Hypothesis Tests for Parameters

In hypothesis testing you met the general structure: state $H_0$ and $H_1$, find a p-value, compare it with the significance level, and conclude in context. This page applies that structure to parameters: a population mean, a proportion, and a Poisson rate. You’ll also learn a second way to decide, using a critical region, and how to calculate the chance of each kind of wrong conclusion.

Key ideas

Critical values and critical regions

The critical region is the set of values of the test statistic that lead you to reject $H_0$. The boundary of the region is the critical value. You can decide a test either way:

• p-value method: reject $H_0$ if the p-value is less than the significance level.
• Critical region method: reject $H_0$ if the observed value lies in the critical region.

The two methods always give the same conclusion. The critical region method is useful because you can work it out before collecting data, and you need it to find error probabilities.

Testing a mean: σ known (normal distribution)

If the population is normal with known $\sigma$ (or $n$ is large, by the central limit theorem), then under $H_0: \mu = \mu_0$,

$$\bar{X} \sim N\left(\mu_0,\ \frac{\sigma^2}{n}\right) \qquad\text{and}\qquad z = \frac{\bar{x} - \mu_0}{\sigma/\sqrt{n}}$$

For a one-tailed test with $H_1: \mu \gt \mu_0$ at the $5\%$ level, the critical value of $z$ is $1.645$, so the critical region is $z \gt 1.645$, or equivalently $\bar{x} \gt \mu_0 + 1.645\dfrac{\sigma}{\sqrt{n}}$. For a two-tailed test at $5\%$, split the $5\%$: reject if $z \lt -1.960$ or $z \gt 1.960$. A GDC’s z-test function gives $z$ and the p-value directly.

Testing a mean: σ unknown (t-test)

If $\sigma$ is unknown, use a one-sample t-test with $s_{n-1}$ and $n - 1$ degrees of freedom, as in the t-test topic. Use your GDC’s t-test function to get the test statistic and p-value. As with confidence intervals: σ known → normal; σ unknown → t, regardless of the sample size. You won’t be asked to calculate critical regions for t-tests.

Paired samples. When each individual is measured twice (before and after, left and right), work out the differences $d$ for each pair and do a one-sample test on them, usually with $H_0: \mu_d = 0$. A matched-pairs test is just a single-sample test on the differences.

Testing a proportion with the binomial distribution

To test a claimed probability $p$, count the successes $X$ in $n$ trials. Under $H_0$, $X \sim B(n, p_0)$. These tests are one-tailed only:

• $H_1: p \lt p_0$: the critical region is $X \le k$ (small values).
• $H_1: p \gt p_0$: the critical region is $X \ge k$ (large values).

Because $X$ is discrete, you usually can’t hit the significance level exactly. The IB rule: choose the critical region that makes $P(X \text{ in critical region})$ as large as possible while still less than the significance level. List cumulative probabilities with the GDC to find it.

Testing a mean with the Poisson distribution

To test a claimed Poisson rate, count the events $X$ in the observed interval. Under $H_0$, $X \sim \mathrm{Po}(m_0)$, where $m_0$ is the mean for the whole interval observed (see the Poisson distribution). These tests are also one-tailed only, and the critical region is found the same way as for the binomial.

Testing whether ρ = 0

For bivariate normal data, you can test whether there’s any linear correlation in the population. The population correlation coefficient is $\rho$; the sample value is $r$ (see scatter plots and correlation).

• $H_0: \rho = 0$ (no linear correlation in the population).
• $H_1: \rho \ne 0$, or $\rho \gt 0$, or $\rho \lt 0$ (decide before looking at the data).

Enter the data in two lists and use the GDC’s linear regression t-test (for example LinRegTTest), choosing the alternative hypothesis. It gives $r$ and the p-value. In exams, the data will be given.

Type I and Type II errors

	$H_0$ true	$H_0$ false
Reject $H_0$	Type I error	correct
Don’t reject $H_0$	correct	Type II error

• Type I error: rejecting $H_0$ when it is true. For a continuous test statistic (normal), $P(\text{Type I}) =$ the significance level. For a binomial or Poisson test, $P(\text{Type I}) = P(X \text{ in critical region} \mid H_0)$, which is a bit less than the significance level.
• Type II error: not rejecting $H_0$ when it is false. To calculate it, you need a specific true value of the parameter: $P(\text{Type II}) = P(X \text{ not in critical region} \mid \text{that true value})$.

Lowering the significance level makes Type I errors less likely but Type II errors more likely. Taking a bigger sample reduces $P(\text{Type II})$ without changing the significance level.

Example 1: the critical region is $\bar{x} \gt 502.47$. Orange: $P(\text{Type I}) = 0.05$ if $\mu = 500$. Blue: $P(\text{Type II}) \approx 0.153$ if actually $\mu = 504$.

