10.1: Estimating the Population Mean
- Page ID
- 105854
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Section 0: Introduction to Confidence Intervals
Let’s recall the definitions of the parameter and statistic.
A parameter is a numerical summary of a population, and a statistic is a numerical summary of a sample. The following table summarizes the differences between the two.
|
Statistic |
Parameter |
|---|---|
|
sample |
population |
|
size |
size |
|
varies |
constant |
|
can be found |
unknown |
A parameter is something that is computed from a population, while unknown it remains constant. A statistic is something that is computed from a sample of size \(n\), while it may vary from sample to sample it can be easily found when needed.
The most common application of a statistic is to estimate the unknown parameter.
We define a point estimate of a parameter as a single value used to estimate the parameter.
The sample mean, sample proportion, and sample variance can be used as point estimates of the corresponding population parameters.
|
Statistic |
Relation |
Parameter |
|---|---|---|
|
\(\bar{x}\) |
is a point estimate of |
\(\mu\) |
|
\(\hat{p}\) |
is a point estimate of |
\(p\) |
|
\(s^2\) |
is a point estimate of |
\(\sigma^2\) |
For example, the average age of the students in the class can be used as a point estimate of the average age of all students in the college. But keep in mind that a point estimate does not have to be a statistic - without any survey, one can guesstimate the average age of all college students to be 25 years old. Is there a difference between the two estimates? Intuitively we know that the sample mean is a better estimate, but why?
We define an unbiased estimate as a point estimate whose long-term average equals the parameter, otherwise the point estimate is called a biased estimate of the parameter.
All the sample statistics in the table above are unbiased point estimates of their corresponding parameters.
Now that we know that we can estimate an unknown parameter with the appropriate sample statistic we have to address the following questions:
- How confident can one be when using a sample statistic as a point estimate? In other words, what are the chances that a random observation of a statistic will be equal to the parameter exactly?
The answer is zero because the sample statistic is a continuous random variable and the probability of any continuous random variable being equal to a single value is always zero. In other words, we have zero confidence that a point estimate is equal to the unknown population parameter and vice versa we have zero confidence that the unknown population parameter is equal to one random observation of the sample statistic.
- If there are zero chances that any point estimate equals exactly the parameter, why do we use the point estimates?
Consider the following example - let’s say we are running late to a meeting, and we receive a call with the question: “How long will it take us to get there?” One of the appropriate and probably the most common answers is “We will be there in, let’s say, 10 minutes” This is an example of a point estimate of our arrival time. According to our discussion, we are 0% confident that we will get there in exactly 600 seconds or 10 minutes. Notice that while there are zero chances that it will take us exactly 10 minutes to arrive, the point estimate did answer the question. So often, we use a point estimate because a problem demands some number, and it is necessary to assume some value to continue with the solution. Soon, we will have situations in which we need a population standard deviation or proportion but instead we will be using a sample standard deviation or a sample proportion.
- Is it possible to increase the confidence level?
Consider the previous example. We are running late to a meeting, and we receive a call with the question: “How long will it take us to get there?” Is there another way to answer this question? How about instead of saying that we will be there in ten minutes which is obviously wrong we say something like “We will be there between 5 and 15 minutes.”? This is an example of an interval estimate of our arrival time. Notice that the chances of our estimate being correct are now significantly higher than zero. In other words, with positive confidence, we estimate our arrival time to be between 5 and 15 minutes. In general, we can increase the confidence level by using intervals instead of a single number!
- Is it possible to increase the confidence level to 100%?
Continuing with the previous example. We are still running late to a meeting, and we receive a call with the question: “How long will it take us to get here?” Is there a way to answer this question with a 100% confidence level? Turns out that it is not as hard as it seems. How about “We will be there sometime between now and tomorrow.”? This is an example of a 100% confidence interval estimate of our arrival time. Why is there 100% confidence? Because the destination is only 5 miles away, thus we can be 100% sure that we will arrive there sometime before the end of the day even if we must walk the rest of the trip. Is there a value in a 100% confidence interval? Not much. In practice, the most practical estimate will be with the confidence level of 95% and we can increase it to 99% or decrease it to 90% depending on the application.
