Why Use a Z Interval for a Mean?
A sample mean is an estimate, not an exact measurement, of a population mean. An interval estimate expresses the uncertainty in that estimate by giving a range of plausible values. When the observations follow a normal distribution and the population standard deviation is known, the sampling distribution of the sample mean is also normal. This makes it possible to construct a two-sided confidence interval with an exact coverage probability under the model.
The word known is important. The procedure below uses the population standard deviation, denoted by \(\sigma\), as a specified model quantity. It is not a procedure for replacing \(\sigma\) with a sample standard deviation. The interval is aligned with the SOA VEE Mathematical Statistics syllabus.
The standard deviation of \(\bar X\) is its standard error. It describes how much sample means vary from sample to sample under the model. Increasing \(n\) reduces this standard error; it does not change the known population standard deviation itself.
Constructing the Two-Sided Interval
For a chosen confidence level \(1-\alpha\), the two-sided equal-tailed interval leaves probability \(\alpha/2\) in each tail of the standard normal distribution. Let \(z_{\alpha/2}\) denote the positive standard normal critical value satisfying \(P(Z>z_{\alpha/2})=\alpha/2\) for \(Z\sim N(0,1)\). Thus the central probability between \(-z_{\alpha/2}\) and \(z_{\alpha/2}\) is \(1-\alpha\).
The interval is centered at \(\bar x\). Its margin of error is \(z_{\alpha/2}\sigma/\sqrt n\), so its total width is \(2z_{\alpha/2}\sigma/\sqrt n\). The confidence level determines the critical value: a higher confidence level requires a larger critical value and produces a wider interval, all else equal.
| Confidence level | \(\alpha\) | Critical value \(z_{\alpha/2}\), approximately |
|---|---|---|
| 90% | 0.10 | 1.645 |
| 95% | 0.05 | 1.960 |
| 99% | 0.01 | 2.576 |
Why the Formula Has the Claimed Coverage
The justification comes from standardizing the sample mean. Under the model, \(\bar X\sim N(\mu,\sigma^2/n)\), so subtracting \(\mu\) and dividing by its standard error gives a standard normal random variable.
By the normal sampling distribution, \(Z=(\bar X-\mu)/(\sigma/\sqrt n)\) has a standard normal distribution.
By the definition of \(z_{\alpha/2}\), \(P(-z_{\alpha/2}<Z<z_{\alpha/2})=1-\alpha\).
Multiplying by the positive number \(\sigma/\sqrt n\) and rearranging gives \(\bar X-z_{\alpha/2}\sigma/\sqrt n<\mu<\bar X+z_{\alpha/2}\sigma/\sqrt n\).
Therefore, before data are observed, the random interval contains the fixed population mean with probability \(1-\alpha\). After the data are observed, the endpoints are fixed. The correct interpretation is that the procedure captures \(\mu\) in \(100(1-\alpha)\%\) of repeated samples when its assumptions hold—not that the fixed \(\mu\) has a \(1-\alpha\) probability of lying in this particular realized interval.
Worked Applications
Worked Example: A 95% Interval From Eight Sensor Readings
A hypothetical calibration test produces eight independent readings, in units, from a normal measurement process. The population standard deviation is known to be 12 units. The readings are 44, 46, 47, 48, 49, 51, 54, and 61. Construct a 95% confidence interval for the population mean.
Method and assumptions. Use the known-\(\sigma\) Z interval. The problem stipulates independent normal readings and gives the population standard deviation, so the interval’s exact-coverage assumptions are met.
Calculate the sample mean. The sum is \(44+46+47+48+49+51+54+61=400\). With \(n=8\), the observed mean is \(\bar x=400/8=50\) units.
Find the standard error. Here \(\sigma=12\), so \(\sqrt n=\sqrt8\approx2.8284\), and \(\sigma/\sqrt n=12/\sqrt8\approx4.2426\) units.
Find the margin of error. For 95% confidence, use \(z_{\alpha/2}=1.96\). Thus the margin is \(1.96(4.2426)\approx8.3155\) units.
Form and check the limits. The lower limit is \(50-8.3155=41.6845\), and the upper limit is \(50+8.3155=58.3155\). Rounded to two decimal places, the interval is (41.68, 58.32) units. It is centered at 50 and has half-width 8.32 units, as required.
Common mistake avoided: using \(12\) as the standard error. The value 12 is the standard deviation of one observation; the standard error of the mean is \(12/\sqrt8\).
Worked Example: A 90% Interval for a Production Process
In a hypothetical process study, four independent measurements from a normal population are 114, 118, 122, and 126 units. The population standard deviation is known to be 8 units. Find a 90% confidence interval for the population mean.
Method and assumptions. Apply the two-sided known-\(\sigma\) Z interval. The observations are independent and normal, and \(\sigma=8\) is specified as known.
