25 CIs for one mean
So far, you have learnt to ask a RQ, design a study, classify and summarise the data, and construct confidence intervals for one proportion. You have also been introduced to confidence intervals. In this chapter, you will learn to
- construct confidence intervals for one mean.
- determine whether the conditions for using the confidence intervals apply in a given situation.
25.1 Sampling distribution for \(\bar{x}\): \(\sigma\) known
In this chapter, we study the situation where a population mean \(\mu\) (the parameter) is estimated by a sample mean \(\bar{x}\) (the statistic). The sample mean is computed from just one of the many possible samples, and each possible sample is likely to produce a different value of \(\bar{x}\). That is, the value of the sample mean varies from sample to sample, called sampling variation.
Remember: studying a sample leads to the following observations:
- Every sample is likely to be different.
- We observe just one of the many possible samples.
- Every sample is likely to yield a different value for the sample statistic.
- We observe just one of the many possible values for the statistic.
Since many values for the sample mean are possible, the possible values of the sample mean vary (called sampling variation) and have a distribution (called a sampling distribution).
Consider rolling dice again. Suppose a die is rolled \(n = 25\) times, and the mean of the sample of \(25\) numbers that are rolled is recorded. Since every face of the die is equally likely to appear on any one roll, the population mean of all possible rolls is \(\mu = 3.5\) (in the middle of the numbers on the faces of the die, so this is also the median).
What will be the sample mean of the numbers in the \(25\) rolls? We cannot be sure, as the sample mean will vary from sample to sample (sampling variation). But suppose we roll a die \(25\) times, to see how the sample mean varies in \(25\) rolls as shown in the animation below for ten samples of \(25\) rolls. The mean of the \(25\) rolls clearly varies, as expected. In the simulation, the sample mean of \(25\) rolls was as low as \(3.08\) and as high as \(3.76\).
The mean for any single sample of \(n = 25\) rolls will sometimes be higher than \(\mu = 3.5\), and sometimes lower than \(\mu = 3.5\), but most of the time the mean should be close to \(3.5\). If thousands of people made one sample of \(25\) rolls each, and computed the mean for their sample, every person would have a sample mean for their sample, and we could produce a histogram of all these sample means (see the animation below).
From the animation above, the sample means vary with an approximate normal distribution (as we saw with the sample proportions). This normal distribution does not describe the data; it describes how the values of sample means vary across all possible samples. Under certain conditions, the values of the sample means can vary with a normal distribution, and this normal distribution has a mean and a standard deviation.
This sampling distribution describes how sample means vary.
The mean of this sampling distribution---the sampling mean---has the value \(\mu\).
The standard deviation of this sampling distribution is called the standard error of the sample means, denoted \(\text{s.e.}(\bar{x})\).
When the population standard deviation \(\sigma\) is known, the standard error happens to be
\[
\text{s.e.}(\bar{x}) = \frac{\sigma}{\sqrt{n}}.
\]
The possible values of the sample means have a sampling distribution described by:
- an approximate normal distribution,
- with a sampling mean whose value is \(\mu\), and
- a standard deviation, called the standard error, of \(\text{s.e.}(\bar{x}) = \sigma/\sqrt{n}\).
However, since the population standard deviation is rarely ever known, let's focus on the case where the value of \(\sigma\) is unknown.
25.2 Sampling distribution for \(\bar{x}\): \(\sigma\) unknown
A sample mean is used to estimate a population mean, but the sample mean varies from sample to sample: sampling variation exists. Since the value of the population standard deviation \(\sigma\) is almost never known, the sample standard deviation \(s\) is used to give an estimate of the standard error of the mean: \(\text{s.e.}(\bar{x}) = s/\sqrt{n}\). With this information, the sampling distribution of the sample mean can be described.
Definition 25.1 (Sampling distribution of a sample mean with $\sigma$ unknown) When the population standard deviation is unknown, the sampling distribution of the sample mean is (when certain conditions are met; Sect. 25.4) described by:
- an approximate normal distribution,
- centred around a sampling mean whose value is \(\mu\),
- with a standard deviation (called the standard error of the mean) \(\text{s.e.}(\bar{x})\), whose value is
\[\begin{equation} \text{s.e.}(\bar{x}) = \frac{s}{\sqrt{n}}, \tag{25.1} \end{equation}\] where \(n\) is the size of the sample, and \(s\) is the sample standard deviation of the observations.
