20 Confidence intervals for one proportion

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 proportion. You will learn to:

  • identify situations where the analysis of one sample proportion is appropriate.
  • form confidence intervals for single proportions.

20.1 Sampling distribution: Known proportion

Suppose a fair, six-sided die373 is rolled 25 times. What proportion of the rolls will produce an even number? That is, what will be the sample proportion of even numbers?

No-one knows exactly what will happen for any individual roll, so no-one knows what proportion will be even for any sample of 25 rolls.

In addition, the proportion of the 25 rolls that will be even will not be the same for every sample of 25 rolls. The sample proportion will vary: there is sampling variation.

Describing how the sample proportion varies from sample to sample is useful (Sect. 15.4.2).

To do this, statistical theory could be used... or thousands of repetitions of a sample of 25 rolls could be performed... or a computer could simulate many sets of 25 rolls.

Let's simulate rolling a die 25 times, using just 10 sets of 25 rolls; see the animation below.

The proportion of rolls that is even varies from set to set. For these 10 sets of \(n=25\) rolls, the percentage of even rolls ranged from \(\hat{p} = 0.32\) even rolls to \(\hat{p} = 0.60\) even rolls.

The sample proportion of even rolls would be expected to vary around \(p = 0.5\), since three of the six faces of the die are even numbers (the population proportion), using the classical approach to probability.

Of course, the sample proportion could be very small or very high by chance, but we wouldn't expect to see that very often.

In this example, the population proportion of even rolls is known to be \(p = 0.5\). Each sample of \(n = 25\) rolls is a sample of all possible sets of \(n=25\) rolls, and the sample proportion of even rolls is denoted by \(\hat{p}\).

For any sample of 25 rolls, the value of \(\hat{p}\) will be unknown until we roll the die. The proportion of even rolls is likely to vary from sample to sample; that is, the sample proportions exhibit sampling variation, and the amount of sampling variation is quantified using a standard error.

\(p\) refers to the population proportion, and \(\hat{p}\) refers to the sample proportion.

The symbol \(\hat{p}\) is pronounced 'pee-hat'.

Suppose a fair die was rolled 25 times, and this was repeated thousands of times (not just 10 times as in the animation above), and the proportion of even rolls was recorded for every sample of 25 rolls.

These thousands of sample proportions \(\hat{p}\), one from every sample of 25 rolls, could be graphed using a histogram; see the animation below.

The shape of the histogram is roughly a normal distribution. This is no accident: statistical theory says this will happen (when certain conditions are met: see Sect. 20.6).

The possible values of the sample proportion \(\hat{p}\) have a sampling distribution which is roughly a normal distribution; the mean and standard deviation of the normal distribution the animation above can even be determined. The possible values of the sample proportion \(\hat{p}\) will have a sampling distribution, described by:

  • an approximate normal distribution;
  • centred around a mean of \(p = 0.5\);
  • with a standard deviation of 0.1 (where this number comes from will be revealed soon).

This distribution is called a sampling distribution, as discussed in Sect. 18.1. The standard deviation of the sampling distribution is called a standard error, since it measures how much a sample statistic (in this case, a sample proportion \(\hat{p}\)) varies from sample to sample.

Since the variation in the sample proportions can be described, a picture of this normal distribution can be drawn (Fig. 20.1). We still don't know exactly what we'll find next roll... but we have some idea of how the sample proportion is likely to vary in sets of 25 rolls.

The normal distribution, showing how the proportion of even rolls varies when a die is rolled 25 times

FIGURE 20.1: The normal distribution, showing how the proportion of even rolls varies when a die is rolled 25 times

The parameter \(p\) and the statistic \(\hat{p}\) are both proportions, but a mean and standard deviation are used to describe the sampling distribution.

The value of \(p\) (the population proportion: the proportion of even numbers on the die) remains the same, but the value of \(\hat{p}\) (the sample proportion: the proportion of even numbers in the sample of 25 rolls) is not the same in every sample of 25 rolls. That is, \(\hat{p}\) varies, and exhibits sampling variation. The variation in \(\hat{p}\) from sample to sample is measured by the standard error of the sample proportion, written as \(\text{s.e.}(\hat{p})\).

