When we make a measurement, we usually want to report the result along with an estimate of the likely error. Sometimes we call it the margin of error. That error is an estimate of the width of the distribution of very many measurements of the same type. Often we simply use the standard deviation, $\sigma$. Now we'd like to go further and express our results using confidence intervals of our choosing. That means we can describe our results with sentences like:
On this page we'll divide confidence intervals into those describing measurements of proportion $(p)$ and of means $(\bar x \; \text{or} \; \mu)$. The concept is the same but the details are different.
In math and the sciences, including social sciences and data science, we often refer to making a "measurement." The measurement in this case is just a piece of data that in part answers the question being asked. Here are some examples:
Study | Measurement |
---|---|
What is the average height of adult men in the U.S.? | Measure the heights of a group of men to a given precision, say the nearest half centimeter. |
Do voters approve of the President's job performance? | Ask people whether they approve and accumulate the data as a proportion who approve. |
Is this six-sided die "loaded"? | Toss the die 1,000 times and record the number of times it lands with each face up. |
The basic idea of a confidence interval isn't a great mystery. This figure shows the standard normal distribution (top graph), defined as the normal distribution with center $\mu = 0$ and standard deviation $\sigma = 1$, and which encloses an area of 1.
That's where we begin. We find the limits (Z-scores) of the region below the graph that encloses the amount of probability we want, say 95% of it. That area is centered on the mean, leaving two "tails" on either side. The Z-scores are found using standard normal tables or a program on a calculator, such as the invNorm( ) program on a TI-84.
The second step is to simply scale that Z-score to the distribution of the data with which we're actually working by multiplying it by the standard error of that data. We map the Z-score of the SND onto the distribution of our data or sample.
We're used to using the standard deviation $(±\sigma)$ to describe the possible error in a statistical estimate of a parameter; here we're just replacing $\sigma$ with a more deliberate choice of error range.
The Z-score of a location, $x$ in a set of data — in the domain of a probability function — is just its distance – left or right – away from the mean, $\bar x$, expressed in units of the standard deviation, $\sigma$:
$$Z = \frac{x - \bar x}{\sigma}$$
Division of a normal distribution into sections of width $\mu ± n\sigma$, where $n$ is 1, 2, 3, ... , $\mu$ is the mean and $\sigma$ is the standard deviation, should be familiar to you. The area under the normal distribution curve, which represents all of the probability for a particular measurement or event, can be conveniantly divided into chunks of width $\mu ± \sigma$, $\mu ± 2\sigma$, and so on.
If we express our mean as $\mu ± \sigma$, then what we're really saying is that we are 68% confident that our estimate of $\mu$ is in the interval $[\mu - \sigma, \mu + \sigma]$. Take a look at the normal curve below and notice that this means there is a 32% chance that the estimate of $\mu$ is outside of this interval.
If we broaden our interval to $\mu ± 2\sigma$ then there is a 95% probability that our estimate of $\mu$ is in the interval $[\mu - 2\sigma, \mu + 2\sigma]$, and only a 5% chance that it is outside of that interval.
That's a way of expressing higher confidence in our estimate, but with the obvious corollary that the range of our prediction has been broadened.
Now we are free to choose any confidence interval we'd like. In practice, the 95% confidence interval is quite a common one, and we'll use it a lot. Certain situations will dictate using narrower or broader intervals.
In this section we'll figure out how to express means and proportions of samples of populations using whatever confidence interval we choose — but that choice ought to be an informed one.
The percentages we often list as the "68-95-99 rule" for $±\sigma, \, ±2\sigma, ± 3\sigma$ are rounded for convenience. The actual percentages are
$$ \begin{align} \mu ± \sigma &\longrightarrow 68.27\% \\[5pt] \mu ± 2\sigma &\longrightarrow 95.45\% \\[5pt] \mu ± 3\sigma &\longrightarrow 99.73\% \end{align}$$
A 95% confidence interval centered around the estimate of a mean looks like the figure below. The blue area constitutes 95% of the total probability of whatever sample we take. There is only a 5% chance, on average, that our mean lies outside of this interval.
We need to have a mathematical way (a "formalism") for expressing and calculating confidence intervals. Check out the new labeling of the normal distribution below.