A mean or a median may be appropriate for describing the data. However, the sampling distribution for the sample mean (under certain conditions) has a normal distribution, and so the mean is appropriate for describing the sampling distribution.
25.3 Confidence intervals for \(\mu\)
We don't know the value of \(\mu\) (the parameter), but we have an estimate: the value of \(\bar{x}\), the sample mean (the statistic). The actual value of \(\mu\) might be a bit larger than \(\bar{x}\), or a bit smaller than \(\bar{x}\); that is, the value of \(\mu\) is \(\bar{x}\), give-or-take a bit.
Furthermore, the values of \(\bar{x}\) vary from sample to sample (sampling variation), and vary with an approximate normal distribution. So, the \(68\)--\(95\)--\(99.7\) rule could be used to construct an approximate \(95\)% interval for the plausible values of \(\mu\) that may have produced the observed values of the sample mean. This is a confidence interval.
A confidence interval (CI) for the population mean is an interval surrounding a sample mean.
In general, a confidence interval (CI) for \(\mu\) is
\[
\bar{x} \pm \overbrace{(\text{multiplier}\times\text{s.e.}(\bar{x}))}^{\text{The `margin of error'}}.
\]
For an approximate \(95\)% CI, the multiplier is about \(2\) (since about \(95\)% of values are within two standard deviations of the mean, from the \(68\)--\(95\)--\(99.7\) rule).
CIs are commonly \(95\)% CIs, but any level of confidence can be used (but a different multiplier is needed). In this book, a multiplier of \(2\) is used when approximate \(95\)% CIs are created manually, and otherwise software is used. Commonly, CIs are computed at \(90\)%, \(95\)% and \(99\)% confidence levels.
The multiplier is not a \(z\)-score here (but is like a \(z\)-score). The multiplier is be a \(z\)-score if the value of the population standard deviation was known (e.g., the situation in Sect. 25.1). When \(\sigma\) is unknown, and the sample standard deviation is used instead, the multiplier is a \(t\)-score.
The values of \(t\)- and \(z\)-multipliers are very similar, and (except for small sample sizes) using an approximate multiplier of \(2\) is reasonable for computing approximate \(95\)% CIs in either case.
Pretend for the moment that the value of \(\mu\) was unknown, and we tossed a die \(25\) times, and found \(\bar{x} = 3.2\) and \(s = 2.5\).
Then,
\[
\text{s.e.}(\bar{x}) = \frac{s}{\sqrt{n}} = \frac{2.5}{\sqrt{25}} = 0.5.
\]
The sample means vary with an approximate normal distribution, centred around the unknown value of \(\mu\), with a standard deviation of \(\text{s.e.}(\bar{x}) = 0.5\) (Fig. 25.1).
FIGURE 25.1: The sampling distribution is a normal distribution; it shows how the sample mean of 25 die rolls varies in samples of size \(n = 25\)
Our estimate of \(\bar{x} = 3.2\) may be a bit smaller than the value of \(\mu\), or a bit larger than the value of \(\mu\); that is, the value of \(\mu\) is in the range '\(\bar{x}\) give-or-take a bit'. A range of \(\bar{x}\) values that are likely to straddle \(\mu\) is given by a CI. An approximate \(95\)% CI is from \(3.2 - (2 \times 0.5)\) and \(3.2 + (2 \times 0.5)\), or from \(2.2\) to \(4.2\). Hence, values of \(\mu\) between \(2.2\) to \(4.2\) could reasonably have produced a sample mean of \(\bar{x} = 3.2\).
Would a \(99\)% CI for \(\mu\) be wider or narrower than the \(95\)% CI? Why?
A wider interval is needed to be more confident that the interval contains the population mean.
25.4 Statistical validity conditions
As with any confidence interval, the underlying mathematics requires certain conditions to be met so that the results are statistically valid (i.e., the sampling distribution is sufficiently like a normal distribution). The CI for one mean will be statistical valid if one of these is true:
- The sample size is at least \(25\), or
- The sample size is smaller than \(25\) and the population data has an approximate normal distribution.
The sample size of \(25\) is a rough figure, and some books give other (similar) values (such as \(30\)).
This condition ensures that the sampling distribution of the sample means has an approximate normal distribution (so that the \(68\)--\(95\)--\(99.7\) rule can be used). Provided the sample size is larger than about \(25\), this will be approximately true even if the distribution of the individuals in the population do not have a normal distribution. That is, when \(n > 25\) the sample means generally have an approximate normal distribution, even if the data themselves do not follow a normal distribution.