In general, the standard error for a sample proportion when the value of \(p\) is known is given by

\[\begin{equation} \text{s.e.}(\hat{p}) = \sqrt{\frac{p \times (1-p)}{n}}, \tag{20.1} \end{equation}\] where \(n\) is the number of rolls, and \(p\) is the population proportion. For this example, there are \(n = 25\) rolls of a die, and the population proportion of even rolls is \(p = 0.5\). Then, the standard error of the sample proportion is

\[\begin{equation} \text{s.e.} (\hat{p}) = \sqrt{\frac{0.5 \times (1-0.5)}{25}} = 0.1. \tag{20.2} \end{equation}\]

This standard error is the standard deviation of the normal distribution in Fig. 20.1.

Recall that the the standard error is just a special standard deviation, that measures how much a sample estimate is likely to vary from sample to sample. In that sense, the standard error of the proportion measures how precisely \(\hat{p}\) estimates the population proportion \(p\).

Almost always, the value of \(p\) is unknown, so when \(\text{s.e.}(\hat{p})\) is computed, the value of \(p\) can't be used. Instead, the best available estimate of \(p\) is used, which is \(\hat{p}\). This situation is studied from Sect. 20.3 onwards.

Definition 20.1 (Sampling distribution of a sample proportion when p is known) When the value of \(p\) is known, the sampling distribution of the sample proportion is described by

  • an approximate normal distribution,
  • centred around a mean of \(p\),
  • with a standard deviation (called the standard error of \(\hat{p}\)) of

\[ \text{s.e.}(\hat{p}) = \sqrt{\frac{ p \times (1-p)}{n}}, \]

when certain conditions are met, where \(n\) is the size of the sample, and \(p\) is the population proportion.

In general, the approximation gets better as the sample size gets larger.

From the die example, the values of \(\hat{p}\) will vary:

  • with an approximate normal distribution;
  • centred around \(p = 0.5\); and
  • with a standard error of approximately \(\text{s.e.}({\hat{p}}) = 0.1\).

This distribution is shown in Fig. 20.1. Based on this picture, how often would a value of \(\hat{p}\) larger than 0.80 be expected?

Figure 20.1 suggests that, while not impossible, 0.80 or greater will be observed rarely.

20.2 Sampling intervals: Known proportion

The possible values of the sample proportions \(\hat{p}\) can be described by an approximate normal distribution, as just discussed. This enables the 68--95--99.7 rule to be applied; for example, about 68% of the time with sets of 25 rolls, the sample proportion of even rolls will be between \(0.5\) give-or-take one standard deviation (that is, give-or-take 0.1). So, about 68% of the time, the proportion of even rolls in a set of 25 rolls will be between:

  • \(0.5 - 0.1 = 0.4\) and
  • \(0.5 + 0.1 = 0.6\).

Similarly, about 95% of the time, the proportion of even rolls will be between \(0.5\) give-or-take two standard deviations, or between:

  • \(0.5 - (2\times0.1) = 0.3\) and
  • \(0.5 + (2\times0.1) = 0.7\).

This interval tell us what values of \(\hat{p}\) are likely to be observed in samples of size 25. Most of the time (i.e., approximately 95% of the time), the value of \(\hat{p}\) is expected to be between 0.30 and 0.70. (For instance, in the animation above, all ten sets of 25 rolls (or 100%) had a sample proportion betweeen 0.30 and 0.70.)

More formally, the sample proportion \(\hat{p}\) is likely to lie within the interval

\[ p \pm (\text{multiplier} \times \text{s.e.}(\hat{p})), \] where \(\text{s.e.}(\hat{p})\) is the standard error of the sample proportion (calculated using Eq. (20.1)).

The symbol '\(\pm\)' means 'plus or minus', or (colloquially) 'give-or-take'.

The multiplier depends on how confident we wish to be that the interval contains the value of \(\hat{p}\).

For a 95% interval---the most common level of confidence---the multiplier is approximately 2, based on the 68--95--99.7 rule: Approximately 95% of observations are within two standard deviations of the value of \(p\) (the mean of the normal distribution in Fig. 20.1).

