Hypothesis Tests Using Normal Approximations

A2 · S2 · 14 min

When the sample is large or the mean count is large, exact binomial and Poisson tests become impractical: testing $72$ successes out of $200$ needs $P(X \ge 72)$ for $B(200, 0.3)$ , a sum of $129$ terms. The syllabus allows you to replace the binomial or Poisson distribution by a normal distribution with the same mean and variance, provided you apply a continuity correction. The logic of the test does not change at all; only the way the probability is calculated does. These tests appear regularly on Paper 6, and the continuity correction is where most marks are lost.

The approximations

Both approximations come from earlier work: the normal approximation to the binomial from S1, and the normal approximation to the Poisson.

Normal approximations for tests

Under $H_0$ :

Exact distribution	Approximation	Condition
$X \sim B(n, p_0)$	$N\big(np_0,\ np_0(1 - p_0)\big)$	$np_0 > 5$ and $n(1 - p_0) > 5$
$X \sim \text{Po}(\lambda_0)$	$N(\lambda_0,\ \lambda_0)$	$\lambda_0 > 15$

Because $X$ is discrete and the normal is continuous, a continuity correction is always applied:

P(X \ge x) \approx P(Y > x - 0.5), \qquad P(X \le x) \approx P(Y < x + 0.5).

The conditions are checked with the parameter value in $H_0$ , because the distribution is the one that holds if $H_0$ is true.

The continuity correction treats each integer $x$ as the interval from $x - 0.5$ to $x + 0.5$ . "At least $72$ " includes the whole bar at $72$ , which starts at $71.5$ ; "at most $29$ " includes the whole bar at $29$ , which ends at $29.5$ . The graph shows $\text{Po}(40)$ as bars with the curve of $N(40, 40)$ ; the shaded area up to $29.5$ approximates $P(X \le 29)$ .

y = 40^floor(x + 0.5) exp(-40) / fact(floor(x + 0.5)) y = exp(-(x - 40)^2 / 80) / sqrt(80 pi) fill 20 29.5 y = exp(-(x - 40)^2 / 80) / sqrt(80 pi)

Carrying out the test

You can decide in either of two equivalent ways.

Compare a probability with the significance level. Calculate the approximate tail probability, with the continuity correction, and compare it with the significance level.

Compare a $z$ -value with a critical value. Standardise the continuity-corrected value and compare it with $1.645$ , $1.96$ , $2.326$ or $2.576$ as appropriate. For an upper-tail test with observed value $x$ ,

z = \frac{(x - 0.5) - \mu}{\sigma},

and for a lower-tail test,

z = \frac{(x + 0.5) - \mu}{\sigma}.

The hypotheses are still written for the binomial proportion $p$ or the Poisson mean $\lambda$ . The normal distribution is only a tool for calculating the probability.

Test using a normal approximation

Define the parameter and state $H_0$ and $H_1$ .
State the exact distribution under $H_0$ , check the condition, and state the approximating normal distribution with its mean and variance.
Apply the continuity correction in the direction of the tail you need.
Standardise and find the probability (or compare $z$ with the critical value).
Compare with the significance level, halved for each tail of a two-tailed test.
Decide about $H_0$ and conclude in context.

Routine: a binomial proportion

In the past, $30\%$ of customers at a supermarket bought organic vegetables. After a promotion, a random sample of $200$ customers includes $72$ who buy organic vegetables. Test at the $5\%$ significance level whether the proportion has increased.

Solution

Let $p$ be the proportion of customers who buy organic vegetables.

$H_0: p = 0.3$ , $\quad H_1: p > 0.3$ .

Under $H_0$ , $X \sim B(200, 0.3)$ . Since $np = 60 > 5$ and $n(1 - p) = 140 > 5$ , use $X \approx Y \sim N(60, 42)$ .

P(X \ge 72) \approx P(Y > 71.5) = P\left(Z > \frac{71.5 - 60}{\sqrt{42}}\right) = P(Z > 1.774) = 1 - 0.9620 = 0.0380

$0.0380 < 0.05$ , so reject $H_0$ .

There is evidence at the $5\%$ level that the proportion of customers who buy organic vegetables has increased.

A Poisson mean

A switchboard receives calls at random at an average rate of $40$ per day. After a website is redesigned, there are $29$ calls on a randomly chosen day. Test at the $5\%$ significance level whether the rate of calls has decreased.

