Two-Sided Z Intervals From Summary Statistics

From Summary Statistics to a Two-Sided Interval

A confidence interval summarizes how precisely a sample estimates a population mean. For a two-sided interval, the goal is to place a lower and an upper limit around the sample mean so that the procedure captures the true mean at a selected long-run rate. The calculation uses four inputs: the sample size, the sample mean, the known population standard deviation, and the confidence level.

This tutorial focuses on carrying out and interpreting that calculation from summary statistics. The distinction between the population standard deviation and the sample standard deviation matters: the Z interval described here requires the population standard deviation to be known.

Definition: Suppose $X_1,\ldots,X_n$ are an independent random sample from a normal population with unknown mean $\mu$ and known standard deviation $\sigma$. A two-sided $100(1-\alpha)\%$ Z confidence interval for $\mu$ is $$ \bar{X}\ \pm\ z_{1-\alpha/2}\frac{\sigma}{\sqrt{n}}, $$ or, equivalently, $$ \left(\bar{X}-z_{1-\alpha/2}\frac{\sigma}{\sqrt{n}},\ \bar{X}+z_{1-\alpha/2}\frac{\sigma}{\sqrt{n}}\right). $$ Here $\bar{X}$ is the sample mean, $0<\alpha<1$, and $z_{1-\alpha/2}$ is the standard normal quantile with cumulative probability $1-\alpha/2$.

The standard error of the sample mean is $\sigma/\sqrt{n}$. It measures the standard deviation of the sampling distribution of $\bar{X}$, not the spread of individual observations. The margin of error is $z_{1-\alpha/2}\sigma/\sqrt{n}$; the two endpoints are the sample mean minus and plus that margin.

If the population is normal, the sampling distribution of $\bar{X}$ is normal for any sample size under the independent random-sampling assumption. For a sufficiently large sample from a nonnormal population, a normal approximation may sometimes be justified by the central limit theorem, subject to appropriate conditions on the population. The interval is not justified by this formula merely because the sample size is known. In particular, substituting a sample standard deviation when $\sigma$ is unknown does not meet the stated known-$\sigma$ assumption.

Why the Critical Value Is Split Between Two Tails

A two-sided confidence level leaves a total probability $\alpha$ outside the central interval. The equal-tailed construction assigns $\alpha/2$ to each tail of the standard normal distribution. Thus the probability between $-z_{1-\alpha/2}$ and $z_{1-\alpha/2}$ is $1-\alpha$.

Under the model above, standardizing the sample mean gives $Z=(\bar{X}-\mu)/(\sigma/\sqrt{n})\sim N(0,1)$. Consequently,

\begin{aligned} P\left(-z_{1-\alpha/2}\leq \frac{\bar{X}-\mu}{\sigma/\sqrt{n}} \leq z_{1-\alpha/2}\right) &= (1-\alpha/2)-(\alpha/2)\\ &=1-\alpha. \end{aligned}

Rearranging the inequalities inside this probability statement gives $\bar{X}-z_{1-\alpha/2}\sigma/\sqrt{n}\leq\mu\leq \bar{X}+z_{1-\alpha/2}\sigma/\sqrt{n}$. This is the interval formula. The derivation shows why the critical value depends on the confidence level and why it uses $\alpha/2$, not $\alpha$.

A Reliable Calculation Process

For a calculation from summary statistics, keep the standard error and the critical value separate until the margin of error is found. A table or calculator may provide the critical normal quantile. Common values are $1.645$ for a 90% interval, $1.960$ for a 95% interval, and $2.576$ for a 99% interval. These are rounded critical values; use the precision provided by the question or calculation tool.

Set the tail probability.
For confidence level $1-\alpha$, calculate $\alpha=1-\text{confidence level}$, then use $\alpha/2$ in each tail.

Find the standard error.
Calculate $\sigma/\sqrt{n}$, checking that $\sigma$ is the population standard deviation and that $n$ is the sample size.

Find the margin of error.
Multiply the standard error by $z_{1-\alpha/2}$.

Form and interpret the endpoints.
Subtract and add the margin to $\bar{x}$, report appropriate units, and interpret the interval for the population mean.