That is, the approximate 95% interval is:

\[ p \pm (2 \times \text{s.e.}(\hat{p}) ). \] For a 90% interval, either tables or a computer would be used to find the correct multiplier, since the 68--95--99.7 rule isn't helpful.

In practice, 95% intervals are the most common, and we'll use a multiplier of \(2\) to find an approximate 95% interval when computing the interval without using software. Software can be used for any other percentage interval (or for an exact 95% interval).

20.3 Sampling distribution: Unknown proportion

In the die example (Sect. 20.1), an equation was given for computing the standard error for the sample proportion for samples of size \(n\), when the value of \(p\) was known.

However, usually the value of \(p\) (the parameter) is unknown; after all, the reason for taking a sample is to estimate the unknown value of \(p\). When \(p\) is unknown, the best available estimate can be used, which is \(\hat{p}\). When the value of \(p\) is unknown, the standard error of the sample proportion (written \(\text{s.e.}(\hat{p})\)) is approximately

\[\begin{equation} \text{s.e.}(\hat{p}) = \sqrt{\frac{ \hat{p} \times (1 - \hat{p})}{n}}. \tag{20.3} \end{equation}\]

Definition 20.2 (Sampling distribution of a sample proportion when p is unknown) When the value of \(p\) is unknown, the sampling distribution of the sample proportion is described by

  • an approximate normal distribution,
  • centred around the (unknown) mean of \({p}\),
  • with a standard deviation (called the standard error of \(\hat{p}\)) of

\[\begin{equation} \text{s.e.}(\hat{p}) = \sqrt{\frac{ \hat{p} \times (1-\hat{p})}{n}}, \tag{20.4} \end{equation}\] when certain conditions are met, where \(n\) is the size of the sample, and \(\hat{p}\) is the sample proportion.

In general, the approximation gets better as the sample size gets larger.

Let's pretend for the moment that the proportion of even rolls of a fair die is unknown (to demonstrate some points). In this case, an estimate of the proportion of even rolls can be found by rolling a die \(n = 25\) times and computing \(\hat{p}\).

Suppose 11 of the \(n = 25\) rolls produced an even number, so that \(\hat{p} = 11/25 = 0.44\). Then (from Definition 20.2),

\[ \text{s.e.}(\hat{p}) = \sqrt{ \frac{ 0.44 \times (1 - 0.44)}{25}} = 0.099277. \] (This is very similar to the value of 0.1, the value of the standard error when the value of \(p\) was known; see Eq. (20.2).)

Hence, the sample proportions would vary with an approximate normal distribution (Fig.20.2), centred around the unknown value of \(p\) with a standard deviation of \(\text{s.e.}(\hat{p}) = 0.099277\).

The normal distribution, showing how the proportion of even rolls varies when a die is rolled 25 times

FIGURE 20.2: The normal distribution, showing how the proportion of even rolls varies when a die is rolled 25 times

Using the 68--95--99.7 rule again:

About 95% of the values of \(\hat{p}\) are expected to be between \(p - 0.199\) and \(p + 0.199\).

Though we are pretending the value of \(p\) is unknown, the value of \(\hat{p}\) is known however. What if the roles of \(p\) and \(\hat{p}\) were 'reversed'? Then,

About 95% of the values of \(p\) are expected to be between \(\hat{p} - 0.199\) and \(\hat{p} + 0.199\).

Since \(\hat{p} = 0.44\), this is equivalent to:

About 95% of the values of \(p\) are expected to be between \(0.24\) and \(0.64\).

This interpretation is not quite correct, but the idea seems reasonable. This is called a confidence interval (or CI), based on ideas from Sect. 20.2.

In summary, using \(\hat{p} = 0.44\) and \(\text{s.e.}(\hat{p}) = 0.0993\), the (approximate) 95% CI is

\[ 0.44 \pm (2 \times 0.0993), \] or from 0.241 to 0.639. This CI straddles the population proportion of \(p = 0.5\), though we would not know this if \(p\) truly was unknown.

In this case, we know the value of the population parameter: \(p = 0.5\).

Usually we do not know the value of the parameter. That's why we are taking a sample; to estimate the value of the population proportion.

