22 CIs for one mean
So far, you have learnt to ask a RQ, identify different ways of obtaining data, design the study, collect the data describe the data, summarise data graphically and numerically, and understand the tools of inference.
In this chapter, you will learn about confidence intervals for one mean. You will learn to:
 produce confidence intervals for one mean.
 determine whether the conditions for using the confidence intervals apply in a given situation.
22.1 Sampling distribution: One mean with population standard deviation 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).
Of course, every sample is likely to be different, and is likely to produce a different sample mean \(\bar{x}\). That is, the value of the sample mean will vary from sample to sample and exhibit sampling variation (which can be quantified using the standard error).
Consider rolling dice again. Suppose a die is rolled \(n=25\) times, and the mean of the 25 numbers that are rolled is recorded. What will be the sample mean of the numbers in the 25 rolls?
The sample mean will vary from sample to sample, (sampling variation). 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, which is also the median).
An example of the mean after repeatedly rolling a die 25 times is shown in the animation below for 10 sets of 25 rolls. The mean of the 25 rolls clearly varies. 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 many people made a set of 25 rolls, and computed the mean for their set, every person would have a sample mean for their set of 25 rolls, and we could produce a histogram of all these sample means; see the animation below.
From the animation above, the sample means appear to vary with an approximate normal distribution (as we saw with the sample proportions). This normal distribution is centred around the population mean \(\mu\). The standard deviation of the normal distribution is the standard error of the sample mean \(\bar{x}\), written as \(\text{s.e.}(\bar{x})\).
When the population standard deviation \(\sigma\) is known, then
\[ \text{s.e.}(\bar{x}) = \frac{\sigma}{\sqrt{n}}. \] So the possible values of the sample means have a sampling distribution described by:
 an approximate normal distribution,
 with mean \(\mu\), and
 a standard deviation, called the standard error, of \(\text{s.e.}(\bar{x}) = \sigma/\sqrt{n}\).
Usually the population mean and the population standard deviation are unknown. Nonetheless, because the sampling distribution has an approximate normal distribution, the 689599.7 rule can be applied: approximately 95% of the sample means are expected to be within two standard errors of \(\mu\).
22.2 Sampling distribution: One mean with population standard deviation unknown
When a sample mean is used to estimate a population mean, the sample mean will vary from sample to sample: sampling variation exists, as we saw in the previous section.
When we do not know the population standard deviation \(\sigma\) (which is almost always the case), we estimate it using the sample standard deviation \(s\). Then, the standard error of the sample mean is \(\displaystyle\text{s.e.}(\bar{x}) = \frac{s}{\sqrt{n}}\). With this information, we can describe the sampling distribution of the sample mean.
Definition 22.1 (Sampling distribution of a sample mean) When the population standard deviation is unknown, the sampling distribution of the sample mean is described by:
 an approximate a normal distribution,
 centred around \(\mu\),
 with a standard deviation (called the standard error of the mean) of
\[\begin{equation} \text{s.e.}(\bar{x}) = \frac{s}{\sqrt{n}}, \tag{22.1} \end{equation}\] when certain conditions are met, where \(n\) is the size of the sample, and \(s\) is the standard deviation of the individual observations in the sample (that is, the sample standard deviation).
22.3 Confidence intervals: One mean
We don't know the value of \(\mu\) (the paremeter), the population mean, 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, \(\mu\) is probably about \(\bar{x}\), giveortake a bit.
Furthermore, we have seen that the values of \(\bar{x}\) vary from sample to sample (sampling variation), and noted that they vary with an approximate normal distribution. So, using the 689599.7 rule, we could create an approximate 95% interval for the plausible values of \(\mu\) that may have given 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, an approximate 95% confidence interval (CI) for \(\mu\) is \(\bar{x}\) giveortake about two standard errors. In general, the confidence interval (CI) for \(\mu\) is
\[ \bar{x} \pm \overbrace{(\text{Multiplier}\times\text{s.e.}(\bar{x}))}^{\text{Called the `margin of error'}}. \] For an approximate 95% CI, the multiplier is, as usual, about \(2\) (since about 95% of values are within two standard deviations of the mean from the 689599.7 rule).
We often find 95% CIs, but we can find a CI with any level of confidence: we just need a different multiplier. We'll just use a multiplier of \(2\) (and hence find approximate 95% CIs), and otherwise use software. Commonly, CIs are computed at 90%, 95% and 99% confidence levels.
The multiplier of 2 is not a \(z\)score here. The multiplier would be a \(z\)score if we knew the value of \(\sigma\); since we don't, the multiplier is a \(t\)score and not a \(z\)score.