Worked Example: A 95% Interval for a Mean Processing Time

A hypothetical laboratory takes an independent random sample of $64$ completed tests. The sample mean processing time is $18.4$ minutes. Assume processing times follow a normal population with known standard deviation $4.0$ minutes. Calculate and interpret a 95% confidence interval for the population mean processing time. Round endpoints to two decimal places.

Method and assumptions. Use the two-sided Z interval because the sample is random and independent, the population is assumed normal, and the population standard deviation is known. Here $n=64$, $\bar{x}=18.4$, $\sigma=4.0$, and $\alpha=1-0.95=0.05$. Each tail has probability $\alpha/2=0.025$, so $z_{0.975}=1.960$.

Standard error. Since $\sqrt{64}=8$,

\frac{\sigma}{\sqrt{n}}=\frac{4.0}{8}=0.5\text{ minutes}.

Margin of error.

1.960(0.5)=0.980\text{ minutes}.

Endpoints. The lower endpoint is $18.4-0.980=17.420$ minutes, and the upper endpoint is $18.4+0.980=19.380$ minutes. Rounded to two decimal places, the interval is $(17.42,\ 19.38)$ minutes.

Interpretation and check. The procedure used to form this interval captures the true mean processing time in 95% of repeated samples under the stated model. The observed interval is centered at $18.4$, and its margin is $0.98$, so the endpoints are symmetric as required. The common mistake avoided here is treating the standard deviation $4.0$ as the standard error; dividing by $\sqrt{64}$ is essential.

Confidence Level, Width, and Interpretation

For a fixed sample size and known $\sigma$, a higher confidence level requires a larger critical value and therefore a wider interval. For instance, a 99% interval uses a larger critical value than a 95% interval. This is the cost of requiring the procedure to capture the mean more often over repeated samples. The sample mean remains the center; the confidence level changes the margin of error.

A confidence level is not a probability assigned to the fixed population mean after the interval has been calculated. In the frequentist interpretation, $\mu$ is fixed, while the sample and the interval endpoints vary from sample to sample. The confidence statement describes the long-run performance of the interval procedure. Once a particular interval has been observed, it either contains $\mu$ or it does not.

Worked Example: A 90% Interval for a Hypothetical Survey Measure

A hypothetical service organization samples $100$ independent account holders. Their average reported monthly usage is $72$ units. Assume the population is normal and its known standard deviation is $15$ units. Find and interpret a 90% confidence interval for the population mean usage. Round endpoints to two decimal places.

Method and assumptions. The known population standard deviation and normal-population model permit the Z interval. The confidence level is $0.90$, so $\alpha=0.10$ and $\alpha/2=0.05$. The standard normal quantile with cumulative probability $0.95$ is $z_{0.95}=1.6449$ (rounded).

Standard error. Since $\sqrt{100}=10$,

\frac{\sigma}{\sqrt{n}}=\frac{15}{10}=1.5\text{ units}.

Margin of error.

1.6449(1.5)=2.46735\text{ units}.

Endpoints. The lower endpoint is $72-2.46735=69.53265$, and the upper endpoint is $72+2.46735=74.46735$. Rounded to two decimal places, the interval is $(69.53,\ 74.47)$ units.

Interpretation and check. Under the stated assumptions, this method produces intervals containing the population mean in 90% of repeated samples. The margin is approximately $2.47$ units on either side of $72$, agreeing with both endpoints. The interval has the same units as the sample mean, as it should.

Using a More Stringent Confidence Level

The same calculation applies at any selected confidence level. A 99% interval leaves $1\%$ outside the central area, or $0.5\%$ in each tail. The corresponding critical value is $z_{0.995}$, approximately $2.5758$. Do not use the 95% critical value simply because it is familiar: the chosen confidence level determines the quantile.

Worked Example: A 99% Interval for a Hypothetical Component Dimension

A manufacturer studies a hypothetical production process by measuring $36$ independently selected components. The sample mean dimension is $250$ millimeters. Assume a normal population with known standard deviation $12$ millimeters. Calculate a 99% confidence interval for the population mean dimension and explain the effect of the confidence level. Round endpoints to two decimal places.