We replace 95% with the expression $1 - \alpha$, where $\alpha$ represents the probability outside of our confidence interval. Sometimes we call that area the tails. For a 75% confidence interval, we'd have $1 - \alpha = 0.75$, or $\alpha = 0.25$. The probability in both of the tails is 25% of the total, or 12.5% each due to the left-right symmetry of the normal distribution. Here's that setup.
Now our confidence intervals will be the Z-scores (labeled $Z_{\frac{\alpha}{2}}$ above). Those can be determined using inverse-normal tables, or by using the inverse-normal function on a TI-84 calculator, invnorm( ).
Now notice that we've derived this description using the standard normal distribution, for which $\mu = 0$ (the mean) and its standard deviation is $\sigma = 1$. In order to "map" this result onto any normal distribution, we just multiply those Z-scores by the standard deviation of the distribution that's useful to us. Remember, however, that the standard deviation can change depending on the size of the sample we take – say of $n$ individuals. so our formula for the appropriate Z-scores – the ones that will bookend our interval of confidence – is
$$MOE = Z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}},$$
where MOE stands for margin of error.
You can use the inverse normal [invNorm( )] function on a TI-84 calculator to calclulate Z-score that goes with a given area or probability under the standard normal curve $(\mu = 0, \; \sigma = 1)$. Here's how to do it:
In the distrib menu [2ND][DISTR], select invNorm( ).
Fill in the parameters: area = the left-tail probability; $\mu$ = the standard deviation of the distribution (zero for the standard normal distribution, SND); and $\sigma$ is the standard deviation (one for the SND).
Just calculate. Notice that the answer will be negative because the program works from left to right, but there are two matching Z-scores for each tail; just use the positive (right-side) one.
Use your calculator or a standard normal table to calculate the Z-scores for the given α levels.
1. 0.10
This area corresponds to an 90% confidence interval, in which each of the two tails accounts for 5% of the probability or area. Enter
invNorm(0.05, 0, 1)
on a TI-84 calculator to get the result, $Z = -1.6449$. The two Z-scores that form part of the MOE are ±1.6449.
2. 0.20
This area corresponds to an 80% confidence interval, in which each of the two tails accounts for 10% of the probability or area. Enter
invNorm(0.10, 0, 1)
on a TI-84 calculator to get the result, $Z = -1.2816$. The two Z-scores that form part of the MOE are ±1.2816.
3. 0.05
This area corresponds to a 95% confidence interval, in which each of the two tails accounts for 2.5% of the probability or area. Enter
invNorm(0.025, 0, 1)
on a TI-84 calculator to get the result, $Z = -1.96$. The two Z-scores that form part of the MOE are ±1.96.
For sample proportion problems, the margin of error is
$$\text{MOE} = Z_{\frac{\alpha}{2}} \cdot \sqrt{\frac{p(1-[p)])}{n}}$$
and the confidence interval is
$$p \in [\hat p - \text{MOE}, \, \hat p + \text{MOE}]$$
where $p$ is the population proportion, $\hat p$ is the sample proportion, $n$ is the number of individuals in the sample and $\text{MOE}$ is the margin of error.
In order for confidence interval calculations to be valid, these conditions must be met for sampling:
The large count conditions must apply:
Suppose that a research firm estimates the percent of adults living in a city who have cell phones less than 1 year old. Four hundred adults are selected randomly and surveyed. Of the 400 people, 132 responded "yes." Use a 95% confidence interval to estimate the proportion of adults in the city who have newer cell phones.
The proportion of new phone owners is
$$\hat p = \frac{132}{400} = 0.33$$
The Z-score for a 95% confidence interval is found using the invNorm( ) function on a TI-84 calculator:
invNorm(0.025, 0, 1) = 1.96
Then the margin of error is
$$ \begin{align} \text{MOE} &= Z_{\frac{\alpha}{2}} \sqrt{\frac{\hat p (1 - \hat p)}{n}} \\[5pt] &= 1.96 \frac{0.33(1-0.33)}{400} \\[5pt] &= 0.0461 \end{align}$$
So the confidence interval is
$$ \begin{align} p ± \text{MOE} &= 0.33 ± 0.0461 \\[5pt] p &\in [0.33 - 0.0461, \, 0.33 + 0.0461] \\[5pt] p &\in [0.28, \, 0.38] \end{align}$$
With 95% confidence, the proportion of our city adults who own cell phones that are less than a year old is between 28% and 38%.