The \(t\) and \(z\)multipliers are very similar, and (except for very small sample sizes) using an approximate multiplier of 2 is reasonable for computing approximate 95% CIs in either case.
We'll let software handle the specifics.
If we collected many samples of a specific size, \(\bar{x}\) and \(s\) would be different for each sample, so the calculated CI would be different for each. Some CIs would straddle the population mean \(\mu\), and some would not; and we never know if the CI computed from our single sample straddles \(\mu\) or not.
Loosely speaking, there is a 95% chance that our 95% CI straddles \(\mu\). For a CI computed from a single sample, we don't know if our CI includes the value of \(\mu\) or not. The CI could also be interpreted as the range of plausible values of \(\mu\) that could have produced the observed value of \(\bar{x}\).
Example 22.1 (School bags) A study of the school bags that 586 children (in Grades 68 in Tabriz, Iran) take to school found that the mean weight was \(\bar{x} = 2.8\) kg with a standard deviation of \(s=0.94\) kg.^{396}
The parameter is the population mean weight of school bags for Iranian children in Grades 68.
Of course, another sample of 586 children would produce a different sample mean: the sample mean varies from sample to sample.
The standard error of the sample mean is
\[ \text{s.e.}(\bar{x}) = \frac{s}{\sqrt{n}} = \frac{0.94}{\sqrt{586}} = 0.03883; \] see Fig. 22.1. The approximate 95% CI for the population mean schoolbag weight is
\[ 2.8\pm(2 \times 0.03883), \] or \(2.8\pm0.07766\). (The margin of error is 0.07766.) This is equivalent to an approximate 95% CI from 2.72 kg to 2.88 kg. This CI has a 95% chance of straddling the population mean bag weight.
Would a 99% CI for \(\mu\) be wider or narrower than the 95% CI? Why?
(Answer is here^{397}.)
Example 22.2 (Black bears) A study of American black bears^{398} found that the mean weight of the \(n = 185\) male bears in their study was \(\bar{x} = 84.9\) kg, with a standard deviation of \(s = 51.1\) kg.
The parameter of interest is the population mean weight of an American black bear, \(\mu\).
Using the sample information, the standard error of the mean is
\[ \text{s.e.}(\bar{x}) = \frac{s}{\sqrt{n}} = \frac{51.1}{\sqrt{185}} = 3.756947, \] so the approximate 95% CI is from \[ 84.9  (2 \times 3.756947) = 77.38611 \] to \[ 84.9 + (2 \times 3.756947) = 92.41389. \] The approximate 95% CI is from 77.4 to 92.4 kg. We would say:
We are approximately 95% confidence that the population mean weight of male American black bears is between 77.4 and 92.4 kg.
The article gives the exact CI as 77.4 to 99.5 kg, agreeing with the CI we calculated.
22.4 Statistical validity conditions: One mean
As with any inference procedure, the underlying mathematics requires certain conditions to be met so that the results are statistically valid. 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 here, and some books give other (similar) values (such as 30). This condition ensures that the distribution of the sample means has an approximate normal distribution so that the 689599.7 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 does not have a normal distribution. That is, when \(n > 25\) the sample means generally have an approximate normal distribution, even if the data themselves don't have a normal distribution.
A mean or a median may be appropriate for describing the data.
However, the CI is about the mean.
Since the sampling distribution for the sample mean (under certain conditions) has a normal distribution, the mean is appropriate for describing the sampling distribution.
In addition to the statistical validity condition, the CI will be
 internally valid if the study was well designed; and
 externally validity if the the sample is a simple random sample and is internally valid.
When \(n > 25\) approximately, we do not require that the data has a normal distribution.
We require that the sample means have a normal distribution, which is approximately true if the statistical validity condition is true.
This is one reason why means are used to describe samples: under certain conditions, sample means have an approximate normal distribution (so the 689599.7 rule applies). In contrast, the distribution of sample medians is far more complicated to describe.
To determine if assuming the population has an approximate normal distribution in the statistical validity condition, the histogram of the sample can be constructed. However, we can't really be sure about the distribution of the population from the distribution of the sample. All we can reasonably do is to identify (from the sample) populations that likely to be very nonnormal (when the CI would be not valid).
Example 22.3 (Assumptions) A study^{399} to examine exposure to radiation for CT scans in the abdomen assessed \(n = 17\) patients. A histogram of the total radiation dose received is shown in Fig. 22.2; the sample mean dose is 26.86 rads.
A CI for the mean radiation dose received could be formed. However, as the sample size is 'small' (less than 25), the population must have a normal distribution for the CI to be statistically valid. 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: the histogram of the sample data doesn't look normally distributed at all.