The goal is to create a formal approach for constructing interval estimates with some level of confidence.
Section 1: The Generalized Empirical Rule
Next, we will develop a procedure to create an interval estimate of the unknown parameter with positive confidence.
Let’s assume that Bobert's commute time is normally distributed with the mean 25 minutes and standard deviation 2 minutes. In other words, if we let \(X\) be a randomly selected commute time of Bobert, then
\(X\sim N(\mu=25, \sigma=2)\)
Now, recall that for a normal random variable the Empirical Rule applies and according to the Empirical rule the following probability statements are true!
- The probability that a random observation is within 1 standard deviation from the mean is 68%.
- The probability that a random observation is within 2 standard deviations from the mean is 95%.
- The probability that a random observation is within 3 standard deviations from the mean is 99.7%.
In our cases, where \(X\) is a random Bobert’s commute time, we can apply and interpret the Empirical rule in the following way:
- The probability that a random Bobert’s commute is between 23 and 27 minutes is 68%.
- The probability that a random Bobert’s commute is between 21 and 29 minutes is 95%.
- The probability that a random Bobert’s commute is between 19 and 31 minutes is 99.7%.
In general, for any continuous random variable \(X\), \((1-\alpha)\cdot100\%\) of the area under the probability density curve is between \(x_{1-\alpha/2}\) and \(x_{\alpha/2}\), thus for \(X\sim N(\mu, \sigma)\), it can be visualized as:
Alternatively, this fact can be expressed as the following probability statement:
\(P\left(x_{1-\alpha/2}<X<x_{\alpha/2}\right)=1-\alpha\)
where
\(x_{\alpha/2}=\mu+z_{\alpha/2}\sigma\)
and
\(x_{1-\alpha/2}=\mu+z_{1-\alpha/2}\sigma=\mu-z_{\alpha/2}\sigma\)
thus
\(P\left(\mu-z_{\alpha/2}\sigma<X<\mu+z_{\alpha/2}\sigma\right)=1-\alpha\)
We will refer to the last statement above as the generalized empirical rule.
Let’s get used to the generalized empirical rule and practice applying it for some common values of alpha.
When \(\alpha=0.32\):
\(1-\alpha=0.68\)
\(\alpha/2=\dfrac{0.32}{2}=0.16\)
\(z_{\alpha/2}=z_{0.16}=0.99\approx1\)
Hence,
\(P\left(\mu-1\sigma<X<\mu+1\sigma\right)=0.68\)
which is equivalent to the first part of the Empirical Rule. We may also interpret this probability statement as if we are 68% confident that a random observation is within \(1\sigma\) from \(\mu\).
When \(\alpha=0.10\):
\(1-\alpha=0.90\)
\(\alpha/2=\dfrac{0.10}{2}=0.05\)
\(z_{\alpha/2}=z_{0.05}=1.65\)
Hence,
\(P\left(\mu-1.65\sigma<X<\mu+1.65\sigma\right)=0.90\)
which goes beyond what the Empirical Rule states. We may also interpret this probability statement as if we are 90% confident that a random observation is within \(1.65\sigma\) from \(\mu\).
Note that the generalized empirical rule allows us to make probability statements and interpret them in a certain way for any choice of \(\alpha\).
|
Probability statement |
Interpretation |
|---|---|
|
\(P\left(\mu-1\sigma<X<\mu+1\sigma\right)=0.68\) |
We are \(68\%\) confident that a random observation \(x_i\) is within \(1\sigma\) from \(\mu\). |
|
\(P\left(\mu-1.65\sigma<X<\mu+1.65\sigma\right)=0.90\) |
We are \(90\%\) confident that a random observation \(x_i\) is within \(1.65\sigma\) from \(\mu\). |
| \(P\left(\mu-z_{\alpha/2}\sigma<X<\mu+z_{\alpha/2}\sigma\right)=1-\alpha\) |
We are \((1-\alpha)\cdot100\%\) confident that a random observation \(x_i\) is within \(z_{\alpha/2}\sigma\) from \(\mu\). |
The most common values of \(\alpha\) are 0.01, 0.05, and 0.10 to produce 99%, 95%, and 90% confidence levels, respectively. This means that to make a statement with a certain level of confidence one just simply has to pick the right alpha. For example, what value of \(\alpha\) will result in 80% confidence? Answer: \(\alpha=0.2\)
Each probability statement about \(X\) made by applying the generalized empirical rule
\(P\left(\mu-z_{\alpha/2}\sigma<X<\mu+z_{\alpha/2}\sigma\right)=1-\alpha\)
can be interpreted in two ways as shown below:
We are \((1-\alpha)\cdot100\%\) confident that
(1) \(x_i\) is within \(z_{\alpha/2}\sigma\) from \(\mu\)
OR
(2) \(\mu\) is within \(z_{\alpha/2}\sigma\) from \(x_i\).
