Hypothesis test for a mean | CAIE 9709 Probability & Statistics 2 Notes

Chapters

Probability & Statistics 2

Mathematics9709

Write hypotheses about the population mean

Put the claimed value in $H_0:\mu=\mu_0$ .
The words greater, lower or different choose $H_1:\mu>\mu_0$ , $H_1:\mu<\mu_0$ or $H_1:\mu\ne\mu_0$ .
Hypotheses use the fixed parameter $\mu$ , not the observed statistic $\bar x$ .

Is the sample mean too high?

μ₀ = 100

critical 103.29

observed 104

104 lies beyond the critical value, inside the rejection region.

The tail is chosen by the alternative hypothesis; the sample mean is then compared with that tail.

Key idea

The alternative hypothesis chooses the tail before the sample result is judged. Never choose the direction because of the observed value.

Measure the gap in standard errors

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

Symbols

$\bar x$ = observed sample mean (same as x)
$\mu_0$ = mean stated by H-zero (same as x)
$\sigma/\sqrt n$ = standard error (same as x)

The observations must come from a random sample.
If the population is normal and σ is known, this z model is exact.
For a large random sample with unknown σ, estimate it using s and use $s/\sqrt n$ .
At 5%, the one-tail critical magnitude is 1.645 and the two-tail magnitude is 1.960. At 1%, the one-tail magnitude is 2.326.

Examiner note

Calculator route: Normal CDF under

H_0

gives the p-value directly. Write the standard error and z-value as evidence for the method marks.

Make a decision, then conclude in context

Worked example

Current-paper large-sample mean test

A random sample of 60 journey times has $\sum t=2750$ and $\sum t^2=127000$ . Test a passenger's claim that the population mean exceeds 45 minutes at 5%.

\bar t=2750/60=45.833

\frac{2750^2}{60}=126041.667

s^2=\frac{127000-126041.667}{59}

Hence $s^2=2875/177=16.2$ to 3 significant figures.

H_0:\mu=45,\qquad H_1:\mu>45

z=\frac{2750/60-45}{\sqrt{(2875/177)/60}}

z=1.602

Since $1.602<1.645$ , do not reject $H_0$ .

Conclusion: There is insufficient evidence, at the 5% level, that the population mean journey time exceeds 45 minutes. (9709/61/M/J/25 Q3)

Your turnIndependent mean-test check [5 marks]

A normal population has known standard deviation 12. A random sample of 36 has mean 104. Test at the 2.5% level whether the population mean is greater than 100.

Show worked answer

H_0:\mu=100,\qquad H_1:\mu>100

z=\frac{104-100}{12/\sqrt{36}}=2.00

The upper 2.5% critical value is 1.960. Since 2.00 exceeds 1.960, reject $H_0$ .

Conclusion: There is sufficient evidence that the population mean is greater than 100.

Common mistake

“Do not reject” does not prove that

H_0

is true. It means the sample does not provide sufficient evidence for

H_1