Why Critical Values Matter
A proportion interval uses an estimated standard error to describe the sampling variation of a sample proportion. The critical value determines how many standard errors extend from the sample estimate to each endpoint. A larger critical value therefore produces a wider interval when the standard error stays the same.
This tutorial focuses on choosing that number. The relevant values come from the standard normal distribution, \(Z\sim N(0,1)\). The choice depends on the central area desired, not on the observed sample proportion or sample size. Those sample details affect the standard error and thus the margin of error, but not the critical value.
For example, a central level of 95% means \(C=0.95\), so \(\alpha=1-0.95=0.05\). Each tail has area \(0.05/2=0.025\). The critical value is the point with 0.025 of the standard normal distribution to its right, or cumulative probability 0.975 to its left.
Finding a Critical Value
Let \(\Phi(z)=P(Z\leq z)\) be the cumulative distribution function of the standard normal distribution. Because the normal curve is symmetric about zero, a central level \(C\) leaves \(\alpha/2=(1-C)/2\) in each tail. Thus the area to the left of the positive critical value is \(1-\alpha/2\):
A cumulative standard normal table usually gives \(\Phi(z)\), the area to the left of \(z\). To find \(z^*\), convert the desired central level to the left-tail cumulative probability, then locate the corresponding \(z\). A calculator or statistical software can find the same quantile. The critical value is positive; the two endpoints around zero are \(-z^*\) and \(z^*\).
Worked Example: Finding the 95% Critical Value
Find the two-sided standard normal critical value for a central level of 95%.
Step 1: Find the total area outside the center. Here \(C=0.95\), so \( \alpha=1-C=1-0.95=0.05 \).
Step 2: Divide the area between the two tails. Each tail has area \( \alpha/2=0.05/2=0.025 \).
Step 3: Convert to a cumulative left-tail probability. The area to the left of the positive critical value is \( 1-0.025=0.975 \). Thus \(z^*=\Phi^{-1}(0.975)\), which is approximately \(1.960\).
The central area is between \(-1.960\) and \(1.960\). The two remaining tail areas are each about 0.025, for a total outside area of about 0.050.
Common central levels and their corresponding critical values are listed below. Values are rounded to three decimal places; small differences in the last digit can result from rounding.
| Central level \(C\) | Total tail area \(\alpha=1-C\) | Area in each tail \(\alpha/2\) | Critical value \(z^*\) |
|---|---|---|---|
| 80% | 0.20 | 0.10 | 1.282 |
| 90% | 0.10 | 0.05 | 1.645 |
| 95% | 0.05 | 0.025 | 1.960 |
| 98% | 0.02 | 0.01 | 2.326 |
| 99% | 0.01 | 0.005 | 2.576 |
| 99.9% | 0.001 | 0.0005 | 3.291 |
Worked Example: Finding a 92% Critical Value
Find the two-sided standard normal critical value for a central level of 92%.
Write \(C=0.92\). The total area outside the center is \( \alpha=1-0.92=0.08 \), so the area in each tail is \( \alpha/2=0.08/2=0.04 \).
The positive critical value has cumulative area \( 1-0.04=0.96 \) to its left. Therefore, \( z^*=\Phi^{-1}(0.96)\approx1.751 \). The corresponding central area is between \(-1.751\) and \(1.751\); approximately 0.04 lies in each tail. This is why a table lookup for 0.96, rather than for 0.92, finds the critical value.
From the Critical Value to Margin of Error
For a proportion interval, the margin of error is the critical value multiplied by the estimated standard error. Using the sample proportion \(\hat{p}\) and sample size \(n\), the usual estimated standard error is \(SE=\sqrt{\hat{p}(1-\hat{p})/n}\). This tutorial uses that quantity to illustrate the role of \(z^*\); checking when the interval approximation is appropriate is a separate step.
The multiplication has a direct effect: holding \(SE\) fixed, a larger \(z^*\) gives a larger \(ME\). A higher central level leaves less probability in the tails, so its positive critical value must be farther from zero. Consequently, higher central levels produce wider intervals when based on the same estimate and standard error.
Worked Example: Calculating and Comparing Margins of Error
Suppose a hypothetical survey has \(n=400\) respondents, of whom 248 give a particular response. Then \(\hat{p}=248/400=0.620\). Find the estimated standard error and compare margins of error at the 90%, 95%, and 99% central levels.
First calculate the estimated standard error:
The standard error is about 0.02427, or 2.427 percentage points. For a 90% central level, \(z^*=1.645\), so \(ME=1.645(0.02427)\approx0.03992\), about 3.992 percentage points.
For a 95% central level, \(z^*=1.960\), giving \(ME=1.960(0.02427)\approx0.04757\), about 4.757 percentage points. For a 99% central level, \(z^*=2.576\), giving \(ME=2.576(0.02427)\approx0.06252\), about 6.252 percentage points.
The estimated standard error is unchanged across these calculations because the sample information is the same. The margin grows as the central level rises because the corresponding critical value grows. These are margins of error in proportion units; multiplying by 100 expresses them in percentage points.
Why the Tail Split Works
The equal split between tails follows from the symmetry of the standard normal distribution. For a positive number \(z^*\), symmetry gives \(P(Z<-z^*)=P(Z>z^*)\). The total probability outside the interval from \(-z^*\) to \(z^*\) is therefore twice the right-tail probability. If the central area is \(C\), the outside area is \(1-C=\alpha\), so each tail must contain \(\alpha/2\).
By the definition of the cumulative distribution function, \(P(Z\leq z^*)=\Phi(z^*)\). Since \(P(Z>z^*)=\alpha/2\), the area to the left is \(1-\alpha/2\). Applying the inverse cumulative distribution function gives \(z^*=\Phi^{-1}(1-\alpha/2)\). This establishes both the tail calculation and the lookup rule.
Common Mistake: Putting All the Tail Area on One Side
A frequent error is to subtract the entire outside area \(\alpha\) from 1 and use that as the table probability. For a 95% central level, this gives \(1-0.05=0.95\), and the resulting quantile is about 1.645. But 1.645 leaves 0.05 in the right tail and, by symmetry, 0.05 in the left tail. The central area is therefore only about 0.90, not 0.95.
For a two-sided critical value, first divide the outside area by two, then use the left-tail cumulative probability \(1-\alpha/2\). At 95%, that is \(1-0.05/2=0.975\), yielding \(z^*\approx1.960\). The value 1.645 is associated with a 90% central level under this two-sided construction. Always check that the two tail areas add to \(1-C\).
Check Your Understanding
Try these questions without looking back.
- For a central level of 98%, what is the total tail area, and what area lies in each tail?
- A standard normal table reports cumulative left-tail areas. What cumulative probability should be used to find the positive critical value for a 90% central level?
- Why does a 99% central level have a larger critical value than a 95% central level?
- If \(SE=0.018\) and \(z^*=1.960\), calculate the margin of error. State the result in proportion units.
- For a two-sided central level of 95%, explain why looking up a cumulative probability of 0.95 gives the wrong critical value.