In the first interpretation, we need to know \(\mu\) to estimate a random observation \(x_i\), and in the second interpretation, we need to know the random observation \(x_i\) to estimate the unknown \(\mu\). We certainly see the potential in the second interpretation!
Section 2: Constructing a Confidence Interval Using a Single Observation
Let's assume now that the average of Bobert's commute is unknown, that is, 𝑋 is normally distributed with the unknown mean (\(\mu\)) and given standard deviation (\(\sigma\)) of 2 minutes. Instead of trying to predict today’s commute we can simply measure it. So, let’s assume that today’s commute (\(x_i\)) was 24 minutes. Can we estimate the unknown mean commute time (\(\mu\)) with 95% confidence?
By the Generalized Empirical Rule for \(\alpha=0.05\):
\(P\left(\mu-2\sigma<X<\mu+2\sigma\right)=95\%\)
Hence, we are 95% confident that:
\(x_i\) is within \(2\sigma\) from \(\mu\)
which is the same as
\(\mu\) is within \(2\sigma\) from \(x_i\)
which is the same as
\(\mu\) is within 4 from 24
which is the same as
\(\mu\) is between 20 and 28
As a result, we obtained a 95% confidence interval for the unknown mean commute time based on a single random observation.
Next, we are going to construct an interval estimate with a 90% confidence.
By the Generalized Empirical Rule for \(\alpha=0.1\):
\(P\left(\mu-1.65\sigma<X<\mu+1.65\sigma\right)=95\%\)
Hence, we are 90% confident that:
\(x_i\) is within \(1.65\sigma\) from \(\mu\)
which is the same as
\(\mu\) is within \(1.65\sigma\) from \(x_i\)
which is the same as
\(\mu\) is within 3.30 from 24
which is the same as
\(\mu\) is between 20.7 and 27.3
As a result, we obtained a 90% confidence interval for the unknown mean commute time based on a single random observation.
Next, we are going to introduce some vocabulary.
Interval estimate is a range of numbers that we think contains the unknown parameter.
Confidence level is the confidence that we have that the interval estimate actually contains the parameter.
Confidence interval is a way to estimate the parameter using an interval estimate along with some confidence level.
Also, let’s recall the properties of an interval.
An interval has the center, the radius, and the left and right boundaries. When constructing a confidence interval, it is common to use a point estimate as the center, the radius is usually called the margin of error and the boundaries are referred to as lower and upper bounds.
To construct an interval means to find its boundaries. In our example,
- the point estimate is a single random observation \(x_i\),
- the margin of error can be found using the provided formula
\(ME=z_{\alpha/2}\sigma\)
- the boundaries can be found by adding and subtracting the ME from and to the point estimate
\(x_i\pm ME\)
Note, that in this context, the values \(-z_{\alpha/2}\) and \(z_{\alpha/2}\) are referred to as the left and right critical values.