20.4 Confidence intervals: Unknown proportion

Suppose thousands of people rolled a die 25 times, and each person found \(\hat{p}\) for their sample, and hence computed the CI for their sample of 25 rolls.

Every sample of 25 rolls could produce a different estimate \(\hat{p}\), and so a different value for \(\text{s.e.}(\hat{p})\), and hence a different 95% CI. However, about 95% of these thousands of confidence intervals from those thousands of repetitions would straddle the true proportion \(p\).

Since we usually don't know the value of \(p\), and we usually only have one sample (and hence one CI), in general we never know whether the single CI computed from our single sample straddles \(p\) or not.

Again, consider letting the computer simulate the situation. Suppose the process of recording the sample proportion of even numbers in \(n = 25\) rolls is repeated 50 times, and for each of those 50 sets of 25 rolls a CI is produced (see the animation below).

Most of those CIs straddle the population proportion of \(p = 0.5\) (shown as solid lines)... but some do not (shown as dashed lines). Of course, since the value of \(p\) is usually unknown, we never know if our CI contains \(p\) or not.

Definition 20.3 (Confidence interval) A confidence interval is an interval in which the population parameter is likely to be contained, if we found many samples the same way.

If a 95% confidence interval (or CI) is computed from each sample, about 95% of the CIs would straddle the parameter of interest. This interval is called a confidence interval.

In general, a CI for the population proportion \(p\) is found using

\[ \hat{p} \pm ( \text{multiplier} \times \text{s.e.}(\hat{p})), \] where the multiplier is 2 for an approximate 95% CI (based on the 68--95--99.7 rule).

Definition 20.4 (Confidence interval for p) A confidence interval (CI) for the unknown value of the parameter \(p\) is \[\begin{equation} \hat{p} \pm ( \text{multiplier} \times \text{s.e.}(\hat{p})), \tag{20.5} \end{equation}\] where

\[ \text{s.e.}(\hat{p}) = \sqrt{\frac{ \hat{p} \times (1 - \hat{p}) }{n}} \] is the standard error of \(\hat{p}\), \(\hat{p}\) is the sample proportion, and \(n\) is the sample size. For an approximate 95% CI, the multiplier is 2.

In general, higher confidence means wider intervals (Fig. 20.3), since wider intervals are needed to be more certain that the interval contains \(\hat{p}\). Try changing the confidence level for the CI in the interaction below.

FIGURE 20.3: Changing the confidence level of the CI changes the width

Example 20.1 (Energy drinks in Canadian youth) A study of young Canadians aged 12--24374 found that 365 of the 1516 respondents reported sleeping difficulties after consuming energy drinks.

The unknown parameter is \(p\), the population proportion of young Canadians reporting sleeping difficulties.

The sample proportion reporting sleeping difficulties after consuming energy drinks is \(\hat{p} = 365/1516 = 0.241\). As usual, the sample proportion would vary from one sample of size \(n = 1516\) to another; sampling variation exists. The standard error (Definition 20.4) quantifies how much the sample proportion is likely to vary from sample to sample:

\[\begin{align*} \text{s.e.}(\hat{p}) &= \sqrt{\frac{\hat{p}\times(1-\hat{p})}{n}}\\ &= \sqrt{\frac{0.241 \times (1-0.241)}{1516}} = 0.01098449, \end{align*}\] or about \(0.011\). So, in samples of size 1516, the approximate 95% CI (Definition 20.4) is between

  • \(0.241 - (2\times 0.01098449) = 0.2190\) and
  • \(0.241 + (2\times 0.01098449) = 0.2627\).

The approximate 95% CI is from 0.219 to 0.263.

This CI may or may not straddle the population proportion \(p\); it is likely that the interval straddles the value of \(p\). In other words, it is plausible that the sample proportion of \(p = 0.241\) may have come from a population with a proportion somewhere between 0.219 and 0.263.

Notice that many decimal places are used in the working, but final answers are rounded.

Example 20.2 (Koalas crossing roads) A study of koalas375 found that 18 of the \(n = 51\) koalas studied in a certain area (over 30 months) had crossed at least one road during that time.

The unknown parameter is \(p\), the population proportion of koalas that had crossed at least one road over the 30 months.

