Statistics

Critical value

A confidence interval reaches out some number of standard errors from the estimate. For a chosen level, such as exactly 95% or 90% or 98%, how many standard errors is that, and where does the number come from?

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Confidence level and interpreting an intervalStatistics Normal calculations: areas and percentilesAlgebra 2

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Critical value

Why this exists

Maricel runs a community center that is thinking of starting a weekend Tagalog class. She asks a random sample of 400 residents of the neighborhood, and 252 say they are interested (made-up data). The estimate and its Standard error are

\begin{aligned} \hat{p} &= \frac{252}{400} = 0.63 \\ \text{SE}_{\hat{p}} &= \sqrt{\frac{0.63 \times 0.37}{400}} \approx 0.0241 \end{aligned}

In Confidence level and interpreting an interval the interval reached out “about 2 standard errors” for “about 95\%”, and 1 and 3 standard errors gave about 68\% and 99.7\%. Those are rounded readings of the normal curve, and they offer only three levels. Maricel’s board wants exactly 95\%, and a grant form asks for 90\%. She needs the exact multiplier for any level she chooses. That multiplier is the critical value.

Definition

For a statistic whose sampling distribution is close to normal, the critical value z^* for confidence level C is the positive number with area C under the standard normal curve between -z^* and z^*.

The area left outside, 1 - C, splits equally into two tails of \displaystyle \frac{1 - C}{2} each, because the curve is symmetric about 0. So z^* is the point with area

\Phi(z^*) = 1 - \frac{1 - C}{2} = \frac{1 + C}{2}

to its left. It is found by reading a normal table backwards, as in Normal calculations: areas and percentiles, or with a calculator’s inverse normal command (often called invNorm) applied to \displaystyle \frac{1 + C}{2} with mean 0 and standard deviation 1.

The critical value turns a standard error into a Point estimates and the margin of error:

\text{margin of error} = z^* \times \text{SE}, \qquad \text{interval: } \hat{p} \pm z^*\,\text{SE}_{\hat{p}}

Intuition

Figure 1The critical value for 95\% confidence. The central area 0.95 leaves 0.025 in each tail, so z^* is the point with 0.975 to its left.

The sampling distribution of \hat{p} is close to normal and centered at p. Measured in standard errors, it is close to the standard normal curve. Choosing a level means choosing how much of that curve the interval must reach across: an estimate that lands in the shaded middle of Figure 1 sits within z^* standard errors of p, and so its interval reaches back to p. An estimate in either tail misses. The level is the shaded area, and z^* is where the shading stops.

More confidence means more shaded area, and the only way to shade more of the curve is to move the cuts outward. A higher level always needs a larger z^*.

Results

The standard critical values. Each comes from the area to its left, \displaystyle \frac{1 + C}{2}:

Level C	Each tail	Area to the left	z^*
80\%	0.10	0.90	1.282
90\%	0.05	0.95	1.645
95\%	0.025	0.975	1.960
98\%	0.01	0.99	2.326
99\%	0.005	0.995	2.576

From a critical value back to a level. If a multiplier z^* is used, the level is the central area

C = \Phi(z^*) - \Phi(-z^*) = 2\Phi(z^*) - 1

The critical value grows with the level. As C rises toward 1, z^* rises without bound. No finite z^* gives 100\%.

The critical value depends on the level only. It does not depend on n, on \hat{p} or on the context. The sample enters the margin only through the standard error.

The normal areas this lesson reads from a table:

z	\Phi(z)	z	\Phi(z)
1.28	0.8997	2.32	0.9898
1.29	0.9015	2.33	0.9901
1.64	0.9495	2.57	0.9949
1.65	0.9505	2.58	0.9951
1.96	0.9750

Why it is true

Why the shaded area is the capture rate. The interval \hat{p} \pm z^*\sigma_{\hat{p}} contains p exactly when |\hat{p} - p| \le z^*\sigma_{\hat{p}}, that is, when the standardized statistic \displaystyle Z = \frac{\hat{p} - p}{\sigma_{\hat{p}}} satisfies -z^* \le Z \le z^*. For a large random sample Z is close to standard normal, so the share of samples whose interval captures p is close to the area between -z^* and z^*, which is C by the definition of z^*. The working interval uses \text{SE}_{\hat{p}} in place of \sigma_{\hat{p}}; for a large sample the two are close.

Why the tails are equal. Any two cuts enclosing area C would give the right capture rate, but the curve is tallest in the middle, so a stretch of the axis near 0 holds more area than the same stretch farther out. The symmetric cuts enclose area C in the shortest stretch of axis, which gives the narrowest interval at that level.