The procedure can be summarized in the form of the table:
|
Unknown Parameter |
\(\mu\), population mean |
|---|---|
|
Point Estimate |
\(x_i\), a random observation |
|
Confidence Level (of the point estimate) |
0% |
|
Confidence Level (for a new interval estimate) |
\((1-\alpha)100\%\) \(\alpha\) \(\alpha/2\) |
| Critical Value(s) | \(z_{\alpha/2}\) |
|
Margin of Error |
\(ME=z_{\alpha/2}\sigma\) |
|
Lower Bound |
\(LB=x_i-ME\) |
|
Upper Bound |
\(UB=x_i+ME\) |
| Interpretation | We are \((1-\alpha)100\%\) confident that the unknown population mean \(\mu\) is between LB and UB. |
In our example, we fill out the table in the following way.
|
Unknown Parameter |
\(\mu\), the average commute time |
|---|---|
|
Point Estimate |
\(x_i=24\), today's commute time |
|
Confidence Level (of the point estimate) |
0% |
|
Confidence Level (for a new interval estimate) |
\((1-\alpha)100\%=90\%\) \(\alpha=0.1\) \(\alpha/2=0.05\) |
| Critical Value(s) | \(z_{\alpha/2}=z_{0.05}=1.65\) |
|
Margin of Error |
\(ME=1.65\cdot2=3.30\) |
|
Lower Bound |
\(LB=24-3.30=20.70\) |
|
Upper Bound |
\(UB=24+3.30=27.30\) |
| Interpretation | We are 90% confident that the average commute time is between 20.70 and 27.30 minutes. |
We are 90% confident that the average commute time is between 20.70 and 27.30 minutes.
Section 3: Constructing a Confidence Interval Using a Sample
Is there a way to improve the accuracy of our existing procedure? Naturally, the smaller the ME the more accurate the interval estimate. The ME formula has only two components: \(\sigma\) and \(\alpha\):
\(ME=z_{\alpha/2}\cdot\sigma\)
\(\sigma\) cannot be chosen as it is the population parameter and therefore must remain constant! If we look at \(\alpha\), we can see that a higher accuracy can be achieved by sacrificing the confidence. In our example above, we obtained the following confidence intervals:
[20,28] is a 95% confidence interval
[20.70,27.30] is a 90% confidence interval
[24,24] is a 0% confidence interval
Easy to see that a 90% confidence interval is narrower than the 95% confidence. This observation is consistent with our understanding of the estimating process. The narrowest confidence interval is the point estimate which has a 0% confidence since there is a 0% chance that the point estimate is correct! So, to answer the question, we can improve the accuracy by lowering the confidence. But is there a way to improve the accuracy without sacrificing the confidence? The short answer is YES!
Let’s take a look at the same problem once again. Based on the previous procedure, we are 90% confident that 
is between 20.7 and 27.3. Is there anything in the problem that we could have done better? Does a sample of size one seem large enough? Intuitively we know that a sample of size greater than 1 should have produces a better result but how exactly?
This time let's assume that Bobert’s commute is normally distributed with the unknown mean and given standard deviation 2 minutes. Instead of measuring it once we surveyed a sample of 10 commutes, and the results are listed below:
25, 21, 23, 22, 22, 26, 24, 27, 24, 26
Can we improve the estimate of the average commute with 90% confidence level?
Note that the average of the sample is
\(\bar{x}_i=\dfrac{25+21+23+22+22+26+24+27+24+26}{10}=24\)
and \(\bar{x}_i\) is the best choice for a point estimate of the unknown population mean.
Previously, we have applied the Generalized Empirical Rule on the normal population, in our example, the population of commute times. However, by the Central Limit Theorem the distribution of sample means is also normally distributed with known parameters \(\mu_\bar{X}\) and \(\sigma_\bar{X}\):
\(\bar{X}\sim N\left(\mu_\bar{X}=\mu, \sigma_\bar{X}=\dfrac{\sigma}{\sqrt{n}}\right)\)
Thus, the Generalized Empirical Rule can also be applied for a normally distributed sample mean, \(\bar{X}\). This results in the following probability statement:
\(P\left(\mu_\bar{X}-z_{\alpha/2}\sigma_\bar{X}<X<\mu_\bar{X}+z_{\alpha/2}\sigma_\bar{X}\right)=1-\alpha\)
Replacing \(\mu_\bar{X}=\mu\) and \(\sigma_\bar{X}=\dfrac{\sigma}{\sqrt{n}}\):
\(P\left(\mu-z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}<X<\mu+z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\right)=1-\alpha\)
Visually, the only difference is in the presence of the square root of the sample size in the denominators. However, this is exactly what will cause the ME to get smaller as the sample size gets larger!