The sample proportion having crossed a road is \(\hat{p} = 18/51 = 0.3529\). The standard error (Definition 20.4) is

\[\begin{align*} \text{s.e.}(\hat{p}) &= \sqrt{ \frac{ \hat{p} \times (1 - \hat{p})}{n} }\\ &= \sqrt{ \frac{0.3529 \times (1 - 0.3529)}{51} }\\ &= 0.06692. \end{align*}\] An approximate 95% CI, then, is \(0.3529 \pm (2 \times 0.06692)\), or

\[ 0.3529 \pm 0.1338. \] The margin of error is \(0.1338\).

Computing the 'plus' and the 'minus' bits, the approximate 95% CI is from 0.219 to 0.487 (after rounding appropriately).

The approximate 95% CI for the population proportion of koalas that crossed at least one road in the last 30 months is from 0.219 to 0.487. That is, it is plausible that the sample proportion of \(\hat{p} = 0.3529\) may have come from a population with a proportion somewhere between 0.219 and 0.487.

The research article reports:

Of the 51 koalas, 18 (35.3%) crossed at least one road. The [...] probability of a koala crossing at least one road during the study was 35.3% (95% CI = 22--48%).

--- Dexter et al.376, p. 70.

This agrees with our calculations.

Example 20.3 (CI) A study of how paramedics administer pain medication,377 and to whom, found that

Forty-five percent of patients reporting pain did not receive analgesia (791/1766) (95% confidence interval [CI], 43%-47%).

--- Lord, Cui, and Kelly378, p. 525

That is, \(\hat{p} = 791/1766 = 0.4479049\) and \(n = 1766\).

Hence,

\[ \text{s.e.}(\hat{p}) = 0.01183326, \] so the approximate 95% CI is from \[ 0.4479049 - (2\times 0.01183326) = 0.4242384 \] to \[ 0.4479049 + (2\times 0.01183326) = 0.47157134, \] or from \(0.424\) to \(0.472\), which agrees with the article.

(Notice that many decimal places were kept in the working, but the final answers were rounded.)

20.5 Interpretation of a CI

The correct interpretation (Definition 20.3) of a 95% CI is the following:

If samples were repeatedly taken many times, and the 95% confidence interval computed for each sample, 95% of these confidence intervals formed would contain the population parameter.

However, most people would think of our 95% CI as having a 95% chance of containing the value of population parameter \(p\). This is not strictly correct (our CI either does or does not contain the value of \(p\)), but is common.

More details on interpreting a CI are given in Sect. 21.2.

20.6 Statistical validity conditions

The histogram in Sect. 20.1, shows the proportion of \(n = 25\) rolls that were even for many samples; it has an approximate normal distribution. Because of this, the 68--95--99.7 rule could be used to form the approximate 95% CIs.

However, the distribution of the sample proportions only looks like a normal distribution under certain conditions. Certain conditions must be true for the calculations to be sensible, or statistically valid.

Definition 20.5 (Statistical validity) A result is statistically valid if the conditions for the underlying mathematical calculations and assumptions to be approximately correct are met. Every confidence interval has statistical validity conditions.

Example 20.4 (Statistical validity analogy) Suppose your doctor asks you to get a blood test, after fasting (refraining from eating) for 12 hours before your blood test.

After leaving the doctor, you proceed to a restaurant for dinner. You start the next day with a hearty breakfast, have lunch at a beach-side cafe, and then go for your blood test. Your blood is extracted, the blood is analysed in the pathology lab, and your doctor is emailed the results of the blood test.

However, since you did not fast as required, the results may or may not be valid. The doctor can still learn something... but not as much as if you had followed the instructions.

Similarly, if the conditions for computing the confidence interval are not met, the results may be suspect.

The statistical validity conditions for creating CI for a single proportion is that:

  • the number of individuals in the group of interest must exceed 5, and
  • the number of individuals in the group not of interest must exceed 5.

These conditions ensure that the sampling distribution of \(\hat{p}\) has an approximate normal distribution, so that the 68--95--99.7 rule (approximately) applies. If this condition is not met, the sampling distribution may not have normal distribution, so the 68--95--99.7 rule (used to create the CI) may be inappropriate, and so the CI may also be inappropriate.