When \(n > 25\) approximately, we do not require that the data has a normal distribution. The sample means need to have a normal distribution, which is approximately true if the statistical validity condition is true.
Example 25.1 (Statistical validity) In the die example (Sect. 25.3), where \(n = 25\), the CI is statistically valid.
Deciding whether the population has a normal distribution is obviously difficult; we do not have access to the whole population. All we can reasonably do is to identify (from the sample) populations that are likely to be very non-normal (when the CI would be not valid).
Example 25.2 (Assumptions) A study (Silverman et al. 1999; Zou, Tuncali, and Silverman 2003) to examine exposure to radiation for CT scans in the abdomen assessed \(n = 17\) patients. As the sample size is 'small' (less than \(25\)), the population data must have a normal distribution for a CI for \(\mu\) to be statistically valid.
A histogram of the total radiation dose received using the sample data (Fig. 25.2) suggests this is very unlikely. Even though the histogram is from sample data, it seems improbable that the data in the sample would have come from a population with a normal distribution.
A CI for the mean of these data will probably be not statistically valid. Other methods (beyond the scope of this course) are possible for computing a CI for the mean.
FIGURE 25.2: The radiation doses from CT scans for 17 people
25.5 Example: cadmium in peanuts
A study of peanuts from the United States (Blair and Lamb 2017) found the sample mean cadmium concentration was \(\bar{x} = 0.076\) ppm with a standard deviation of \(s = 0.0460\) ppm, from a sample of \(290\) peanuts gathered from a variety of regions at various times (a representative sample). The parameter is \(\mu\), the population mean cadmium concentration in peanuts.
Every sample of \(n = 290\) peanuts is likely to produce a different sample mean, so sampling variation in \(\bar{x}\) exists and can be measured using the standard error:
\[
\text{s.e.}(\bar{x}) = \frac{s}{\sqrt{n}} = \frac{0.0460}{\sqrt{290}} = 0.002701\text{ ppm}.
\]
The approximate \(95\)% CI is \(0.0768 \pm (2 \times 0.002701)\), or \(0.0768 \pm 0.00540\), which is from \(0.0714\) to \(0.0822\) ppm.
(The margin of error is \(0.00540\).)
We write:
The sample mean cadmium concentration of peanuts is \(\bar{x} = 0.0768\) ppm (s.e.: \(0.00270\); \(n = 290\)), with an approximate \(95\)% CI from \(0.0714\) to \(0.0822\) pmm.
If we repeatedly took samples of size \(290\) from this population, about \(95\)% of the \(95\)% CIs would contain the population mean (but our CI may or may not contain the value of \(\mu\)). The plausible values of \(\mu\) that could have produced \(\bar{x} = 0.0768\) are between \(0.0714\) and \(0.0822\) ppm. Alternatively, we are about \(95\)% confident that the CI of \(0.0714\) to \(0.0822\) ppm straddles the population mean.
Since the sample size is larger than \(25\), the CI is statistically valid.
25.6 Chapter summary
To compute a confidence interval (CI) for a mean, compute the sample mean, \(\bar{x}\), and identify the sample size \(n\).
Then compute the standard error, which quantifies how much the value of \(\bar{x}\) varies across all possible samples:
\[
\text{s.e.}(\bar{x})
=
\frac{ s }{\sqrt{n}},
\]
where \(s\) is the sample standard deviation.
The margin of error is (Multiplier\(\times\)standard error), where the multiplier is \(2\) for an approximate \(95\)% CI (using the \(68\)--\(95\)--\(99.7\) rule).
Then the CI is:
\[
\bar{x} \pm \left( \text{Multiplier}\times\text{standard error} \right).
\]
The statistical validity conditions should also be checked.
25.7 Quick review questions
- True or false: The value of \(\bar{x}\) varies from sample to sample.
- True or false: A CI for \(\mu\) is never statistically valid if the histogram of the data has a non-normal distribution.
- A sample of data produces \(s = 8\) and \(n = 20\).
Which one of the following is definitely true?
25.8 Exercises
Answers to odd-numbered exercises are available in App. E.
Exercise 25.1 A study of American black bears (Bartareau 2017) found the mean weight of the \(n = 185\) male bears was \(\bar{x} = 84.9\) kg, with a standard deviation of \(s = 51.1\) kg.
- Write down the parameter of interest.
- Compute the standard error of the mean.
- Compute the approximate \(95\)% CI.