Why a higher level needs a larger z^*. \Phi is increasing, since every step to the right adds area. A larger C makes \displaystyle \frac{1 + C}{2} larger, and the point with more area to its left lies farther right.

Worked examples

Example 1

The exact multiplier for 95%

Find z^* for 95\% confidence from the table, check that it encloses 0.95, and give Maricel’s 95\% interval.

Solution

The level leaves 1 - 0.95 = 0.05 outside, split as 0.025 in each tail. The upper cut has the central 0.95 and the lower tail to its left:

0.025 + 0.95 = 0.975

The table gives \Phi(1.96) = 0.9750, so z^* = 1.96. By symmetry \Phi(-1.96) = 1 - 0.9750 = 0.0250, and the area between the cuts is 0.9750 - 0.0250 = 0.9500, as required. This is the “about 2” of Confidence level and interpreting an interval, made exact.

The margin is 1.96 \times 0.0241 \approx 0.047, so the interval is

\begin{aligned} 0.63 - 0.047 &= 0.583 \\ 0.63 + 0.047 &= 0.677 \end{aligned}

Maricel can report that she is 95\% confident that between 58.3\% and 67.7\% of the neighborhood’s residents are interested in a weekend Tagalog class.

Example 2

90% and 99%

Derive z^* for 90\% and for 99\% from the table, and give both intervals.

Solution

For 90\% each tail holds \displaystyle \frac{0.10}{2} = 0.05, so z^* has 0.95 to its left. The table has \Phi(1.64) = 0.9495 and \Phi(1.65) = 0.9505; 0.95 lies halfway between, so z^* \approx 1.645. Technology gives 1.6449.

For 99\% each tail holds 0.005, so z^* has 0.995 to its left. The table has \Phi(2.57) = 0.9949 and \Phi(2.58) = 0.9951, so z^* is about 2.575; technology gives 2.5758, which rounds to 2.576.

The margins and intervals:

\begin{aligned} 90\%{:}\quad & 1.645 \times 0.0241 \approx 0.040, \quad 0.590 \text{ to } 0.670 \\ 99\%{:}\quad & 2.576 \times 0.0241 \approx 0.062, \quad 0.568 \text{ to } 0.692 \end{aligned}

Figure 2 sets the three intervals side by side.

Figure 2One survey (made-up data), three levels. The center and the standard error stay fixed; only z^* changes, and the interval grows with it.

Example 3

Which area goes in

Ligaya, who keeps the center’s records, wants z^* for 95\% and asks a calculator for the point with 0.95 to its left. It returns 1.645. What level does an interval using 1.645 actually have?

Solution

The point with 0.95 to its left has 0.05 to its right, and by symmetry -1.645 has 0.05 to its left. The central area is

\Phi(1.645) - \Phi(-1.645) = 0.95 - 0.05 = 0.90

so 1.645 is the critical value for 90\%, not 95\%. The level is a central area, and the area to the left of z^* is the central area plus one tail, \displaystyle \frac{1 + C}{2}. For 95\% that is 0.975, which returns 1.96.

Example 4

Any level: 80% and 98%

The board asks how the interval looks at 80\% and at 98\%. Find both critical values and both intervals.

Solution

At 80\% each tail holds 0.10, so the area to the left is 0.90. The table has \Phi(1.28) = 0.8997 and \Phi(1.29) = 0.9015; 0.90 is much nearer the first, and technology gives z^* = 1.282. The margin is 1.282 \times 0.0241 \approx 0.031, giving 0.599 to 0.661.

At 98\% each tail holds 0.01, so the area to the left is 0.99. The table has \Phi(2.32) = 0.9898 and \Phi(2.33) = 0.9901, and technology gives z^* = 2.326. The margin is 2.326 \times 0.0241 \approx 0.056, giving 0.574 to 0.686.

Across the five levels from 80\% to 99\% the critical value runs 1.282, 1.645, 1.96, 2.326, 2.576: it rises with the level, and so does every margin.

Example 5

From a margin back to a level

Dalisay writes the center’s newsletter. From Maricel’s sample she prints “63\% of residents are interested, give or take 3.5 percentage points.” What confidence level does that margin carry?