Just like before, this probability statement can be interpreted in two ways as shown below:
We are \((1-\alpha)\cdot100\%\) confident that
(1) \(\bar{x}_i\) is within \(z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\) from \(\mu\)
OR
(2) \(\mu\) is within \(z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\) from \(\bar{x}_i\).
In the first interpretation, we need to know \(\mu\) to estimate the average of a random sample, \(\bar{x}_i\), and in the second interpretation, we need to know the average of a random sample, \(\bar{x}_i\), to estimate the unknown population mean, \(\mu\). We again will use the second interpretation to develop a formal procedure for estimating the unknown mean using the mean of a random sample.
The table below summarizes the differences between using a single random observation (on the left) and the mean of a random sample of size n (on the right).
| Single observation | Sample of size \(n\) | |
|---|---|---|
|
Unknown Parameter |
\(\mu\), population mean |
|
|
Point Estimate |
\(x_i\), a random observation |
\(\bar{x}_i\), the mean of a random sample of size \(n\) |
|
Confidence Level (of the point estimate) |
0% |
|
|
Confidence Level (for a new interval estimate) |
\((1-\alpha)100\%\) \(\alpha\) \(\alpha/2\) |
|
| Critical Value(s) | \(z_{\alpha/2}\) | |
|
Margin of Error |
\(ME=z_{\alpha/2}\sigma\) |
\(ME=z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\) |
|
Lower Bound |
\(LB=x_i-ME\) |
\(LB=\bar{x}_i-ME\) |
|
Upper Bound |
\(UB=x_i+ME\) |
\(UB=\bar{x}_i+ME\) |
| Interpretation | We are \((1-\alpha)100\%\) confident that the unknown population mean \(\mu\) is between LB and UB. | |
Again, it is easy to see that the main difference is in the formula for the margin of error. With the square root of the sample size in the denominator, we found a way to decrease the margin of error without sacrificing the confidence!
In our example, we are going to use this template to construct a 90% confidence interval for the unknown population mean using the average of a random sample of size 10.
|
Unknown Parameter |
\(\mu\), population mean |
|---|---|
|
Point Estimate |
\(\bar{x}_i=24\), the mean of the sample of size \(n=10\) |
|
Confidence Level (of the point estimate) |
0% |
|
Confidence Level (for a new interval estimate) |
\((1-\alpha)100\%=90\%\) \(\alpha=0.1\) \(\alpha/2=0.05\) |
| Critical Value(s) | \(z_{\alpha/2}=z_{0.05}=1.65\) |
|
Margin of Error |
\(ME=1.65\cdot\dfrac{2}{\sqrt{10}}=0.73\) |
|
Lower Bound |
\(LB=24-0.73=23.27\) |
|
Upper Bound |
\(UB=24+0.73=24.73\) |
| Interpretation |
We are 90% confident that the average commute time is between 23.27 and 24.73 minutes.
|
We are 90% confident that the average commute time is between 23.27 and 24.73 minutes.
Let’s list all the assumptions that were made in order for the designed procedure to work:
- The population standard deviation, \(\sigma\), is known.
- The sample is simple random.
- The CLT must be applicable, that is at least one of the following must be true:
- The sample size, \(n\), is greater than 30.
- The population is normally distributed.
We have developed a procedure for finding a confidence interval for the unknown population mean with known population standard deviation by using the average of a random sample.
Section 4: The Accuracy and Margin of Error
Let’s take another look at the following formula.
\(ME=z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\)
Since \(\sigma\) is a constant and cannot be changed, a higher accuracy or smaller ME can be achieved by increasing \(\alpha\) and sacrificing the confidence OR by increasing the sample size without any sacrifice.