Computing a CI for the mean of these data will probably be statistically invalid. Other methods (beyond the scope of this course) are possible for computing a confidence interval for the mean.
Example 22.4 (School bags) In Example 22.1, an approximate 95% CI was formed for the mean weight of school bags for Iranian children.
Since the sample size was \(n = 586\), the CI is statistically valid.
We do not have to assume that the distribution of school bag weights has a normal distribution in the population, as the sample size is (much) larger than 25.
Example 22.5 (Black bears) In Example 22.2, the approximate 95% CI was formed from a sample of size \(n = 185\) male bears.
This CI is statistically valid, since the sample size is much larger than 25.
We do not have to assume that the distribution of the weights of male black bears has a normal distribution in the population, as the sample size is (much) larger than 25.
22.5 Example: NHANES
Previously, we asked this question about the NHANES data:
Among Americans, is the mean direct HDL cholesterol different for current smokers and nonsmokers?
The response variable is direct HDL cholesterol concentration. The parameter is \(\mu\), the population mean HDL cholesterol concentration.
What is the population mean direct HDL cholesterol concentration?
From the data (using jamovi or SPSS), the sample mean is \(\bar{x} = 1.3649\) mmol/L; the standard deviation is \(s=0.39926\) mmol/L; and the sample size is \(n= 8474\).
The value of \(\bar{x}\) will vary from sample to sample; sampling variation exists. The standard error is:
\[
\text{s.e.}(\bar{x}) =
\frac{s}{\sqrt{n}}
=
\frac{ 0.39926}
{\sqrt{8474}} =
0.00434\text{ mmol/L}.
\]
The approximate 95% CI uses a multiplier of \(2\),
so the marginoferror is
\[
2\times 0.0043
=
0.00867.
\]
The approximate 95% CI is \(1.365\),
giveortake \(0.00867\);
or
from \(1.356\) to \(1.374\) mmol/L.
Based on the sample of size \(n= 8474\), a 95% CI for the population mean direct HDL cholesterol levels of Americans is between \(1.356\) and \(1.374\) mmol/L.
If many samples of the same size were found in the same way, and computed the CI from each, about 95% of the CIs would contain \(\mu\) (but this particular CI may or may not contain the value of \(\mu\)). We could also say that the CI gives a range of plausible values for \(\mu\), or that we are about 95% confident that this CI straddles the value of \(\mu\).
The statistical validity condition should also be checked to ensure the CI is statistically valid.
Since the sample size is much larger than 25, this CI for mean direct HDL cholesterol is statistically valid, even though the histogram of direct HDL cholesterol for individuals is skewed right (Fig. 22.3). Recall: the distribution of the sample means should be normally distributed, not the distribution of the data.
22.6 Example: Cadmium in peanuts
A study of peanuts from the United States^{400} found the sample mean cadmium concentration was 0.0768 ppm with a standard deviation of 0.0460 ppm, from a sample of size 290 peanuts gathered from a variety of regions at various times (attempting to find 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; that is, the sample means show sampling variation. The sampling variation 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.)
If we repeatedly took samples of size 290 from this population, about 95% of the 95% CIs would contain the population mean (but this 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 0.0822ppm. 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.
22.7 Quick review questions
True or false: The value of \(\bar{x}\) varies from sample to sample.
True or false: A CI for \(\mu\) is statistically valid only if the histogram of the data has an approximate normal distribution.

Suppose \(s = 8\) and \(n=20\). Which one of the following is true?
Progress:
22.8 Exercises
Selected answers are available in Sect. D.21.
Exercise 22.1 A study^{401} of the lung capacity of children in East Boston measured the forced expiratory volume (FEV) of children in the area.
The sample contained \(n = 45\) elevenyearold 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 elevenyearold females from East Boston.
Exercise 22.2 A study of lead smelter emissions near children's public playgrounds^{402} 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 sevenday 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 22.3 A study^{403} of the brushing time for 60 young adults (aged 1822 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 22.4 A study of paramedics^{404} 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?
 How many paramedics would be needed if the mean guess was to be estimated with an precision of giveortake 50 ml?
 How many paramedics would be needed if the mean guess was to be estimated with an precision of giveortake 25 ml?
 How many times greater does the sample size need to be to halve the width of the margin of error?
Exercise 22.5 In Sect. 22.5, the approximate 95% CI for the mean direct HDL cholesterol was given as \(1.356\) to \(1.374\) mmol/L. Which (if any) 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 22.6 An article^{405} 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 689599.7 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 689599.7 rule)?