Hypothesis tests: correlation and Normal mean – Year 13 A level maths revision notes to download as PDF

Worked example 1

Ten pupils' revision hours and scores give \(r=0.72\). Test at 5% for positive correlation.

\(H_0:\rho=0\), \(H_1:\rho>0\). The critical value is \(0.5494\). Since \(0.72>0.5494\), reject \(H_0\): there is sufficient evidence of positive correlation between revision time and score.

Worked example 2

Two-tailed, 5%, \(n=10\), \(r=0.62\). Then negative: \(n=12\), \(r=-0.55\), \(H_1:\rho<0\).

The two-tailed critical value is \(0.6319\) and \(0.62<0.6319\), so there is insufficient evidence to reject \(H_0\). For the negative case the critical value is \(-0.4973\) and \(-0.55<-0.4973\), so reject \(H_0\). (With \(S_{xy}=14.5\), \(S_{xx}=S_{yy}=17.5\), \(r=0.829\).)

Worked example 3

\(X\sim N(80,12^2)\), \(n=9\). Find \(P(\bar X>84)\).

\(\bar X\sim N(80,4^2)\), so \(P(Z>1)=0.1587\).

Worked example 4

Packets should have mean 500 g, standard deviation 8 g. A sample of 16 has mean 496.5 g. Test at 5% whether the mean is lower.

\(H_0:\mu=500\), \(H_1:\mu<500\). Under \(H_0\), \(\bar X\sim N(500,2^2)\). \(P(\bar X\le496.5)=P(Z\le-1.75)=0.0401<0.05\), so reject \(H_0\): there is evidence the mean mass is less than 500 g.

Worked example 5

\(n=25\), \(\sigma=10\), \(H_0:\mu=50\), \(H_1:\mu>50\), 5%. Find the critical region.

\(\bar X\sim N(50,2^2)\); critical value \(50+1.645\times2=53.29\). The critical region is \(\bar x\ge53.3\), and a sample mean of 54 lies inside it.

Worked example 6

\(n=36\), \(\sigma=15\), \(H_0:\mu=100\), \(H_1:\mu\ne100\), sample mean 104.5.

Standard error 2.5, \(z=1.8\); two-tailed \(p\)-value \(2\times0.0359=0.0719\). At 5% do not reject \(H_0\); at 10% reject. For a two-tailed test at 5% split the level (2.5% in each tail, critical \(z=\pm1.96\)).

Worked example 7: A one-tailed test of correlation with a calculator

The heights and arm spans (cm) of 6 students are given. Test, at the 5% level, for positive correlation if \(r=0.93\) and the critical value for \(n=6\) is 0.7293.

\(H_0:\rho=0\), \(H_1:\rho>0\), where \(\rho\) is the population correlation between height and arm span. Since \(0.93>0.7293\) we reject \(H_0\). There is sufficient evidence, at the 5% level, of positive correlation between height and arm span. Notice that with only 6 points the critical value is large, because a small sample can show a strong correlation by chance.

Worked example 8: Standard error and sample size

\(X\sim N(100,20^2)\). Compare the standard deviation of \(\bar X\) for samples of size 4, 25 and 100.

\(\sigma_{\bar X}=\dfrac{20}{\sqrt n}\). For \(n=4\): \(10\). For \(n=25\): \(4\). For \(n=100\): \(2\). To halve the standard deviation of the sample mean you must multiply the sample size by 4. A larger sample gives a sample mean that is more tightly clustered around \(\mu\).

Worked example 9: Decision using a critical region for a lower tail

\(X\sim N(\mu,5^2)\), \(n=25\), test \(H_0:\mu=60\) against \(H_1:\mu<60\) at the 5% level. Find the critical region and decide for \(\bar x=58.7\).

\(\bar X\sim N(60,1^2)\). Critical value \(=60-1.6449\times1=58.355\). Critical region: \(\bar x\le58.36\). Since \(58.7>58.36\), we do not reject \(H_0\): there is insufficient evidence at the 5% level that the mean is less than 60.

Worked example 10: A two-tailed test using the p-value

\(\sigma=8\), \(n=64\), \(H_0:\mu=70\), \(H_1:\mu\ne70\), \(\bar x=71.8\). Test at the 5% level.

\(\bar X\sim N(70,1^2)\) since \(8/\sqrt{64}=1\). \(z=1.8\). \(P(\bar X>71.8)=0.0359\), so the two-tailed \(p\)-value is \(0.0719\). Since \(0.0719>0.05\) we do not reject \(H_0\). (For a one-tailed test with \(H_1:\mu>70\) the \(p\)-value 0.0359 would lead to rejection: the form of \(H_1\) matters.)

Worked example 11: A test written out in full

A bakery claims its loaves have mean mass 800 g with standard deviation 20 g. A customer weighs 10 loaves and finds a mean of 787 g. Test the claim that the mean is lower, at 5%.

Let \(\mu\) be the mean mass of a loaf. \(H_0:\mu=800\), \(H_1:\mu<800\). Under \(H_0\), \(\bar X\sim N\left(800,\dfrac{400}{10}\right)=N(800,6.325^2)\). \(z=\dfrac{787-800}{6.325}=-2.055\) and \(p=P(Z<-2.055)=0.0199\). Since \(0.0199<0.05\) we reject \(H_0\). There is sufficient evidence at the 5% level to suggest that the mean mass of the bakery loaves is less than 800 g. Notice the four steps: hypotheses, distribution, comparison, conclusion in context.

Worked example 12: Choosing between one-tailed and two-tailed

For each claim, state whether a one-tailed or two-tailed test is needed: (a) "the new recipe makes the biscuits heavier"; (b) "the mean has changed since the machine was serviced"; (c) "there is a negative correlation between age and speed".

(a) One-tailed, \(H_1:\mu>\mu_0\), because the claim gives a direction. (b) Two-tailed, \(H_1:\mu\ne\mu_0\), because "changed" allows an increase or a decrease. (c) One-tailed, \(H_1:\rho<0\). The wording of the question, not the data, decides. A common mistake is to look at the sample, see that the mean is higher, and then pick a one-tailed test: this makes the result look more significant than it is.

Worked example 13: Using a sample to estimate how often a result happens

A machine produces rods of mean length 20.0 cm and standard deviation 0.3 cm. Find the probability that a sample mean of 9 rods is more than 20.2 cm, and the probability that a single rod is more than 20.2 cm.

For a single rod, \(P(X>20.2)=P(Z>0.667)=0.2525\). For the mean of 9 rods, the standard error is \(0.3/3=0.1\), so \(P(\bar X>20.2)=P(Z>2)=0.0228\). The same difference from the mean is far less likely for a sample mean than for a single value, which is why large samples give good evidence.