Solution

Let $\lambda$ be the mean number of calls per day. $H_0: \lambda = 40$ , $\quad H_1: \lambda < 40$ .

Under $H_0$ , $X \sim \text{Po}(40)$ . Since $40 > 15$ , use $X \approx Y \sim N(40, 40)$ .

P(X \le 29) \approx P(Y < 29.5) = P\left(Z < \frac{29.5 - 40}{\sqrt{40}}\right) = P(Z < -1.660) = 1 - 0.9515 = 0.0485

$0.0485 < 0.05$ , so reject $H_0$ .

There is evidence at the $5\%$ level that the rate of calls has decreased.

Using the $z$ -value instead: $-1.660 < -1.645$ , so the result is in the critical region. Same conclusion.

When the continuity correction changes the answer

Because a test is a yes-or-no decision, a small change in the probability can change the conclusion. The continuity correction is not a fine detail; it is part of the method, and leaving it out can give the wrong answer.

A two-tailed test where the correction matters

A coin is tossed $100$ times and shows $60$ heads. Test at the $5\%$ significance level whether the coin is biased.

Solution

Let $p$ be the probability of a head. $H_0: p = 0.5$ , $\quad H_1: p \ne 0.5$ .

Under $H_0$ , $X \sim B(100, 0.5)$ . Since $np = n(1 - p) = 50 > 5$ , use $X \approx Y \sim N(50, 25)$ .

$60$ is above the mean, so use the upper tail.

P(X \ge 60) \approx P(Y > 59.5) = P\left(Z > \frac{59.5 - 50}{5}\right) = P(Z > 1.9) = 1 - 0.9713 = 0.0287

The test is two-tailed, so compare with $0.025$ : $0.0287 > 0.025$ . Do not reject $H_0$ .

There is insufficient evidence at the $5\%$ level that the coin is biased.

Without the continuity correction, $z = \dfrac{60 - 50}{5} = 2$ and $P(Z > 2) = 0.0228 < 0.025$ , which would wrongly lead to rejecting $H_0$ . (The exact binomial probability is $0.0284$ , confirming the corrected value.)

Critical regions with a normal approximation

To find a critical region, set the continuity-corrected boundary equal to the critical value and solve for the count.

For an upper-tail test at level $\alpha$ with critical value $z_\alpha$ , the critical region is $X \ge c$ where $c$ is the smallest integer with

\frac{(c - 0.5) - \mu}{\sigma} \ge z_\alpha.

For a lower-tail test, it is $X \le c$ where $c$ is the largest integer with

\frac{(c + 0.5) - \mu}{\sigma} \le -z_\alpha.

Finding a critical region

A switchboard receives calls at random at an average rate of $40$ per day. Using a single day's count and the $1\%$ significance level, find the critical region for a test of whether the rate has increased.

Solution

$H_0: \lambda = 40$ , $\quad H_1: \lambda > 40$ . Under $H_0$ , $X \sim \text{Po}(40) \approx Y \sim N(40, 40)$ .

We need the smallest integer $c$ with

\frac{c - 0.5 - 40}{\sqrt{40}} \ge 2.326 \;\Rightarrow\; c \ge 40.5 + 2.326\sqrt{40} = 40.5 + 14.71 = 55.21.

So $c = 56$ , and the critical region is $X \ge 56$ .

Round up for an upper-tail region: $55$ would give a $z$ -value below $2.326$ .

Exam-hard: scaling, testing and a critical region

Emails arrive in an office at random at an average rate of $2.5$ per hour. After the office joins a new mailing list, $30$ emails arrive during a randomly chosen $8$ -hour working day.

(a) Use a suitable approximation to test at the $2.5\%$ significance level whether the rate of emails has increased.

(b) Find the critical region for this test.

Solution

(a) For $8$ hours the mean under the old rate is $8 \times 2.5 = 20$ .

Let $\lambda$ be the mean number of emails per $8$ -hour day. $H_0: \lambda = 20$ , $\quad H_1: \lambda > 20$ .

Under $H_0$ , $X \sim \text{Po}(20)$ . Since $20 > 15$ , use $X \approx Y \sim N(20, 20)$ .

P(X \ge 30) \approx P(Y > 29.5) = P\left(Z > \frac{29.5 - 20}{\sqrt{20}}\right) = P(Z > 2.124) = 1 - 0.9832 = 0.0168

$0.0168 < 0.025$ , so reject $H_0$ . There is evidence at the $2.5\%$ level that the rate of emails has increased.