This table collects some common critical Z values (Z^{*})for solving proportion problems. Often you'll use some of them so much you'll memorize them. You might also consider storing one or two in a memory slot on your calculator.
Notice the confidence intervals for Z-scores of 3.0 and 1.0, which are 97.5% and 68%, respectively. These are familiar divisions of the normal probability distribution — the 68-95-97.5 rule.
Z^{*} |
Confidence interval |
---|---|
3.0 | 99.7% |
2.576 | 99% |
1.96 | 95% |
1.645 | 90% |
1.282 | 80% |
1.15 | 75% |
1.0 | 68% |
A random sample of 1000 community college students taken from all CC students in the United States reveals that females constitute 60% of the student body. Calculuate the 90% confidence interval around this result.
The proportion of new phone owners is
$$\hat p = 0.60$$
The Z-score for a 90% confidence interval is found using the invNorm( ) function on a TI-84 calculator:
invNorm(0.05, 0, 1) = 1.6449
Then the margin of error is
$$ \begin{align} \text{MOE} &= Z_{\frac{\alpha}{2}} \sqrt{\frac{\hat p (1 - \hat p)}{n}} \\[5pt] &= 1.6449 \frac{0.60(1-0.60)}{1000} \\[5pt] &= 0.0255 \end{align}$$
So the confidence interval is
$$ \begin{align} p ± \text{MOE} &= 0.60 ± 0.0255 \\[5pt] p &\in [0.60 - 0.0255, \, 0.60 + 0.0255] \\[5pt] p &\in [0.574, \, 0.626] \end{align}$$
With 90% confidence, the proportion of community college students appears to be skewed toward females, because 0.50 or 50% is not included in the confidence interval. We ought to beware, though, that there is a 10% chance of the true proportion lying outside of this interval. Nothing is absolutely certain.
Let's say we want to determine the proportion of M&Ms^{®} that are blue. In which of the following situations are we able to construct a confidence interval?
We take a sample of 25 candies and find 8 to be blue.
We don't know the size of the parent population, so we don't know what fraction of that population a sample of 25 represents.
We look at one bag containing about 60 candies and observe that 12 are blue.
This is a sample of only one bag – presumably chosen randomly. We'd hope for a sample of many bags, up to 10% of some larger population of bags, from which we'd calculate an average proportion of blue M&Ms and its associated standard error.
We want to determine the proportion of blue M&Ms in one small bag (about 60 candies). To do so we take a simple random sample (SRS) of 25 candies from the bag and find 9 to be blue.
25 candies out of 60 is about 42% of the total population of 60. That's not a sample from which we can make reliable inferences. We need to keep the number sampled down to 10% of the total or fewer.
We randomly select 100 candies from a day's worth of factory production and find 23 to be blue.
This is a valid sample. The number, 100, is surely less than 10% of all candies produced in a day, and it's a random sample. The proportion of 0.23, coupled with the relatively large sample, ensures that we'll pass the large-count conditions.
A marketing firm is hired to estimate the proportion of adults in a large metropolitan area who have and use cell phones. They randomly select 500 residents of the area and ask a binary question: "Do you own a cell phone?" 402 of 500 people responded "yes." Using a 95% confidence level (&alpha = 0.05), calculate the mean and 95% confidence interval for the true proportion of metro-area residence who own cell phones.
First we should look at the sample quality. 500 is a large sample, but surely less than 10% of the adults in a large metro area. The sample is an SRS, and we can see without doing a calculation that the large-count conditions will hold.
