Significance tests for a mean

A pizza chain advertises a mean delivery time of 30 minutes. A group of customers suspects the real mean is longer. A confidence interval tells you what values of the mean are plausible; a significance test answers a sharper question: do the data give convincing evidence against a specific claimed value? This lesson carries the testing logic you used for proportions over to means.

Hypotheses about a mean

A significance test starts with two competing claims about the parameter $\mu$.

• The null hypothesis $H_0: \mu = \mu_0$ is the "nothing unusual" claim, where $\mu_0$ is the claimed or historical value.
• The alternative hypothesis $H_a$ is what you suspect or hope to find evidence for. It is one of $\mu > \mu_0$, $\mu < \mu_0$ or $\mu \ne \mu_0$.

Hypotheses are always about the population parameter $\mu$, never about $\bar{x}$. And the direction of $H_a$ must come from the research question, decided before you look at the data. For the pizza chain, $H_0: \mu = 30$ and $H_a: \mu > 30$, where $\mu$ is the true mean delivery time in minutes.

The test statistic

The test asks: if $H_0$ were true, how surprising would a sample mean like ours be? You measure "how far" in standard errors.

• For $H_a: \mu > \mu_0$, the P-value is the area to the right of $t$.
• For $H_a: \mu < \mu_0$, it is the area to the left of $t$.
• For $H_a: \mu \ne \mu_0$, it is twice the area in the tail beyond $|t|$.

The conditions are the same as for the one-sample $t$ interval: Random, 10% and Normal/Large Sample ($n \ge 30$, or a graph of the data shows no strong skewness or outliers).

Making a conclusion

Compare the P-value to the significance level $\alpha$ (usually $0.05$ unless stated otherwise).

• If P-value $\le \alpha$: reject $H_0$. There is convincing evidence for $H_a$.
• If P-value $> \alpha$: fail to reject $H_0$. There is not convincing evidence for $H_a$.

Always write the conclusion in context, and always refer to the alternative hypothesis.

A complete test: State, Plan, Do, Conclude

Here is a table excerpt for this lesson's examples and practice.

$df$	tail 0.10	tail 0.05	tail 0.025	tail 0.01	tail 0.005
14	1.345	1.761	2.145	2.624	2.977
19	1.328	1.729	2.093	2.539	2.861
35	1.306	1.690	2.030	2.438	2.724
39	1.304	1.685	2.023	2.426	2.708

Interpreting a P-value

A P-value is a conditional probability. For the pizza example: "Assuming the true mean delivery time is 30 minutes, there is about a $0.0188$ probability of getting a sample mean of $32.4$ minutes or more, just by chance, in a random sample of $20$ deliveries."

That's not the probability that $H_0$ is true. The calculation assumes $H_0$ is true.

Two-sided tests and confidence intervals

When the question asks whether the mean is different from a claimed value, use a two-sided alternative.

Errors in context

As with proportions, a test can be wrong in two ways.

• A Type I error is rejecting $H_0$ when it is actually true. Its probability is $\alpha$.
• A Type II error is failing to reject $H_0$ when $H_a$ is actually true.

For the soda machine, a Type I error would mean concluding the machine's mean fill differs from 12 ounces when it really is 12, so the company might stop production for no reason. A Type II error would mean missing a real problem with the machine. Larger samples reduce the chance of a Type II error, which increases the power of the test.

Practice

Practice 1

A random sample of $36$ students has a mean reaction-time score of $\bar{x} = 103.2$ with $s = 9.6$. Compute the test statistic for $H_0: \mu = 100$.

Your answer

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

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Practice 2

For the test in the previous problem, the alternative is $H_a: \mu > 100$ and $t = 2.0$ with $df = 35$. Find the P-value to 4 decimal places.

Your answer

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

CheckShow hint

Practice 3

A city says its average household water use is 280 gallons per day. An environmental group believes the true average is lower and collects data from a random sample of households. Which hypotheses should the group test?

A$H_0: \bar{x} = 280$, $H_a: \bar{x} < 280$B$H_0: \mu = 280$, $H_a: \mu < 280$C$H_0: \mu = 280$, $H_a: \mu \ne 280$D$H_0: \mu < 280$, $H_a: \mu = 280$

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Practice 4

In a two-sided test of $H_0: \mu = 50$ versus $H_a: \mu \ne 50$, a random sample of $20$ gives $t = -1.85$. Find the P-value to 4 decimal places.

Your answer

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

CheckShow hint

Practice 5

A one-sample $t$ test of $H_0: \mu = 8$ versus $H_a: \mu > 8$ gives a P-value of $0.031$. Which conclusion is correct at $\alpha = 0.05$?

AFail to reject $H_0$; there is convincing evidence that $\mu = 8$.BReject $H_0$; the data prove that $\mu > 8$.CFail to reject $H_0$; there is not convincing evidence that $\mu > 8$.DReject $H_0$; there is convincing evidence that $\mu > 8$.

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Practice 6

A 95% confidence interval for the mean time (in minutes) customers spend in a store is $(18.2,\ 23.6)$. Based only on this interval, what can you conclude about a two-sided test of $H_0: \mu = 25$ at $\alpha = 0.05$?

AReject $H_0$, because 25 is not in the interval.BFail to reject $H_0$, because 25 is not in the interval.CReject $H_0$, because the interval is wider than 5 minutes.DNothing, because tests and intervals answer unrelated questions.

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Practice 7

A sleep researcher claims teens average $8$ hours of sleep on school nights. A random sample of $15$ teens averages $\bar{x} = 7.3$ hours with $s = 1.1$ hours, and a dotplot shows no strong skew or outliers. For $H_0: \mu = 8$ versus $H_a: \mu < 8$, find the $t$ statistic to 2 decimal places.

Your answer

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

CheckShow hint

Practice 8

Continue the sleep study from the previous problem ($t \approx -2.46$, $df = 14$). What is the conclusion at $\alpha = 0.01$?

AReject $H_0$, because the P-value is about $0.027$, which is less than $0.01$.BReject $H_0$, because $t$ is negative.CFail to reject $H_0$; the P-value is about $0.014$, which is greater than $0.01$, so there is not convincing evidence that teens average less than 8 hours.DFail to reject $H_0$; the P-value is about $0.014$, so the data show that teens average exactly 8 hours.

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