(b) The one-tailed critical value at $2.5\%$ is $1.96$ .

\frac{c - 0.5 - 20}{\sqrt{20}} \ge 1.96 \;\Rightarrow\; c \ge 20.5 + 1.96\sqrt{20} = 20.5 + 8.765 = 29.27

The critical region is $X \ge 30$ , which is consistent with rejecting $H_0$ in (a).

(c) The mean $20$ is greater than $15$ , so $\text{Po}(20)$ is close enough to symmetrical for a normal approximation. A continuity correction is needed because the number of emails is a discrete variable and it is being approximated by a continuous one.

Common mistakes

Leaving out the continuity correction. It is part of the method, it earns a mark, and without it a borderline test can reach the wrong conclusion.
Correcting in the wrong direction. $P(X \ge 72)$ becomes $P(Y > 71.5)$ ; $P(X \le 29)$ becomes $P(Y < 29.5)$ . Think about which bars must be included.
Using the variance as the standard deviation. $N(40, 40)$ has standard deviation $\sqrt{40} = 6.325$ , not $40$ .
Using the sample proportion in the variance. For a test, the variance is $np_0(1 - p_0)$ with the value from $H_0$ . (Confidence intervals use $p_s$ ; tests use $p_0$ .)
Not checking the conditions. State $np_0 > 5$ and $n(1 - p_0) > 5$ , or $\lambda_0 > 15$ , with the numbers.
Rounding a critical value the wrong way. For an upper-tail region round $c$ up; for a lower-tail region round $c$ down.
Writing hypotheses about the normal variable. The hypotheses concern $p$ or $\lambda$ , not $Y$ or $\mu$ of the approximating normal.

Exam tip

The marks usually go: hypotheses; correct normal distribution with mean and variance; continuity correction; standardising; comparison; conclusion in context.
Write the approximation explicitly, with the condition: " $np = 60 > 5$ and $nq = 140 > 5$ , so $X \approx N(60, 42)$ ".
Show the corrected value in the $z$ -formula, such as $\dfrac{71.5 - 60}{\sqrt{42}}$ . Examiners look for the $0.5$ .
Keep $z$ to at least 3 decimal places; read $\Phi$ from the tables to 4 decimal places.
If you compare $z$ with a critical value, say which critical value and why: " $1.774 > 1.645$ , the critical value for a one-tailed test at $5\%$ ".
Questions often say "use a suitable approximation". This is your cue to state which approximation and to justify it.

Summary

Use $B(n, p_0) \approx N(np_0, np_0(1 - p_0))$ when $np_0 > 5$ and $n(1 - p_0) > 5$ .
Use $\text{Po}(\lambda_0) \approx N(\lambda_0, \lambda_0)$ when $\lambda_0 > 15$ .
Always apply a continuity correction: $X \ge x$ becomes $Y > x - 0.5$ ; $X \le x$ becomes $Y < x + 0.5$ .
Compare the approximate tail probability with the significance level, or the corrected $z$ with the critical value.
The variance uses the value in $H_0$ , not a sample estimate.
Critical regions: solve the corrected inequality and round up (upper tail) or down (lower tail).
Hypotheses and conclusions refer to $p$ or $\lambda$ , in context.

Practice questions

Question

It is claimed that $40\%$ of students walk to school. In a random sample of $150$ students, $72$ walk to school. Use a suitable approximation to test at the $5\%$ significance level whether the proportion is greater than $40\%$ .
The number of complaints received by a company per month has the distribution $\text{Po}(25)$ . After a change in policy, $16$ complaints are received in a month. Use a suitable approximation to test at the $5\%$ significance level whether the rate of complaints has decreased.
A coin is tossed $400$ times and shows $221$ heads. Test at the $5\%$ significance level whether the coin is biased.
A test of $H_0: p = 0.2$ against $H_1: p < 0.2$ is carried out at the $5\%$ significance level using a random sample of $300$ . Use a normal approximation to find the critical region.
A machine breaks down at random at an average rate of $4.5$ times per week. After a service, there are $18$ breakdowns in $6$ weeks. Test at the $1\%$ significance level whether the rate of breakdowns has decreased.
In a test using the approximation $\text{Po}(30) \approx N(30, 30)$ , explain why the condition on $\lambda$ is needed and why a continuity correction is used.
A die is thrown $180$ times. Using a normal approximation, find the least number of sixes that would give evidence at the $5\%$ significance level that the die is biased towards six.
The number of visitors to a website per hour has the distribution $\text{Po}(50)$ . A two-tailed test at the $5\%$ significance level is to be carried out using the number of visitors in one hour. (a) Use a normal approximation to find the critical region. (b) In a particular hour there are $64$ visitors. State the conclusion of the test.