Worked examples

Example 1: A z-test with a critical region

A machine fills bags of flour with masses that are normally distributed with mean $500$ g and standard deviation $6$ g. After a repair, the manager suspects that the mean has increased. A random sample of $16$ bags has a mean mass of $503.1$ g. Assume the standard deviation is still $6$ g.

• (a) Find the critical region for $\bar{x}$ for a test at the $5\%$ level.
• (b) Carry out the test.
• (c) Find the probability of a Type II error if the true mean is actually $504$ g.

Solution.

(a) $H_0: \mu = 500$ and $H_1: \mu \gt 500$. Under $H_0$, $\bar{X} \sim N\left(500, \dfrac{6^2}{16}\right)$, with standard deviation $\dfrac{6}{4} = 1.5$.

The critical value $c$ has $P(\bar{X} \gt c) = 0.05$. Inverse normal with area $0.95$, $\mu = 500$, $\sigma = 1.5$:

$$c = 502.467\ldots$$

The critical region is $\bar{x} \gt 502.47$ g (to 2 d.p.).

(b) $503.1$ is in the critical region, so reject $H_0$. There is significant evidence at the $5\%$ level that the mean mass has increased.

Check with the p-value: $z = \dfrac{503.1 - 500}{1.5} = 2.07$ (3 s.f.), and $P(\bar{X} \ge 503.1) = 0.0194 \lt 0.05$. Same conclusion.

(c) A Type II error means $\bar{x}$ is not in the critical region even though $\mu = 504$. Now $\bar{X} \sim N(504, 1.5^2)$:

$$P(\text{Type II}) = P(\bar{X} \le 502.467\ldots) = 0.153 \text{ (3 s.f.)}$$

Example 2: A test for a proportion

A website says that $30\%$ of students at a school walk to school. A student thinks the true proportion is lower. She asks a random sample of $25$ students and $4$ of them walk.

• (a) Find the critical region for a test at the $5\%$ level, and state the probability of a Type I error.
• (b) Carry out the test.

Solution.

(a) Let $X$ be the number who walk. $H_0: p = 0.3$ and $H_1: p \lt 0.3$. Under $H_0$, $X \sim B(25, 0.3)$. The critical region is $X \le k$; list the cumulative probabilities:

$k$	$2$	$3$	$4$
$P(X \le k)$	$0.00896$	$0.0332$	$0.0905$

$P(X \le 3) = 0.0332 \lt 0.05$ but $P(X \le 4) = 0.0905 \gt 0.05$, so the critical region is $X \le 3$. The probability of a Type I error is $P(X \le 3) = 0.0332$ (3 s.f.).

Under $H_0$, $X \sim B(25, 0.3)$. The orange bars ($X \le 3$) form the critical region, with total probability $0.0332 \lt 0.05$.

(b) The observed value $4$ is not in the critical region, so don’t reject $H_0$. There is insufficient evidence at the $5\%$ level that fewer than $30\%$ of students walk to school. (The p-value is $P(X \le 4) = 0.0905 \gt 0.05$, which agrees.)

Example 3: A test for a Poisson mean

A help desk has historically received calls at an average rate of $4$ per hour. After an advertising campaign, the manager thinks the rate has increased. In a $3$-hour period, $20$ calls are received.

• (a) Find the critical region for a test at the $5\%$ level, and carry out the test.
• (b) Find the probability of a Type I error.
• (c) The true rate is actually $6$ calls per hour. Find the probability of a Type II error.

Solution.

(a) Let $X$ be the number of calls in $3$ hours. $H_0: m = 12$ (that is, $4$ per hour) and $H_1: m \gt 12$. Under $H_0$, $X \sim \mathrm{Po}(12)$, and the critical region is $X \ge c$.

$c$	$18$	$19$	$20$
$P(X \ge c)$	$0.0630$	$0.0374$	$0.0213$

The critical region is $X \ge 19$. Since $20 \ge 19$, reject $H_0$: there is significant evidence at the $5\%$ level that the call rate has increased.

(b) $P(\text{Type I}) = P(X \ge 19 \mid m = 12) = 0.0374$ (3 s.f.)

(c) At $6$ calls per hour, the mean for $3$ hours is $18$, so $X \sim \mathrm{Po}(18)$. A Type II error means $X$ is not in the critical region:

$$P(\text{Type II}) = P(X \le 18 \mid m = 18) = 0.562 \text{ (3 s.f.)}$$

Even with a $50\%$ higher rate, this test misses the change more than half the time, because $3$ hours is a short period. A longer observation period would reduce this.

Example 4: Testing ρ = 0 with technology

A teacher records the number of hours $8$ randomly chosen students spent studying and their test scores. The data can be assumed to come from a bivariate normal distribution.