Compute the center. The sum is \(114+118+122+126=480\), so \(\bar x=480/4=120\) units.
Compute the standard error. Since \(\sqrt4=2\), the standard error is \(8/2=4\) units.
Choose the critical value and margin. A 90% interval has \(\alpha=0.10\), leaving \(0.05\) in each tail. Use \(z_{\alpha/2}=1.645\). The margin is \(1.645(4)=6.58\) units.
Calculate and check the interval. The limits are \(120-6.58=113.42\) and \(120+6.58=126.58\). The interval is (113.42, 126.58) units. Its midpoint is 120, and its width is \(13.16=2(6.58)\) units.
Common mistake avoided: using the 95% critical value merely because it is familiar. The requested 90% confidence level calls for 1.645, not 1.96; using 1.96 would produce a wider interval with a different coverage level.
Worked Example: A 99% Interval for a Small Normal Sample
A hypothetical laboratory records nine independent measurements: 7, 8, 8, 9, 10, 10, 11, 12, and 15. Assume the measurements come from a normal population with known standard deviation 1.2 units. Construct a 99% confidence interval for the population mean.
Method and assumptions. Use the exact known-standard-deviation Z procedure. The sample is independent, its population is normal, and the value \(\sigma=1.2\) is known. The small sample size does not invalidate this procedure because normality is assumed directly.
Find the mean. The sum is \(7+8+8+9+10+10+11+12+15=90\). Therefore, with \(n=9\), \(\bar x=90/9=10\) units.
Find the standard error. Since \(\sqrt9=3\), the standard error is \(1.2/3=0.4\) units.
Find the margin. For 99% confidence, \(z_{\alpha/2}=2.576\). The margin of error is \(2.576(0.4)=1.0304\) units.
Calculate and check the limits. The lower endpoint is \(10-1.0304=8.9696\), and the upper endpoint is \(10+1.0304=11.0304\). Rounded to two decimal places, the interval is (8.97, 11.03) units. It is symmetric around the sample mean and uses a larger critical value than the 95% interval would.
Common mistake avoided: assuming a small sample by itself rules out a Z interval. Here the model explicitly assumes normal observations and supplies the known population standard deviation; both conditions support this exact procedure.
Interpretation, Limits, and Exam Traps
A confidence interval combines a center and an uncertainty allowance. For this procedure, the center is always the observed sample mean, while the uncertainty allowance depends on the critical value, known population standard deviation, and sample size. Keep the units consistent: \(\bar x\) and \(\sigma\) must use the same measurement units, and the standard error and interval endpoints then use those units as well.
The exact normal-theory coverage relies on the stated model. If the population is not normal, the standardized sample mean need not have an exact standard normal distribution, especially for a small sample. A large-sample approximation is a different justification and should not be silently substituted for the exact assumptions here. Likewise, if the population standard deviation is not known, inserting a sample standard deviation into this formula does not make it a known-\(\sigma\) Z interval.
- Do not confuse \(\sigma\) with the standard error. Divide \(\sigma\) by \(\sqrt n\) before multiplying by the critical value.
- Match the critical value to the confidence level. A two-sided interval allocates \(\alpha/2\) to each tail.
- Do not give a probability statement about the fixed parameter after observing the data. Confidence describes the long-run performance of the interval procedure.
- Check the arithmetic structure. The interval must be centered at \(\bar x\), and its total width must be twice the margin of error.
Check Your Understanding
For each problem, identify the confidence level and known standard deviation, calculate the sample mean and standard error, then form and check the interval. Unless stated otherwise, assume the observations are independent and normal. Round interval endpoints to two decimal places.
- A hypothetical set of four independent readings is 18, 20, 22, and 24 units. The known population standard deviation is 5 units. Find a 95% interval for the population mean. (Answer: (16.10, 25.90) units)
- Nine independent normal observations have values 30, 31, 32, 33, 34, 35, 36, 37, and 38 units. The known population standard deviation is 6 units. Find a 90% interval for the mean. (Answer: (30.71, 37.29) units)
- Sixteen independent normal measurements have a total of 4,000 units. The known population standard deviation is 12 units. Find a 95% interval for the population mean. (Answer: (244.12, 255.88) units)
- Twenty-five independent measurements are from a normal population with known standard deviation 5 units, and their observed mean is 40 units. Find a 95% interval. In checking the method, explain why using 5 directly as the standard error is incorrect. (Answer: (38.04, 41.96) units; the standard error is \(5/\sqrt{25}=1\) unit, not 5 units.)
- A study has six independent observations from a population that is stated only to have mean \(\mu\) and variance \(16\); no normality assumption is given. Is the exact normal-theory Z interval justified by these facts alone? State which key assumption is missing. (Answer: No; the population’s normality is not specified, so the exact standard normal sampling distribution for the mean is not established.)