Solution

The margin is z^* \times \text{SE}, so the multiplier she used is

z^* = \frac{0.035}{0.0241} \approx 1.45

The level is the central area for that cut:

2\Phi(1.45) - 1 \approx 2(0.9265) - 1 = 0.853

so her margin carries about 85\% confidence. A margin with no level beside it hides this choice; the same sample supports \pm 0.035 at 85\% or \pm 0.047 at 95\%.

Example 6

A claimed value at three levels

Last year a smaller survey suggested that 58\% of residents were interested. Is 0.58 plausible at 90\%, 95\% and 99\%? How much wider is the 99\% interval than the 90\% one?

Solution

From Figure 2: 0.58 is below 0.590 and below 0.583, so it is not plausible at 90\% or at 95\%. It is above 0.568, so it is plausible at 99\%. The verdict depends on the level, which is why the level is chosen before looking at the claim.

Both intervals share the same center and the same standard error, so their widths are in the ratio of their critical values:

\frac{2.576}{1.645} \approx 1.566

The 99\% interval is about 57\% wider. The extra confidence is bought with width alone, since the data are unchanged.

Example 7

Changing the level of someone else’s interval

A center across town reports a 95\% interval of 0.41 to 0.49 for the share of its residents interested in a similar class, built as \hat{p} \pm 1.96\,\text{SE} from a random sample. Find the standard error, about how many residents were asked, and the 90\% interval from the same sample.

Solution

The center is the midpoint, \displaystyle \hat{p} = \frac{0.41 + 0.49}{2} = 0.45, and the margin is half the width, 0.04. Dividing out the multiplier,

\text{SE} = \frac{0.04}{1.96} \approx 0.0204

Solving \displaystyle \text{SE}^2 = \frac{\hat{p}(1 - \hat{p})}{n} for n, with the unrounded \displaystyle \text{SE} = \frac{0.04}{1.96},

n = 0.45 \times 0.55 \times \left(\frac{1.96}{0.04}\right)^2 \approx 594.25

so about 594 residents. The 90\% margin uses the same standard error with the smaller multiplier:

1.645 \times 0.0204 \approx 0.0336

so the 90\% interval runs from about 0.416 to 0.484. No raw data were needed: the critical value is the only part that changes with the level.

Boundaries

z^* gives the stated level only when the sampling distribution of the statistic is close to normal. For a proportion, that takes a random sample, a sample small against the population, and enough successes and failures, the subject of Conditions for one proportion. When the conditions fail, the true capture rate can fall short of C whatever z^* is used.
z^* comes from the standard normal curve. For an interval for a mean, where the population standard deviation is estimated from the sample, the multiplier is a t^* from the The t distribution and degrees of freedom instead, taken up for means later in the course.
The critical value here is two-sided: the level is a central area with equal tails. A one-sided bound puts all of 1 - C in one tail, and its multiplier is different.
z^* exists for every level strictly between 0 and 1. At C = 100\% there is no finite critical value, and the interval would be every possible value.
Table readings are rounded to the nearest listed z, so a critical value from a table can differ from technology in the third decimal place.

Where this goes next

With the conditions from Conditions for one proportion added, the interval \displaystyle \hat{p} \pm z^*\sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} becomes the full One sample interval for a proportion and sample size, where z^* also tells how large a sample a chosen margin needs. Intervals for means keep the same shape, an estimate plus or minus a critical value times a standard error, with t^* in place of z^* in the One sample interval for a mean and sample size.

Exercises

Use technology for critical values, to three decimal places, unless a part says otherwise.

Finding critical values

Find z^* for 75\% confidence. State the tail area and the area to the left.
Find z^* for 92\% confidence.
Find z^* for 99.5\% confidence.
Find z^* for 60\% confidence.
An interval uses the multiplier z^* = 2.17. What is its confidence level?
An interval uses the multiplier z^* = 1. What is its confidence level?

Margins and intervals

A random sample of 250 residents at another center finds 95 who have studied a second language. Find \hat{p}, the standard error, and the 95\% and 99\% intervals.
Dalisay surveys a random sample of 600 residents, and 210 attended a center event last year. Find the 80\% and 98\% intervals.
A 95\% margin of error for a sample proportion is 0.052. Without the data, find the 90\% and 99\% margins from the same sample.
A report gives a 99\% interval of 0.22 to 0.30, built as \hat{p} \pm 2.576\,\text{SE}. Find \hat{p}, the standard error, about how many people were sampled, and the 95\% interval.
A standard error is 0.02 and the reported margin is 0.03. What confidence level was used?
Give Maricel’s interval at 70\% confidence.