In addition to the statistical validity condition, the CI will be:

Example 20.5 (Energy drinks in Canadian youth) In Example 20.1, the approximate 95% CI was from 0.192 to 0.236. This confidence interval for the sample proportion will be statistically valid if:

  • the number of youth in the sample who experienced sleeping difficulties exceeds 5; and
  • the number of youth in the sample who didn't experience sleeping difficulties exceeds 5.

The number of youth experiencing sleeping difficulties was 365, which is more than five. The number of youth not experiencing sleeping difficulties was \(1516 - 365 = 1151\), which is also more than five. Hence, the CI is statistically valid.

In addition, the CI will be internally valid if the study was well designed, and will be externally valid if the sample is a simple random sample from the population and is internally valid.

Consider Example 20.2, about koalas crossing roads.

Is the CI likely to be statistically, internally and externally valid?

Statistically: The number of koalas that had crossed a road is 18, and the number that had not crossed a road is \(51 - 18 = 33\). Both of these exceed 5, so the CI will be statistically valid.

Internally: We do not have enough information about how the study was conducted to know. But presumably the scientist conduct the study well.

Externally: It seems unlikely that a random sample of koalas would be obtained, so probably not.

Example 20.6 (Statistical validity) Consider an artificial situation to estimate the proportion of die rolls that show as a one. The population proportion (using the classical approach to probability) is 1/6, or about 0.167.

If we repeatedly rolled a die in sets of \(n = 20\) rolls, say 5000 times, the proportion of rolls that showed as one could be recorded for each sample of 20 rolls. Then, a histogram of the sample proportions could be produced.

Using a computer to simulate this, a histogram of the sample proportions is shown in the top panel of Fig. 20.4. The normal distribution does a poor job of describing the sampling distribution (the distribution is not even symmetric). The statistical validity conditions do not seem satisfied.

Alternatively, we could repeatedly roll a die in sets of \(n = 60\) rolls, say 5000 times, and record the proportion of rolls that show as one for each sample of 60 rolls.

Then, a histogram of the proportion of ones for those sets of 60 rolls could be produced. Using a computer to simulate this, a histogram of these proportions is shown in the bottom panel of Fig. 20.4. The normal distribution does a reasonable job of describing the sampling distribution. The statistical validity conditions seem satisfied.

The sampling distribution of the proportion of ones rolled, for sets of 20 rolls (top panel) and sets of 60 rolls (bottom panel)

FIGURE 20.4: The sampling distribution of the proportion of ones rolled, for sets of 20 rolls (top panel) and sets of 60 rolls (bottom panel)

20.7 Summary: Finding a CI for \(p\)

The procedure for computing a confidence interval (CI) for a proportion is:

  • Compute the sample proportion, \(\hat{p}\), and identify the sample size \(n\).
  • Compute the standard error, which quantifies how much the value of \(\hat{p}\) varies from one sample to the next:

\[ \text{s.e.}(\hat{p}) = \sqrt{\frac{ \hat{p} \times (1-\hat{p})}{n}}. \]

  • Find the multiplier: this is \(2\) for an approximate 95% CI using the 68--95--99.7 rule. (Note: (Multiplier\(\times\)standard error) is called the margin of error.)
  • Compute:

\[ \hat{p} \pm \left( \text{Multiplier}\times\text{standard error} \right). \]

You must use proportions in this formula, not percentages (that is, values between 0 and 1 (like 0.23) and not 23%).

Example 20.7 (NHANES data) For the NHANES data, first seen in Sect. 12.9, the unknown parameter is \(p\), the population proportion of Americans that currently smoke.

In the study, 1466 out of the 3211 respondents who reported their smoking status said they currently smoked: \(\hat{p}= 1466\div 3211 = 0.4566\).

What is the population proportion \(p\) that currently smoke? We don't know, and the estimate of \(p\) from every sample is likely to be different. The standard error is \(\text{s.e.}(\hat{p}) = 0.00879\), so the approximate 95% CI for \(p\) is \(0.4566\pm 0.01758\), or from 0.439 to 0.474. (Check the calculations!)

