Tails, normal approximations and mean tests | CAIE 9709 Probability & Statistics 2 Notes

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Probability & Statistics 2

Mathematics9709

Translate the claim into the sign of H₁

The sign in H₁ chooses the tail

H₁: θ < θ₀

← lower tail

decrease

H₁: θ ≠ θ₀

← both tails →

different or changed

H₁: θ > θ₀

upper tail →

increase

Two-tailed level α normally means α/2 in each tail.

Read H₁ before touching the calculator: its sign selects the tail or tails.

$\theta$ stands for the parameter being tested: often $p$ , $\lambda$ or $\mu$ .
“lower”, “decreased” or “fewer” gives $H_1:\theta<\theta_0$ and a lower tail.
“higher”, “increased” or “more” gives $H_1:\theta>\theta_0$ and an upper tail.
“different”, “changed” or “incorrect” gives $H_1:\theta\ne\theta_0$ and two tails.

Key idea

In a two-tailed CAIE test, compare the relevant one-sided tail with

\alpha/2

. At a 5% level, each tail normally receives 2.5%.

Approximate a large discrete count with continuity correction

Under $H_0$ , replace a suitable binomial or Poisson distribution by a normal distribution. All parameters must come from the null model.

Y\sim\mathrm{N}(\mu,\sigma^2)

Here the continuous variable Y approximates the discrete count X, and $p_0$ or $\lambda_0$ is the value stated by $H_0$ .

Binomial: $\mu=np_0$ and $\sigma^2=np_0(1-p_0)$ .

Poisson: $\mu=\sigma^2=\lambda_0$ .

For a binomial approximation, check that $np_0>5$ and $n(1-p_0)>5$ .
Move a discrete boundary by 0.5 before standardising: $X\le78$ becomes $Y<78.5$ .
For an upper tail $X\ge r$ , use the boundary $r-0.5$ .

Worked example

Large binomial, lower-tailed test

Test $H_0:p=0.25$ against $H_1:p<0.25$ when 78 successes are observed in 400 trials, at the 5% level.

Y\sim\mathrm{N}(100,75)

z=\frac{78.5-100}{\sqrt{75}}=-2.48

P(Y<78.5)\approx0.0065

Answer

\text{Reject }H_0.

Conclusion: There is sufficient evidence that the success probability is below 0.25.

Common mistake

Do not use the observed sample proportion to build the null mean and variance. A hypothesis test asks what would happen if the claimed parameter were true.

For a population mean, standardise the sample mean

If the population is normal with known variance, or the sample is large, the sample mean has standard error $\sigma/\sqrt n$ .

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

With a large sample and unknown variance, use the unbiased estimate $s^2$ in place of $\sigma^2$ .
Use the critical z-value for the stated tail and level: for example, 1.645 for an upper 5% test and 1.960 for an upper 2.5% test.
A two-tailed 5% test has 2.5% in each tail, so it rejects for $z<-1.960$ or $z>1.960$ .
A normal population makes $\bar X$ normal directly; the Central Limit Theorem is not needed in that case.

Worked example

Current-paper mean test

An inspector tests $H_0:\mu=10$ against $H_1:\mu>10$ using a random sample of 50 lengths from a normal population. The sample mean is 10.03 and the standardised test statistic is 1.995.

1.995>1.960

Decision: Reject $H_0$ at the 2.5% level.

Conclusion: There is sufficient evidence that the mean length is greater than 10 cm.

(9709/63/M/J/25 Q2)

Your turnPopulation mean test [5 marks]

A population is normal with standard deviation 4. A random sample of 64 has mean 51.2. Test at the 5% level whether the population mean is greater than 50.

Show worked answer

H_0:\mu=50,\qquad H_1:\mu>50

z=\frac{51.2-50}{4/\sqrt{64}}=2.40

2.40>1.645

Decision: Reject $H_0$ .

Conclusion: There is sufficient evidence that the population mean is greater than 50.

Translate the claim into the sign of H₁

The sign in H₁ chooses the tail

H₁: θ < θ₀

← lower tail

decrease

H₁: θ ≠ θ₀

← both tails →

different or changed

H₁: θ > θ₀

upper tail →

increase

Two-tailed level α normally means α/2 in each tail.

Read H₁ before touching the calculator: its sign selects the tail or tails.

$\theta$ stands for the parameter being tested: often $p$ , $\lambda$ or $\mu$ .
“lower”, “decreased” or “fewer” gives $H_1:\theta<\theta_0$ and a lower tail.
“higher”, “increased” or “more” gives $H_1:\theta>\theta_0$ and an upper tail.
“different”, “changed” or “incorrect” gives $H_1:\theta\ne\theta_0$ and two tails.

Key idea

In a two-tailed CAIE test, compare the relevant one-sided tail with

\alpha/2

. At a 5% level, each tail normally receives 2.5%.

Approximate a large discrete count with continuity correction

Under $H_0$ , replace a suitable binomial or Poisson distribution by a normal distribution. All parameters must come from the null model.

Y\sim\mathrm{N}(\mu,\sigma^2)

Here the continuous variable Y approximates the discrete count X, and $p_0$ or $\lambda_0$ is the value stated by $H_0$ .

Binomial: $\mu=np_0$ and $\sigma^2=np_0(1-p_0)$ .

Poisson: $\mu=\sigma^2=\lambda_0$ .

For a binomial approximation, check that $np_0>5$ and $n(1-p_0)>5$ .
Move a discrete boundary by 0.5 before standardising: $X\le78$ becomes $Y<78.5$ .
For an upper tail $X\ge r$ , use the boundary $r-0.5$ .

Worked example

Large binomial, lower-tailed test

Test $H_0:p=0.25$ against $H_1:p<0.25$ when 78 successes are observed in 400 trials, at the 5% level.

Y\sim\mathrm{N}(100,75)

z=\frac{78.5-100}{\sqrt{75}}=-2.48

P(Y<78.5)\approx0.0065

Answer

\text{Reject }H_0.

Conclusion: There is sufficient evidence that the success probability is below 0.25.

Common mistake

Do not use the observed sample proportion to build the null mean and variance. A hypothesis test asks what would happen if the claimed parameter were true.

For a population mean, standardise the sample mean

If the population is normal with known variance, or the sample is large, the sample mean has standard error $\sigma/\sqrt n$ .

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

With a large sample and unknown variance, use the unbiased estimate $s^2$ in place of $\sigma^2$ .
Use the critical z-value for the stated tail and level: for example, 1.645 for an upper 5% test and 1.960 for an upper 2.5% test.
A two-tailed 5% test has 2.5% in each tail, so it rejects for $z<-1.960$ or $z>1.960$ .
A normal population makes $\bar X$ normal directly; the Central Limit Theorem is not needed in that case.

Worked example

Current-paper mean test

An inspector tests $H_0:\mu=10$ against $H_1:\mu>10$ using a random sample of 50 lengths from a normal population. The sample mean is 10.03 and the standardised test statistic is 1.995.

1.995>1.960

Decision: Reject $H_0$ at the 2.5% level.

Conclusion: There is sufficient evidence that the mean length is greater than 10 cm.

(9709/63/M/J/25 Q2)

Your turnPopulation mean test [5 marks]

A population is normal with standard deviation 4. A random sample of 64 has mean 51.2. Test at the 5% level whether the population mean is greater than 50.

Show worked answer

H_0:\mu=50,\qquad H_1:\mu>50

z=\frac{51.2-50}{4/\sqrt{64}}=2.40

2.40>1.645

Decision: Reject $H_0$ .

Conclusion: There is sufficient evidence that the population mean is greater than 50.