We have
$$\hat p = \frac{402}{500} = 0.804$$
The critical Z value for a 95% confidence interval is $Z^* = 1.96$, so our margin of error is
$$ \begin{align} \text{MOE} &= Z_{\frac{\alpha}{2}} \sqrt{\frac{\hat p (1 - \hat p)}{n}} \\[5pt] &= ±1.96 \, \sqrt{\frac{0.804(1-0.804)}{500}} \\[5pt] &= ±0.035 \end{align}$$
So our 95% confidence interval would be best expressed as $[0.804 - 0.035, 0.804 + 0.035]$ or $[0.769, 0.839]$. So we can say with 95% confidence that the true proportion of residents who own cell phones is between 77% and 84%.
The chief financial officer (CFO) of a company wants to estimate the proportion of accounts due that are more than 30 days overdue. She surveys 500 of more than 1 million accounts and finds that 280 are more than 30 days overdue. Calculate a 90% confidence interval for the true proportion of accounts due that are late. Interpret the confidence interval.
First we should look at the sample quality. 500 is a large sample, but less than 10% of one million accounts due. The sample is an SRS, and we can see without doing a calculation that the large-count conditions will hold, both because of the sample size and the fairly large point estimate of the proportion,
$$\hat p = \frac{280}{500} = 0.56$$
The critical Z value for a 90% confidence interval is $Z^* = 1.645$, so our margin of error is
$$ \begin{align} \text{MOE} &= Z_{\frac{\alpha}{2}} \sqrt{\frac{\hat p (1 - \hat p)}{n}} \\[5pt] &= ±1.645 \, \sqrt{\frac{0.56(1-0.56)}{500}} \\[5pt] &= ±0.036 \end{align}$$
So our 90% confidence interval would be best expressed as $[0.56 - 0.036, 0.56 + 0.036]$ or $[0.524, 0.596]$. So we can say with 90% confidence that the true proportion of accounts due at this firm that are late is between 52% and 60%. That seems like a large proportion!
A campaign staff believes that 35% of all voters favor their candidate.
How large of a simple random sample is required so that the margin of error of a polling estimate seeking to explore that belief is no more than ±3% at the 95% confidence level?
We know that the margin of error is
$$\text{MOE} = Z_{\frac{\alpha}{2}} \sqrt{\frac{\hat p (1 - \hat p)}{n}}$$
Solving for the number of people sampled or surveyed (n) gives:
$$n = \frac{Z_{\frac{\alpha}{2}}}{\text{MOE}^2} \hat p (1 - \hat p)$$
which gives $n = 495$.
So in order to achieve a margin of error of 0.03 (3%) with a 95% confidence level and a point estimate of the proportion of $\hat p = 0.35$, we'd need to survey at least 495 people in a random sample.
Suppose that we want to take a poll such that we are 99% confident within ±3% of the true value. Calculate the sample size that would be required to achieve this result.
This calculation is the same as the one above, except that $Z_{\frac{\alpha}{2}} = 2.576$, yielding n = 651. It makes sense that the sample size would have to be larger to obtain a higher degree of confidence.
A simple random sample (SRS) of 200 college students is chosen from the student population of a large university. Of those sampled, 35 identified themselves as minority students. Calculate the 90% confidence interval for the true proportion of self-identified minority students at this university.
First we should look at the sample quality. 200 is a fairly large sample, but certainly less than 10% of the student population of a large university. The sample is an SRS, and we can check the large-count conditions by calculating the point estimate of the proportion:
$$\hat p = \frac{35}{200} = 0.175$$
The large count conditions are
$$ \begin{align} n \hat p &-= 200(0.175) = 35 \gt 10 \\[5pt] n (1 - \hat p) &= 200(0.825) = 165 \gt 10 \end{align}$$
... so we're good to go.
The critical Z value for a 90% confidence interval is $Z_{\frac{\alpha}{2}} = 1.645$, so our margin of error is
$$ \begin{align} \text{MOE} &= Z_{\frac{\alpha}{2}} \sqrt{\frac{\hat p (1 - \hat p)}{n}} \\[5pt] &= ±1.645 \, \sqrt{\frac{0.175(1-0.175)}{200}} \\[5pt] &= ±0.044 \end{align}$$
So our 90% confidence interval would be best expressed as $[0.175 - 0.044, 0.175 + 0.044]$ or $[0.131, 0.219]$. So we can say with 90% confidence that the true proportion of students at this university who self identify as a minority is between 13% and 22%.