Therefore, if we want to construct a \((-\alpha)\cdot100\%\) confidence interval with the margin of error no more than \(E_0\), how large should weuld choose the sample size? Since the margin of error can’t be greater than \(E_0\) we have the following inequality:
\(ME=z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}<E_0\)
By isolating the sample size, we obtain the following solution:
\(n>\left(z_{\alpha/2}\cdot\dfrac{\sigma}{E_0}\right)^2\)
This means that to find the required sample size we need to know \(\alpha\), \(\sigma\), and \(E_0\), the desired bound on the margin of error.
For example, let’s find the sample size required to construct a 90% confidence interval with the margin of error no more than 0.5, if the population standard deviation is 2?
To apply the formula, we must know \(E_0\), \(\sigma\), and \(z_{\alpha/2}\). From the problem, we know that \(E_0=0.5\) and \(\sigma=2\). For a confidence level of 90%, \(\alpha=0.1\), therefore (z_{\alpha/2}=z_{0.05}=1.65\).
Finally, we get:
\(n>\left(z_{\alpha/2}\cdot\dfrac{\sigma}{E_0}\right)^2=\left(1.65\cdot\dfrac{2}{0.5}\right)^2=43.56\uparrow44\)
Remember in such problems to always round the sample size UP!
We just learned how to choose the sample size to achieve a desired accuracy when constructing a confidence interval for the unknown population mean with known population standard deviation using the average of a random sample.
Section 5: One Mean Z Procedure
Consider the following example:
An incubation period is a time between when you contract a virus and when your symptoms start. Assume that the population of incubation periods for a novel coronavirus is normally distributed with the population standard deviation of 2 days. By surveying randomly selected local hospitals, a researcher was able to obtain the following sample of incubation periods of 10 patients:
6, 2, 4, 3, 3, 7, 5, 8, 5, 7
- Use the sample to construct and interpret a 90% confidence interval for the average incubation period of the novel coronavirus.
First, let’s check if all necessary assumptions are satisfied:
- The sample is simple random.
- The population is normally distributed.
- The population standard deviation, \(\sigma\), is known and \(\sigma=2\).
Also, note that the average of the sample is
\(\bar{x}_i=\dfrac{6+2+4+3+3+7+5+8+5+7}{10}=5\)
Once the assumptions are verified, we may apply the procedure by following the template:
| Example | Template | |
|---|---|---|
|
Unknown Parameter |
\(\mu\), the average incubation period of the novel coronavirus |
\(\mu\), population mean |
|
Point Estimate |
\(\bar{x}_i=5\), the average incubation period in the sample of size \(n=10\) |
\(\bar{x}_i\), the mean of a random sample of size \(n\) |
|
Confidence Level (of the point estimate) |
0% |
0% |
|
Confidence Level (for a new interval estimate) |
\((1-\alpha)100\%=90\%\) \(\alpha=0.1\) \(\alpha/2=0.05\) |
\((1-\alpha)100\%\) \(\alpha\) \(\alpha/2\) |
| Critical Value(s) | \(z_{\alpha/2}=z_{0.05}=1.65\) | \(z_{\alpha/2}\) |
|
Margin of Error |
\(ME=1.65\cdot\dfrac{2}{\sqrt{10}}=1.04\) |
\(ME=z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}\) |
|
Lower Bound |
\(LB=5-1.04=3.96\) |
\(LB=\bar{x}_i-ME\) |
|
Upper Bound |
\(UB=5+1.04=6.04\) |
\(UB=\bar{x}_i+ME\) |
| Interpretation | We are 90% confident that the average incubation period of the novel coronavirus is between 3.96 and 6.04 days. | We are \((1-\alpha)100\%\) confident that the unknown population mean \(\mu\) is between LB and UB. |
Now we can interpret the results: We are 90% confident that the average incubation period of the novel coronavirus is between 3.96 and 6.04 days.
In addition, we may also answer the following question.
- How large must the sample be so that the margin of error is less than 0.25?
To answer this question, we will use the formula
\(n>\left(z_{\alpha/2}\cdot\dfrac{\sigma}{E_0}\right)^2\)
in which we can plug in all the values
\(n>\left(1.65\cdot\dfrac{2}{0.25}\right)^2=174.24\uparrow175\)
In conclusion, 175 is the smallest sample size for which the margin of error will be less than 0.25 for a 90% confidence level.