Reasoning

Explain why z^* for 95\% is the 97.5th percentile of the standard normal distribution and not the 95th.
Explain, using areas under the curve, why z^* for 99\% must be larger than z^* for 95\%, and what happens to z^* as C approaches 100\%.
Ligaya says the critical value for 95\% should be larger for a sample of 100 than for a sample of 1{,}000. Explain what is wrong, and what does change with n.
Is the critical value for 50\% confidence 0? Find it and say what it means.
From one sample, explain why the 99\% interval contains the whole 90\% interval.
Maricel wants a 95\% interval half as wide as hers from the same data, and suggests halving z^*. Find the level an interval with half of 1.96 actually has, and say what would keep 95\% and halve the width.

Answers

Each tail is \displaystyle \frac{0.25}{2} = 0.125, the area to the left is 0.875, and z^* \approx 1.150.
Each tail is 0.04, the area to the left is 0.96, and z^* \approx 1.751.
Each tail is 0.0025, the area to the left is 0.9975, and z^* \approx 2.807.
Each tail is 0.20, the area to the left is 0.80, and z^* \approx 0.842.
2\Phi(2.17) - 1 \approx 2(0.9850) - 1 = 0.970, so about 97\%.
2\Phi(1) - 1 \approx 2(0.8413) - 1 = 0.683, about 68\%, the exact form of the 1-standard-error level.
\displaystyle \hat{p} = \frac{95}{250} = 0.38; \displaystyle \text{SE} = \sqrt{0.38 \times 0.62 / 250} \approx 0.0307. At 95\%: margin 1.96 \times 0.0307 \approx 0.060, interval 0.320 to 0.440. At 99\%: margin 2.576 \times 0.0307 \approx 0.079, interval 0.301 to 0.459.
\displaystyle \hat{p} = \frac{210}{600} = 0.35; \displaystyle \text{SE} = \sqrt{0.35 \times 0.65 / 600} \approx 0.0195. At 80\%: margin 1.282 \times 0.0195 \approx 0.025, interval 0.325 to 0.375. At 98\%: margin 2.326 \times 0.0195 \approx 0.045, interval 0.305 to 0.395.
The standard error is \displaystyle \frac{0.052}{1.96} in both, so the 90\% margin is \displaystyle 0.052 \times \frac{1.645}{1.96} \approx 0.044 and the 99\% margin is \displaystyle 0.052 \times \frac{2.576}{1.96} \approx 0.068.
\hat{p} = 0.26, margin 0.04, \displaystyle \text{SE} = \frac{0.04}{2.576} \approx 0.0155; \displaystyle n = 0.26 \times 0.74 \times \left(\frac{2.576}{0.04}\right)^2 \approx 797.95, about 798. The 95\% margin is 1.96 \times 0.0155 \approx 0.030, so the interval is 0.230 to 0.290.
\displaystyle z^* = \frac{0.03}{0.02} = 1.5, and 2\Phi(1.5) - 1 \approx 0.866, about 86.6\%.
The area to the left is 0.85, so z^* \approx 1.036; margin 1.036 \times 0.0241 \approx 0.025; interval 0.605 to 0.655.
The 95\% is the central area. The point z^* has that central area plus the lower tail of 0.025 to its left, 0.975 in all, so it is the 97.5th percentile. The 95th percentile, 1.645, encloses only 90\% in the middle.
A central area of 0.99 is more than 0.95, and the cuts at \pm 1.96 enclose only 0.95; to enclose more, the cuts must move outward, since the area between them grows as they spread. As C approaches 100\% the tails must shrink to nothing, and the curve never reaches the axis, so z^* grows without bound.
z^* depends only on the level: 1.96 for 95\% at any n. What changes with n is the standard error, \displaystyle \sqrt{\hat{p}(1 - \hat{p})/n}, which is smaller for the larger sample, so the margin 1.96 \times \text{SE} shrinks.
No. Each tail is 0.25, so z^* has 0.75 to its left, and z^* \approx 0.674, the upper quartile. Half of all samples give an estimate within 0.674 standard errors of p. Only a level of 0 would give z^* = 0.
Both intervals have the same center \hat{p} and the same standard error. The 99\% multiplier 2.576 is larger than 1.645, so each end of the 99\% interval lies farther out than the matching end of the 90\% interval.
Half of 1.96 is 0.98, and 2\Phi(0.98) - 1 \approx 0.673: the narrower interval has only about 67\% confidence. Keeping 95\% needs a standard error half as large, which takes a sample four times as large, since the standard error falls like \displaystyle \frac{1}{\sqrt{n}}.