When our sample data suggest a mean rather than a proportion, there are two kinds of problems we can encounter:
When we calculate a confidence interval, we're usually trying to make an estimate of a population parameter and to place it between some limits – plus or minus our margin of error. Thus we don't really know very much about the parameter and its variance. We only really know about our sample.
If we know the popuulation mean and our sample is of sufficient size and quality (n < 0.1N, n ≥ 30), we can assume that Z-scores are distributed normally and we can use them to calculate confidence intervals.
Those Z-scores are
$$Z = \frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n}}},$$
where $\sigma / \sqrt{n}$ is the standard error. If the standard deviation of the population is not known, then we can't use it to properly calculate normally-distributed Z-scores, so we need another method. What we do usually have is the standard deviation of the sample. We can substitute the sample standard deviation for the population $\sigma$ in these cases, but with the small cost that the Z-scores (which we'll call t values) are distributed according to a distribution called the t-distribution.
The critical t value, $t^*$, is
$$t_{\frac{\alpha}{2}} = \frac{\bar X - \mu}{\frac{s}{\sqrt{n}}},$$
where $s$ is the sample standard deviation, which we ought to know because we took the sample. These t-values are distributed according to the well-known t-distribution.
The t-distribution differs from the normal distribution in some important ways. First, the normal distribution, $N(\mu, \sigma)$, is a function of two parameters, the mean, $\mu$, and the standard deviation, $\sigma$. But the t-distribution, $t(\mu, \sigma, df)$ is also a function of the size of the sample, n, or more precisely the number of degrees of freedom, $df = n - 1$. Here are normal and t-distributions (df = 3) on the same graph.
The important shape differences are (1) the t-distribution is a little shorter than the normal distribution, and (2) the t-distribution has wider "tails," The latter has significant consequences because we're often asking about the area beyond some point into the tails. Now let's keep everything the same, but increase the sample size to give df = 19.
Notice from the graph above, that when we increase the sample size, and thus the number of degrees of freedom, the t-distribution
The yellow box in the graph above is expanded below, just so you can see that there's still a small difference in the width of the tails of the distributions.
Now let's do a couple of examples so you can see how this all works.
$$\text{MOE} = Z^* \, \frac{\sigma}{\sqrt{m}}$$
This is the unusual case, where we know the standard deviation of the variable of interest across the population.
$$\text{MOE} = t^* \, \frac{s_x}{\sqrt{n}}$$
This is the usual case, in which we substitute the sample standard deviation, $s_x$, for the population standard deviation, and use the t-distribution instead of the normal distribution.
The T-distribution is similar to the normal distribution except that the latter is a function of the known mean $\mu$ and its standard deviation $\sigma$. The T-distribution does not assume knowledge of those parameters, and is instead a function of the number of degrees of freedom of the problem — the number of data points minus one.
For a small number of degrees of freedom, the "wings" of the T-distribution are wider than those of the Normal distribution. But as the number of degrees of freedom increases, the two distributions converge.
Move the slider under the graph to change the number of degrees of freedom between 1 and 21 to see that convergence.
* Note on the simulation: Calculation of the T-distribution requires evaluation of the Gamma function, $\Gamma(X)$, for which I've used a bit of an approximation. That approximation breaks down slightly for larger degrees of freedom, and that's why the T-distribution curve (black) pops up above the Normal distribution for larger numbers of degrees of freedom; it shouldn't. My point was more to illustrate the differences in the wings, so that's fine with me.
In order for confidence interval calculations to be valid, these conditions must be met for sampling:
A count of Giardia lamblia, the parasite that causes Giardiasis, in one-hundred 1 mL samples of water yields a mean of 30 parasites per mL of water with a standard deviation of 8.3. Several counts of a 0.5L sample yielded a near-normal sampling distribution. Calculate the 95% confidence interval for this distribution and determine if a count of 40 in another sample lies outside of that interval.
invNorm(0.025, 0, 1) = 1.96
Then the margin of error is
$$ \begin{align} \text{MOE} &= Z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}} \\[5pt] &= 1.96 \frac{8.3}{\sqrt{100}} \\[5pt] &= 1.63 \end{align}$$
So the confidence interval is
$$ \begin{align} \mu ± \text{MOE} &= 30 ± 1.63 \\[5pt] \mu &\in [30 - 1.63, \, 30 + 1.63] \\[5pt] \mu &\in [28.37, \, 31.63] \end{align}$$
Thus a value of 40 parasites per mL would be outside of the 95% confidence limits, and something we'd want to look into further.