Hours, $x$	$2$	$5$	$1$	$8$	$4$	$6$	$3$	$7$
Score, $y$	$62$	$70$	$58$	$81$	$66$	$74$	$69$	$72$

Test at the $1\%$ level whether there is a positive correlation between study time and score.

Solution. $H_0: \rho = 0$ and $H_1: \rho \gt 0$.

GDC linear regression t-test (lists for $x$ and $y$, alternative $\rho \gt 0$): $r = 0.938$ (3 s.f.) and p-value $= 0.000286$ (3 s.f.).

Since $0.000286 \lt 0.01$, reject $H_0$. There is significant evidence at the $1\%$ level of a positive correlation between hours of study and test score in the population. (This is evidence of an association, not proof that studying causes higher scores.)

Common mistakes

Using the wrong mean in a Poisson test. If the rate is $4$ per hour and you observe for $3$ hours, $H_0$ is $m = 12$. Rescale before finding the critical region.

Choosing a discrete critical region that’s too big. The probability of the critical region must be less than the significance level. For $B(25, 0.3)$ at $5\%$, $X \le 4$ has probability $0.0905$, which is too large.

Saying P(Type I) is always the significance level. That’s only true for continuous tests. For binomial and Poisson tests, $P(\text{Type I})$ is the actual probability of the critical region under $H_0$, such as $0.0332$.

Finding a Type II error under H0. $P(\text{Type II})$ uses the true (alternative) parameter value, applied to the complement of the critical region. Find the critical region using $H_0$ first, then switch distributions.

Using a z-test when σ is unknown. If the standard deviation comes from the sample, use a t-test, even for large samples. For paired data, test the differences, not the two columns separately.

Concluding “H0 is true”. If you don’t reject $H_0$, say there is insufficient evidence for $H_1$. The data don’t prove $H_0$.

Practice

1. (Warm-up) A food inspector tests $H_0$: “the restaurant’s kitchen meets safety standards” against $H_1$: “it doesn’t”. Describe a Type I error and a Type II error in context.

Solution

Type I error: concluding that the kitchen doesn’t meet the standards when it actually does (rejecting a true $H_0$).

Type II error: concluding there’s not enough evidence of a problem when the kitchen actually fails the standards (not rejecting a false $H_0$).

2. (Warm-up) A population is normal with $\sigma = 10$. A test of $H_0: \mu = 80$ against $H_1: \mu \lt 80$ uses a sample of $25$ at the $1\%$ level. Find the critical region for $\bar{x}$.

Solution

Under $H_0$, $\bar{X} \sim N(80, 2^2)$, since $\dfrac{10}{\sqrt{25}} = 2$. The critical value has $P(\bar{X} \lt c) = 0.01$. Inverse normal:

$$c = 80 - 2.3263\ldots \times 2 = 75.347\ldots$$

The critical region is $\bar{x} \lt 75.3$ (3 s.f.).

3. (Warm-up) A coin is tossed $20$ times to test $H_0: p = 0.5$ against $H_1: p \gt 0.5$, where $p$ is the probability of heads, at the $5\%$ level. Find the critical region and the probability of a Type I error.

Solution

Under $H_0$, $X \sim B(20, 0.5)$. $P(X \ge 14) = 0.0577 \gt 0.05$ and $P(X \ge 15) = 0.0207 \lt 0.05$. The critical region is $X \ge 15$, and $P(\text{Type I}) = 0.0207$ (3 s.f.).

4. (Core) Bolts are made with lengths that are normally distributed with standard deviation $0.4$ mm. The target mean is $12.0$ mm. A sample of $10$ bolts has mean $12.25$ mm. Test at the $5\%$ level whether the mean length differs from $12.0$ mm.

Solution

$H_0: \mu = 12.0$ and $H_1: \mu \ne 12.0$ (two-tailed). $\sigma$ is known, so use a z-test:

$$z = \frac{12.25 - 12.0}{0.4/\sqrt{10}} = 1.976\ldots$$

The critical values are $\pm 1.960$. Since $1.976 \gt 1.960$, $z$ is in the critical region, so reject $H_0$. (p-value $= 0.0481 \lt 0.05$.) There is significant evidence at the $5\%$ level that the mean bolt length is not $12.0$ mm.

5. (Core) Eight adults had their systolic blood pressure (mmHg) measured before and after a $6$-week exercise program.

Adult	A	B	C	D	E	F	G	H
Before	$142$	$150$	$138$	$155$	$147$	$160$	$145$	$152$
After	$138$	$147$	$139$	$149$	$141$	$155$	$144$	$146$

Assuming the differences are normally distributed, test at the $5\%$ level whether the program reduces blood pressure.