Method and assumptions. Use the two-sided Z interval under the normal-population, independent-sampling, and known-standard-deviation assumptions. Here $\alpha=1-0.99=0.01$, so each tail has probability $0.005$, and $z_{0.995}=2.5758$.

Standard error. Since $\sqrt{36}=6$,

\frac{\sigma}{\sqrt{n}}=\frac{12}{6}=2\text{ millimeters}.

Margin of error.

2.5758(2)=5.1516\text{ millimeters}.

Endpoints. The lower endpoint is $250-5.1516=244.8484$ millimeters, and the upper endpoint is $250+5.1516=255.1516$ millimeters. Rounded to two decimal places, the interval is $(244.85,\ 255.15)$ millimeters.

Interpretation and check. The 99% procedure is designed to capture the population mean in 99% of repeated samples under the model. Its margin, about $5.15$ millimeters, is larger than it would be for a 95% interval with the same data because the more stringent confidence level uses a larger critical value. The two endpoints are $250\pm5.1516$, confirming that the interval is centered at the sample mean.

Common Exam Trap: Using the Wrong Tail Probability

For a two-sided 95% interval, the total area outside the central region is $0.05$, and each tail contains $0.025$. Using $z_{0.95}=1.645$ instead of $z_{0.975}=1.960$ is a one-tail calculation mistakenly inserted into a two-sided interval.

For example, with $\bar{x}=18.4$, $\sigma=4$, and $n=64$, the standard error is $0.5$. The incorrect margin would be $1.645(0.5)=0.8225$, giving $(17.5775,\ 19.2225)$. But the probability between $-1.645$ and $1.645$ under a standard normal distribution is approximately $0.90$, not $0.95$. The correction is to split the $0.05$ total tail probability equally, use $z_{0.975}=1.960$, and obtain $(17.42,\ 19.38)$ after rounding to two decimals.

A second useful check is that increasing $n$ reduces the standard error, while increasing the confidence level increases the critical value. Also check the units: $\sigma$ and the margin of error have the same units as the measured quantity, and $n$ and the critical value are unitless.

Key takeaway: For a two-sided Z interval with known population standard deviation, calculate $\bar{x}\pm z_{1-\alpha/2}\sigma/\sqrt{n}$. Split the excluded probability between both tails, preserve the units, and interpret the confidence level as the long-run coverage of the method.

Check Your Understanding

For each problem, state the interval and interpret it in context. Use the stated model and round numerical endpoints to two decimal places.

A hypothetical call center samples $25$ independent calls. The mean waiting time is $41$ seconds, and the normal population has known standard deviation $5$ seconds. Find a 95% confidence interval for the population mean waiting time. (Answer: $(39.04,\ 42.96)$ seconds.)
A hypothetical sensor process has a normal population with known standard deviation $16$ units. For an independent sample of $64$ sensors, the mean reading is $120$ units. Find a 90% interval. (Answer: $(116.71,\ 123.29)$ units.)
A hypothetical study records $100$ independent measurements from a normal population with known standard deviation $2.0$. The sample mean is $8.6$. Find a 98% interval, using $z_{0.99}=2.3263$. (Answer: $(8.13,\ 9.07)$.)
A hypothetical warehouse samples $49$ independently selected packages from a normal population with known standard deviation $7$ kilograms. The sample mean is $52$ kilograms. Find a 90% interval and its margin of error. (Answer: Interval $(50.36,\ 53.64)$ kilograms; margin of error $1.64$ kilograms.)
A hypothetical machine produces measurements from a normal population with known standard deviation $6$. An independent sample of $36$ measurements has mean $30$. Find a 95% interval and its total width. (Answer: Interval $(28.04,\ 31.96)$; width $3.92$.)

What You'll Learn

From Summary Statistics to a Two-Sided Interval

Why the Critical Value Is Split Between Two Tails

A Reliable Calculation Process

Worked Example: A 95% Interval for a Mean Processing Time

Confidence Level, Width, and Interpretation

Worked Example: A 90% Interval for a Hypothetical Survey Measure

Using a More Stringent Confidence Level

Worked Example: A 99% Interval for a Hypothetical Component Dimension

Common Exam Trap: Using the Wrong Tail Probability

Check Your Understanding

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