For the conclusions to be statistically valid, the number of smokers must exceed 5, and the number of non-smokers must exceed 5. Both are true. The CI appears to be statistically valid.

We write:

Based on the sample, we are approximately 95% confident that the interval from 0.429 to 0.474 straddles the population proportion of smokers in the USA.

20.8 Example: Female coffee drinkers

A study of 360 female college students in the United States379 found that 61 drank coffee daily.

The unknown parameter is \(p\), the population proportion of female college students in the United States that drink coffee daily.

The sample size is \(n = 360\), and the sample proportion of daily coffee drinkers is \(\hat{p} = 61/360 = 0.16944\). Another sample of 360 students from the same population is likely to produce a different sample proportion \(\hat{p}\) of daily coffee drinkers: the sample proportion has sampling variation. The size of this sampling variation is quantified using a standard error; from (20.4):

\[ \text{s.e.}(\hat{p}) = \sqrt{ \frac{ 0.16944 \times (1 - 0.16944)}{360}} = 0.01977. \]

An approximate 95% CI is \(0.1694 \pm (2 \times 0.01977)\), or \(0.1694 \pm 0.03954\). That is, the margin of error is \(0.03954\).

Computing the 'plus' and the 'minus' bits, the approximate 95% CI is from \(0.1694 - 0.03954 = 0.12986\) to \(0.1694 + 0.03954 = 0.20894\). Round appropriately, the approximate 95% CI is from \(0.130\) to \(0.209\).

The plausible values for \(p\) that may have led to this value of \(\hat{p} = 0.1694\) are between 0.130 and 0.209. (This CI may or may not contain the true proportion \(p\).)

This CI is statistically valid. We cannot comment on the internal validity: we would need details of how the study was conducted.

The CI is externally valid if the sample is simple random sample of some population, and the study is internally valid. The CI is approximately externally valid if the sample is somewhat representative of some population, and the study is internally valid.

20.9 Quick review questions

  1. True or false: \(p\) is called a parameter.
  2. True or false: The value of \(p\) will vary from sample to sample.
  3. True or false: The standard error refers to the sampling variation in \(p\).
  4. Suppose \(n=50\) and \(\hat{p} = 0.4\). What is the standard error?

Progress:

20.10 Exercises

Selected answers are available in Sect. D.19.

Exercise 20.1 A study of salt intake in the United Kingdom380 found that 2,182 out of the 6,882 people sampled in 2007 'generally added salt at the table'.

Find an approximate 95% CI for the population proportion of Britons that generally add salt at the table.

Exercise 20.2 A study of the eating habits of university students in Canada381 found that 8 students out of 154 met the recommendation for eating a sufficient number of servings of grains each day.

  1. Find an approximate 95% CI for the population proportion of Canadian students that meet the recommendation for eating a sufficient number of servings of grains each day.
  2. Would these results be likely to apply to Australian university students? Why or why not?

Exercise 20.3 A meta-study of hiccups382 found that, of 864 patients examined (across many different studies) who had hiccups, 708 were male.

  1. Find an approximate 95% CI for the true proportion of people with hiccups who are male.
  2. Check if the statistical validity conditions are met or not.
  3. Draw a sketch of how the sample proportion varies from sample to sample for samples of size 864.

Exercise 20.4 We wish to estimate the population proportion of Australians that smoke.

  1. Suppose we wish our 95% CI to be give-or-take \(0.05\). How many Australians would we need to survey?
  2. Suppose we wish our 95% CI to be give-or-take \(0.025\); that is, we wish to halve the width of the interval above. How many Australians would we need to survey?
  3. How many times as many Australians are needed to halve the width of the interval?

Exercise 20.5 A study of turbine failures383 ran 42 turbines for around 3000 hours, and found that nine developed fissures (small cracks). Find a 95% CI for the true proportion of turbines that would develop fissures after 3000 hours of use. Are the statistical validity conditions satisfied?

The study also ran 39 turbines for around 400 hours, and found that zero developed fissures. Find a 95% CI for the true proportion of turbines that would develop fissures after 400 hours of use. Are the statistical validity conditions satisfied?