Answers

$H_0: p = 0.4$ , $H_1: p > 0.4$ . Under $H_0$ , $X \sim B(150, 0.4)$ ; $np = 60 > 5$ , $nq = 90 > 5$ , so $X \approx N(60, 36)$ . $P(X \ge 72) \approx P\left(Z > \dfrac{71.5 - 60}{6}\right) = P(Z > 1.917) = 1 - 0.9724 = 0.0276 < 0.05$ . Reject $H_0$ : there is evidence at the $5\%$ level that more than $40\%$ of students walk to school.
$H_0: \lambda = 25$ , $H_1: \lambda < 25$ . $25 > 15$ , so $X \approx N(25, 25)$ . $P(X \le 16) \approx P\left(Z < \dfrac{16.5 - 25}{5}\right) = P(Z < -1.7) = 1 - 0.9554 = 0.0446 < 0.05$ . Reject $H_0$ : there is evidence at the $5\%$ level that the rate of complaints has decreased.
$H_0: p = 0.5$ , $H_1: p \ne 0.5$ . $X \sim B(400, 0.5) \approx N(200, 100)$ . $P(X \ge 221) \approx P\left(Z > \dfrac{220.5 - 200}{10}\right) = P(Z > 2.05) = 1 - 0.9798 = 0.0202$ . Two-tailed: $0.0202 < 0.025$ . Reject $H_0$ : there is evidence at the $5\%$ level that the coin is biased.
$X \sim B(300, 0.2) \approx N(60, 48)$ . Need the largest $c$ with $\dfrac{c + 0.5 - 60}{\sqrt{48}} \le -1.645$ , so $c \le 59.5 - 1.645\sqrt{48} = 59.5 - 11.40 = 48.10$ . So $c = 48$ and the critical region is $X \le 48$ .
Mean for $6$ weeks $= 27$ . $H_0: \lambda = 27$ , $H_1: \lambda < 27$ . $27 > 15$ , so $X \approx N(27, 27)$ . $P(X \le 18) \approx P\left(Z < \dfrac{18.5 - 27}{\sqrt{27}}\right) = P(Z < -1.636) = 1 - 0.9491 = 0.0509 > 0.01$ . Do not reject $H_0$ : there is insufficient evidence at the $1\%$ level that the rate has decreased.
A Poisson distribution is skewed for small $\lambda$ ; when $\lambda$ is large (greater than $15$ ) it is close enough to symmetrical and bell-shaped to be approximated by a normal distribution. The continuity correction is needed because a discrete distribution (whole-number counts) is being approximated by a continuous one, so each integer is represented by the interval from $0.5$ below it to $0.5$ above it.
$H_0: p = \tfrac{1}{6}$ , $H_1: p > \tfrac{1}{6}$ . $X \sim B\left(180, \tfrac{1}{6}\right) \approx N(30, 25)$ . Need $\dfrac{x - 0.5 - 30}{5} \ge 1.645$ , so $x \ge 30.5 + 8.225 = 38.725$ . The least number of sixes is $39$ .
(a) $X \sim \text{Po}(50) \approx N(50, 50)$ . Each tail has $2.5\%$ , critical values $\pm 1.96$ . Lower: $\dfrac{c + 0.5 - 50}{\sqrt{50}} \le -1.96 \Rightarrow c \le 49.5 - 13.86 = 35.64$ , so $X \le 35$ . Upper: $\dfrac{c - 0.5 - 50}{\sqrt{50}} \ge 1.96 \Rightarrow c \ge 50.5 + 13.86 = 64.36$ , so $X \ge 65$ . The critical region is $X \le 35$ or $X \ge 65$ . (b) $64$ is not in the critical region. Do not reject $H_0$ : there is insufficient evidence at the $5\%$ level that the mean number of visitors per hour has changed.

The approximations

Carrying out the test

When the continuity correction changes the answer

Critical regions with a normal approximation

Practice questions

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