Solution

This is paired data, so use the differences $d = \text{before} - \text{after}$:

$$4,\ 3,\ -1,\ 6,\ 6,\ 5,\ 1,\ 6$$

$H_0: \mu_d = 0$ and $H_1: \mu_d \gt 0$. $\sigma$ is unknown, so use a one-sample t-test on the differences ($\bar{d} = 3.75$, $s_{n-1} = 2.6049\ldots$, $7$ degrees of freedom).

GDC t-test: $t = 4.07$ and p-value $= 0.00237$ (3 s.f.).

Since $0.00237 \lt 0.05$, reject $H_0$. There is significant evidence at the $5\%$ level that the program reduces systolic blood pressure on average.

6. (Core) A highway had an average of $2.5$ accidents per month. After a new speed limit, there were $3$ accidents in the next $3$ months. Test at the $5\%$ level whether the accident rate has decreased, and state the probability of a Type I error.

Solution

Let $X$ be the number of accidents in $3$ months. $H_0: m = 7.5$ and $H_1: m \lt 7.5$. Under $H_0$, $X \sim \mathrm{Po}(7.5)$.

$P(X \le 2) = 0.0203 \lt 0.05$ and $P(X \le 3) = 0.0591 \gt 0.05$, so the critical region is $X \le 2$.

The observed value $3$ is not in the critical region, so don’t reject $H_0$. There is insufficient evidence at the $5\%$ level that the accident rate has decreased.

$P(\text{Type I}) = P(X \le 2) = 0.0203$ (3 s.f.).

7. (Core) A biologist measures the water temperature $x$ (°C) and a fish activity score $y$ at $7$ sites. The data can be assumed to be bivariate normal.

$x$	$1.2$	$2.0$	$2.8$	$3.5$	$4.1$	$5.0$	$5.6$
$y$	$31$	$29$	$33$	$30$	$35$	$32$	$36$

• (a) Test $H_0: \rho = 0$ against $H_1: \rho \ne 0$ at the $5\%$ level.
• (b) Another biologist had decided before collecting the data to test $H_1: \rho \gt 0$. What would her p-value and conclusion be?

Solution

(a) GDC linear regression t-test ($\rho \ne 0$): $r = 0.673$ and p-value $= 0.0976$ (3 s.f.). Since $0.0976 \gt 0.05$, don’t reject $H_0$: there is insufficient evidence at the $5\%$ level of a linear correlation.

(b) With $H_1: \rho \gt 0$, the p-value is half as big: $0.0488$ (3 s.f.). Since $0.0488 \lt 0.05$, she would reject $H_0$ and conclude there is significant evidence of a positive correlation. This is why the alternative hypothesis must be chosen before seeing the data.

8. (Challenge) In question 3, suppose the coin is actually biased with $p = 0.7$. Find the probability of a Type II error.

Solution

The critical region is $X \ge 15$. A Type II error means $X \le 14$ when $X \sim B(20, 0.7)$:

$$P(\text{Type II}) = P(X \le 14) = 0.584 \text{ (3 s.f.)}$$

With only $20$ tosses, the test misses this bias more often than not.

9. (Challenge) IQ-style test scores are normally distributed with $\sigma = 15$. A researcher tests $H_0: \mu = 100$ against $H_1: \mu \gt 100$ at the $5\%$ level.

• (a) With a sample of $36$, find the critical region for $\bar{x}$ and the probability of a Type II error if the true mean is $106$.
• (b) Find the smallest sample size for which the probability of a Type II error is less than $0.1$ when the true mean is $106$.

Solution

(a) Under $H_0$, $\bar{X} \sim N\left(100, 2.5^2\right)$, since $\dfrac{15}{\sqrt{36}} = 2.5$. Critical value: $c = 100 + 1.6448\ldots \times 2.5 = 104.11\ldots$, so the critical region is $\bar{x} \gt 104.1$ (to 1 d.p.).

If $\mu = 106$: $P(\text{Type II}) = P(\bar{X} \le 104.11\ldots)$ with $\bar{X} \sim N(106, 2.5^2)$, which is $0.225$ (3 s.f.).

(b) The critical value is $c = 100 + 1.6448\ldots \times \dfrac{15}{\sqrt{n}}$. We need $P(\bar{X} \le c) \lt 0.1$ when $\mu = 106$, so $c$ must be more than $1.2815\ldots$ standard deviations below $106$:

$$\begin{aligned} 100 + 1.6448\ldots \frac{15}{\sqrt{n}} &\lt 106 - 1.2815\ldots \frac{15}{\sqrt{n}} \\ (1.6448\ldots + 1.2815\ldots)\frac{15}{\sqrt{n}} &\lt 6 \\ \sqrt{n} &\gt 7.3159\ldots \\ n &\gt 53.52\ldots \end{aligned}$$

The smallest sample size is $n = 54$.