Consider a pain relief treatment for which we can measure pain (say by interviewing a patient) on a ten-point scale: 1 = no pain, 10 = extreme pain. Let's say that the following pain levels were recorded in one experiment.
8.4 | 9.2 | 8.0 | 7.7 |
8.4 | 9.2 | 8.0 | 7.7 |
8.4 | 9.2 | 8.0 | 7.7 |
8.4 | 9.2 | 8.0 | 7.7 |
invNorm(0.025, 0, 1) = 1.96
Then the margin of error is
$$ \begin{align} \text{MOE} &= Z_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}} \\[5pt] &= 1.96 \frac{8.3}{\sqrt{100}} \\[5pt] &= 1.63 \end{align}$$
So the confidence interval is
$$ \begin{align} \mu ± \text{MOE} &= 30 ± 1.63 \\[5pt] \mu &\in [30 - 1.63, \, 30 + 1.63] \\[5pt] \mu &\in [28.37, \, 31.63] \end{align}$$
Thus a value of 40 parasites per mL would be outside of the 95% confidence limits, and something we'd want to look into further.
Adult mayflies live between 30 minutes and one day (0.5h to 24 h), depending on the species. Data for one species was collected by tracking the movement of ten mayflies. The measured lifespans (in hours) were
17.68 | 13.69 | 11.22 | 11.05 | 13.86 |
14.47 | 14.50 | 13.47 | 10.04 | 13.10 |
Calculate a 95% confidence interval for the mean lifetime of thiis species of mayfly.
First let's enter the data into a calculator list and use that to calculate the average (our point estimate) and its standard deviation.
Then run 1-variable statistics on the data to get the mean and standard deviation, $\bar x ± \sigma = 13.31 ± 2.17$.
Now we can use those and the sample size (n = 10) to calculate the margin of error and confidence interval:
The critical t value is
invNorm(0.025, 9) = -2.2622
$$ \begin{align} \text{MOE} &= t_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}} \\[5pt] &= 2.2622 \, \frac{2.17}{\sqrt{10}} \\[5pt] &= 1.55 \end{align}$$
So the 95% confidence interval is [13.31 - 1.55, 13.31 + 1.55] = [11.76, 14.86].
For a sample of 10 dairy cows that had recently given birth, the mean protein content in 40 oz. (1.18 L) of milk was 3.42 g with a sample standard deviation of σ = 0.41 g. Calculate a 95% confidence interval for the mean protein content of the milk.
As is typical, we don't know the standard deviation of the parent distribution of cows, so we'll use the sample standard deviation and the t-distribution (with 9 degrees of freedom) rather than the normal distribution.
The critical t-value for a confidence level of 95% (α = 0.05) and nine degrees of freedom can be obtained from a t-table or by using a TI-84 calculator:
invT(0.025, 9) = -2.2622
The margin of error is
$$ \begin{align} \text{MOE} &= t_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}} \\[5pt] &= 2.2622 \, \frac{0.41}{\sqrt{10}} \\[5pt] &= 0.293 \end{align}$$
So the 95% confidence interval is [3.42 - 0.293, 3.42 + 0.293] = [3.13, 3.71].
A random survey of 1500 teen-agers (ages 13-18) showed that the average number of selfies (defined as any picture in which the phone owner both appeared in and took the photo) taken each year was 422 with a standard deviation of 183. Use this information to form a conclusion about all teens with a 90% confidence interval.
We don't know the standard deviation of the parent distribution of teens (why would we?), so we'll use the sample standard deviation, σ = 183, and the t-distribution (with 1499 degrees of freedom) rather than the normal distribution.
