O3StatisticsStretch

Hypothesis Tests for a Normal Population Mean

A sample mean varies from sample to sample. A test asks whether the observed mean is unusually far from a claimed population mean, allowing for that sampling variation. The key decision is to use the spread of sample means, not the spread of individual observations.

45 min Statistical Hypothesis Testing

The big picture

Imagine repeatedly taking 25 observations from the same normal population and plotting their averages. These averages still centre on the population mean, but their spread is only one fifth as large as the spread of individual values. That narrower reference distribution makes a moderate change in the mean detectable. The test measures how unusual the observed average would be if the null claim were true; it does not calculate the probability that the null claim is true.

What you'll be able to do

Use $\bar X\sim N(\mu,\sigma^2/n)$ for an independent random sample from a normal population
State hypotheses about the population mean and choose the tail before seeing the result
Use p-values or critical regions to reach a contextual conclusion
State the normal-population and known-variance assumptions required by AQA 7357

From individual values to the sample mean

If independent observations come from $X\sim N(\mu,\sigma^2)$ , their mean has distribution $\bar X\sim N(\mu,\sigma^2/n)$ . Its standard deviation, called the standard error, is $\sigma/\sqrt n$ .

For this AQA test, the population is normal and the population variance is known, given or assumed. A large sample by itself is not the justification expected here for testing a non-normal population. Do not replace the given population variance with an estimated sample variance and carry on with the same test.

Random sampling and independence matter: 25 measurements of the same object do not automatically behave like 25 independently sampled objects. Dependence can make the stated standard error too optimistic.

\bar X\sim N\left(\mu,\frac{\sigma^2}{n}\right),\qquad\operatorname{sd}(\bar X)=\frac{\sigma}{\sqrt n}

Divide the variance by n, or the standard deviation by the square root of n.

Example 1Bottle fills are normal with variance

36\text{ ml}^2

. What distribution does the mean of 9 independent fills have?

1The individual standard deviation is

\sqrt{36}=6

ml.

2The mean has variance

36/9=4\text{ ml}^2

and standard error

6/3=2

ml.

Answer

\bar X\sim N(\mu,4)

; the second parameter is the variance, not the standard deviation.

A one-tailed test, completely worked

Define the population parameter first. “The machine is underfilling” gives $H_1:\mu<\mu_0$ and a lower-tail test. Choose that alternative from the research question, not because the observed result happens to be low.

Assume $H_0$ and standardise using $z=\dfrac{\bar x-\mu_0}{\sigma/\sqrt n}$ . For a lower-tail test the p-value is $P(Z\leq z)$ ; for an upper-tail test it is $P(Z\geq z)$ .

Example 1A machine claims a mean fill of 500 ml. Fills are normal with known standard deviation 8 ml. A random sample of 16 gives mean 495 ml. Test for underfilling at the 5% level.

1Let

\mu

be the population mean fill.

H_0:\mu=500

H_1:\mu<500

2Under

H_0

\bar X\sim N(500,64/16)=N(500,4)

, so its standard error is 2 ml.

z=(495-500)/2=-2.5

. The lower-tail p-value is

P(Z\leq-2.5)=0.00621

0.00621<0.05

, so reject

H_0

AnswerThere is sufficient evidence at the 5% level that the population mean fill is below 500 ml.

Tip — Write the reference distribution and the probability inequality before using the calculator. A probability with the wrong tail answers a different test.

Two tails and a critical region

A claim that the mean has “changed” gives $H_1:\mu\neq\mu_0$ . For a symmetric normal test, the two-sided p-value is $2P(Z\geq |z|)$ . A test at significance level $\alpha$ assigns $\alpha/2$ to each tail.

Read the heading of any critical-value table. A table may already label its values by the total two-tailed level. Do not halve a level a second time. At 5% two-tailed, the standard normal critical values are approximately -1.960 and 1.960.

Example 1Rod lengths are normal with known standard deviation 4 mm. Test a claimed mean of 100 mm using a random sample of 25 with mean 101.2 mm, at 5% two-tailed.

H_0:\mu=100

H_1:\mu\neq100

. Under

H_0

, the standard error is

4/\sqrt{25}=0.8

mm.

z=(101.2-100)/0.8=1.5

. The two-sided p-value is

2P(Z>1.5)=0.1336

0.1336>0.05

, so do not reject

H_0

4Equivalently, critical sample means are

100\pm1.960(0.8)

, approximately

98.432

and

101.568

. The observed mean lies between them.

AnswerThere is insufficient evidence at 5% that the population mean length differs from 100 mm.

Conclusions, errors and assumptions

“Do not reject” means the data do not provide enough evidence against the null at this level. It does not prove the null, show the means are identical, or establish that the machine is accurate enough for its practical purpose.

A significance level controls the probability of rejecting a true null under the model. It is not the probability that the particular conclusion is wrong. A statistically significant change can still be too small to matter practically.

For an upper-tailed 5% test with standard error $s_e$ , the critical sample mean is $\mu_0+1.645s_e$ . Increasing $n$ reduces $s_e$ : the same change in the mean then produces a larger magnitude test statistic.

State the population, direction and significance level in the final sentence. Retain sufficient precision in intermediate probabilities, especially near a decision boundary.

Example 1A student uses the known standard deviation 10 directly in the denominator, even though the sample size is 100. What changes?

1The correct denominator is

10/\sqrt{100}=1

2The student’s test statistic has one tenth of the correct magnitude.

3The error makes an unusual sample mean appear much less unusual.

AnswerUse the standard error of the mean, not the standard deviation of an individual.

Think like an examiner

Define the population mean and write hypotheses about it, rather than the observed sample mean.
State normality, random independent sampling and known population variance as applicable.
Make the probability or critical-value comparison explicit.
Finish with evidence in context, allowing uncertainty.

Common misconceptions

Using

\sigma/n

as the standard error.

Use

\sigma/\sqrt n

; the variance is divided by

n

Testing a non-normal population solely because the sample is large.

AQA 7357 specifies this test for a normal population with known, given or assumed variance.

A p-value of 0.03 means a 3% chance that the null is true.

It is a probability of data at least as extreme, calculated assuming the null.

Key results to remember

z=\frac{\bar x-\mu_0}{\sigma/\sqrt n}

Null mean in the numerator; standard error in the denominator.

p_{\text{two-sided}}=2P(Z\geq |z|)

Two equally extreme tails for the symmetric normal test.

Stretch yourself

Normal packet masses have known standard deviation 6 g. Find the upper critical sample mean for a 1% upper-tail test of a mean of 250 g when n = 36.

Hint — Use the 0.99 standard normal quantile $2.3263$ .

Key takeaways

The mean has variance $\sigma^2/n$ and standard error $\sigma/\sqrt n$ .
Use the null population mean to calculate the reference distribution.
Choose tails from the question, and handle two-sided levels using the table heading.
A non-significant result leaves uncertainty; it does not prove equality.

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

Related lessons

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Hypothesis Tests for a Normal Population Mean

45 min Statistical Hypothesis Testing