- Write a conclusion.
- Is the CI statistically valid?
Exercise 25.2 A study of the school bags that \(586\) children (in Grades 6--8 in Tabriz, Iran) take to school found the mean weight was \(\bar{x} = 2.8\) kg with a standard deviation of \(s = 0.94\) kg (Dianat et al. 2014).
- Write down the parameter of interest.
- Compute the standard error of the mean.
- Compute the approximate \(95\)% CI.
- Write a conclusion.
- Is the CI statistically valid?
Exercise 25.3 A study of the lung capacity of children in East Boston (Tager et al. 1979; Kahn 2005) measured the forced expiratory volume (FEV) of children in the area. The sample contained \(n = 45\) eleven-year-old girls. For these children, the mean lung capacity was \(\bar{x} = 2.85\) litres and the standard deviation was \(s = 0.43\) litres. Find an approximate \(95\)% CI for the population mean lung capacity of eleven-year-old females from East Boston.
Exercise 25.4 A study of lead smelter emissions near children's public playgrounds (Taylor et al. 2013) found the mean lead concentration at one playground (Memorial Park, Port Pirie, in South Australia) to be \(6956.41\) micrograms per square metre, with a standard deviation of \(7571.74\) micrograms of lead per square metre, from a sample of \(n = 58\) wipes taken over a seven-day period. (As a reference, the Western Australian Government recommends a maximum of \(400\) micrograms of lead per square metre.)
Find an approximate \(95\)% CI for the mean lead concentration at this playground. Would these results apply to other playgrounds?
Exercise 25.5 A study (Ian D. M. Macgregor and Rugg-Gunn 1985) of the brushing time for \(60\) young adults (aged 18--22 years old) found the mean brushing time was \(33.0\) seconds, with a standard deviation of \(12.0\) seconds. Find an approximate \(95\)% CI for the mean brushing time for young adults.
Exercise 25.6 A study of paramedics (B. Williams and Boyle 2007) asked participants (\(n = 199\)) to estimate the amount of blood loss on four different surfaces. When the actual amount of blood spill on concrete was \(1000\) ml, the mean guess was \(846.4\) ml (with a standard deviation of \(651.1\) ml).
- What is the approximate \(95\)% CI for the mean guess of blood loss?
- Are the participants good at estimating the amount of blood loss on concrete?
- Is this CI likely to be valid?
Exercise 25.7 Using data from the NHANES study (CDC 1996), the approximate \(95\)% CI for the mean direct HDL cholesterol is \(1.356\) to \(1.374\) mmol/L. Which (if either) of these interpretations are acceptable? Explain why are the other interpretations are incorrect.
- In the sample, about \(95\)% of individuals have a direct HDL concentration between \(1.356\) to \(1.374\) mmol/L.
- In the population, about \(95\)% of individuals have a direct HDL concentration between \(1.356\) to \(1.374\) mmol/L.
- About \(95\)% of the samples are between \(1.356\) to \(1.374\) mmol/L.
- About \(95\)% of the populations are between \(1.356\) to \(1.374\) mmol/L.
- The population mean varies so that it is between \(1.356\) to \(1.374\) mmol/L about \(95\)% of the time.
- We are about \(95\)% sure that sample mean is between \(1.356\) to \(1.374\) mmol/L.
- It is plausible that the sample mean is between \(1.356\) to \(1.374\) mmol/L.
Exercise 25.8 An article (Grabosky and Bassuk 2016) describes the diameter of Quercus bicolor trees planted in a lawn as having a mean of \(25.8\) cm, with a standard error of \(0.64\) cm, from a sample of \(19\) trees. Which (if any) of the following is correct?
- About \(95\)% of the trees in the sample will have a diameter between \(25.8 - (2\times 0.64)\) and \(25.8 + (2\times 0.64)\) (based on using the \(68\)--\(95\)--\(99.7\) rule).
- About \(95\)% of these types of trees in the population will have a diameter between \(25.8 - (2\times 0.64)\) and \(25.8 + (2\times 0.64)\) (based on using the \(68\)--\(95\)--\(99.7\) rule)?
Exercise 25.9 In a study of \(n = 30\) five-year-old children (Watanabe et al. 1995), the mean time for the children to eat a cookie was \(61.3\) s, with a standard deviation of \(29.4\) s.
- What is an approximate \(95\)% CI for the population mean time for a five-year-old child to eat a cookie?
- Is the CI statistically valid?