The critical t-value for a confidence level of 90% (α = 0.10) and 1499 degrees of freedom can be obtained from a t-table or by using a TI-84 calculator:
invT(0.05, 1499) = -1.6459
The margin of error is
$$ \begin{align} \text{MOE} &= t_{\frac{\alpha}{2}} \frac{\sigma}{\sqrt{n}} \\[5pt] &= 1.6459 \, \frac{183}{\sqrt{1500}} \\[5pt] &= 7.8 \end{align}$$
So the 95% confidence interval is [183 - 8, 183 + 8] = [175, 191]. We can conclude with 95% confidence that the average teen takes between 175 and 191 selfies per year.
A committe is set up to study whether there might be significant waste of prospective jurors' time in the judicial system. They take a random survey from jury records, asking 83 jurors how long they waited in the courthouse building before being called for service. The sample average was 7 hours with a standard deviation of 4 hours. How many people would the survey need to include to have a margin of error of 1 hour at the 95% confidence level?
These problems can be a little tricky because we have to make some assumptions in order to err on the side of caution. If we rearrange the margin of error expression to solve for sample number, we get
$$\text{MOE} = t^* \, \frac{s_x}{\sqrt{n}} \, \longrightarrow \, n = \bigg( \frac{t^* s_x}{\text{MOE}} \bigg)^2$$
Our MOE will be 1 hour, and we'll choose $t^*$ using 82 degrees of freedom. That's a conservative choice because the sample number we calculate will be larger.
invT(0.025, 82) = -1.9893
We'll also use the sample standard deviation (4 hours) obtained from the original study, again so that our calculated sample number is larger than what might actually be necessary. Then we have
$$ \begin{align} n &= \bigg( \frac{t^* s_x}{\text{MOE}} \bigg)^2 \\[5pt] &= \bigg( \frac{1.9893(4)}{1} \bigg)^2 \\[5pt] &= 63.32 \end{align}$$
We'll round that up to a sample size of 64 individuals.
A university bookstore wants to estimate the average amount that a student spends on books each year. They'd like to determine the total to within \$100 with 90% confidence. They can roughly estimate that the annual amount ranges between \$100 and \$1,900. How many students should be included in a random sample in order to make this estimate?
First we rearrange the margin of error expression for a normal distribution to solve for sample number, we get
$$\text{MOE} = Z_{\frac{\alpha}{2}} \, \frac{\sigma}{\sqrt{n}} \, \longrightarrow \, n = \bigg( \frac{Z_{\frac{\alpha}{2}} \sigma}{\text{MOE}} \bigg)^2$$
Our MOE will be 100, and we'll choose $Z_{\frac{\alpha}{2}}$ for a 90% confidence interval:
invNorm(0.05, 0, 1) = -1.6449
Here we can estimate the standard deviation by dividing the range of the data by 4:
$$\sigma \approx \frac{2000 - 100}{4} = 475$$
This rough rule is known as the range rule and is only a rough estimate of σ. Now we can estimate $n$:
$$ \begin{align} n &= \bigg( \frac{Z_{\frac{\alpha}{2}} \sigma}{\text{MOE}} \bigg)^2 \\[5pt] &= \bigg( \frac{1.6449(475)}{100} \bigg)^2 \\[5pt] &= 61.05 \end{align}$$
We'll round that down to a sample size of 61 individuals.
There are two ways we can interpret a confidence interval in the context of the data and measurements. Let's assume here a 90% confidence interval. The first is
We have 90% confidence that any measurement of the same sample size will fall within the confidence interval.
That is, if we repeated the sample measurement a large number of times, then we expect that, on average, 90 of 100 measurements would lie within the confidence interval.
The second way is a little more subtle.
The confidence intervals around 90% of point estimates will include the "true" mean.
We can show that graphically like this.
Each dot represents a point estimate of the mean, and the bar running through it from left to right is the confidence interval. In this case one of ten (the red one) has a confidence interval that does not capture the mean. On average, we'd expect 90% of estimates to capture the true mean.
Here is the basic work flow for calculating a confidence interval.
A corollary is a proposition that follows from and is often appended to one that has already been proved.
Example: We can prove that the sum of the measures of an interior angle of any triangle is 180˚. It follows (is a corollary) that the measure of each angle of an equilateral triangle is 60˚.
xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012-2019, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.