.
A 95% confidence interval for the mean time, in seconds, to complete a task is \(41.22<\mu<48.58\). It was found from a random sample of 16 times from a normal population, using \(s_{n-1}\). Find the sample mean, the margin of error and \(s_{n-1}\).
Have a go first. Then open the working one line at a time.
\(\bar x=\dfrac{41.22+48.58}{2}=44.90\)
The interval is centred on the sample mean.
\(\dfrac{48.58-41.22}{2}=3.68\)
Half the width.
df = 15, 95% (upper tail 0.025): \(t=2.131\)
From the calculator: inverse t with 15 degrees of freedom.
\(3.68=2.131\times\dfrac{s_{n-1}}{\sqrt{16}}\) gives \(s_{n-1}=6.91\)
\(s_{n-1}=\dfrac{3.68\times4}{2.131}\).
The masses of packets of rice are normally distributed with \(\sigma=8\) g. A sample has mean 60 g. Find the 95% confidence interval for the mean mass when the sample size is 25 and when it is 100. Compare the widths.
Have a go first. Then open the working one line at a time.
\(56.9<\mu<63.1\), width 6.27
From the calculator: z Interval, Stats, \(\sigma=8\), \(\bar x=60\), \(n=25\), C Level 0.95.
\(58.4<\mu<61.6\), width 3.14
Same \(\sigma\), \(\bar x\) and C Level, with \(n=100\).
\(\dfrac{6.27}{3.14}=2\)
Four times the sample size halves the width, because the width is proportional to \(\dfrac{1}{\sqrt n}\).