Practise Tests for correlation and for the mean of a normal distribution. 3 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy2 marks
A researcher collects data on the time spent on a phone each evening and the number of hours slept that night for a random sample of teenagers. The researcher wants to test whether there is negative correlation between the time spent on a phone each evening and the number of hours slept that night.
Write down suitable hypotheses for the test, defining the parameter you use.[2]
Show the answer and mark scheme
Answer: \(H_0: \rho = 0, \ H_1: \rho \lt 0\), where \(\rho\) is the population product moment correlation coefficient between the time spent on a phone each evening and the number of hours slept that night.
B1 for \(H_0: \rho = 0\) with \(\rho\) the population correlation coefficient
B1 for \(H_1: \rho \lt 0\)
Question 2Medium7 marks
A teacher records the number of hours of revision and the test mark for a random sample of 18 students. The product moment correlation coefficient between hours of revision and test mark is \(r = 0.42\). Critical values of the product moment correlation coefficient for a sample of size 18 are 0.4000 (one-tailed test at 5%) and 0.4683 (one-tailed test at 2.5%, or two-tailed test at 5%).
(a) Test, at the 5% level of significance, whether there is evidence of positive correlation between hours of revision and test mark. State your hypotheses clearly.[4]
(b) A second teacher tests the same data at the 5% level for evidence of correlation, without specifying positive or negative. State the hypotheses and the conclusion of this test.[2]
(c) Explain why the two conclusions do not contradict each other.[1]
Show the answer and mark scheme
(a)Answer: \(H_0: \rho = 0\), \(H_1: \rho \gt 0\); \(0.42 \gt 0.4000\), so reject \(H_0\): there is evidence of positive correlation between hours of revision and test mark.
B1 for \(H_0: \rho = 0\), \(H_1: \rho \gt 0\)
M1 for comparing 0.42 with 0.4000
A1 for reject \(H_0\) (significant)
A1 for a conclusion in context
Worked solution: \(H_0: \rho = 0\), \(H_1: \rho \gt 0\), where \(\rho\) is the population correlation coefficient. Critical value 0.4000. Since \(0.42 \gt 0.4000\), reject \(H_0\): there is evidence of positive correlation between hours of revision and test mark.
(b)Answer: \(H_0: \rho = 0\), \(H_1: \rho \ne 0\); \(0.42 \lt 0.4683\), so do not reject \(H_0\): insufficient evidence of correlation.
B1 for \(H_1: \rho \ne 0\)
B1 for \(0.42 \lt 0.4683\), not significant: insufficient evidence of correlation between hours of revision and test mark
Worked solution: \(H_0: \rho = 0\), \(H_1: \rho \ne 0\). Critical value 0.4683 (two-tailed, 5%). Since \(0.42 \lt 0.4683\), do not reject \(H_0\): there is insufficient evidence of correlation.
(c)Answer: They test different alternative hypotheses: a one-tailed test puts all 5% in one tail, so it needs weaker evidence in that direction. The result is borderline, significant for one test but not the other.
B1 for the tests have different alternative hypotheses (one-tailed puts all of the 5% in the upper tail, so its critical value is lower); the evidence here is borderline
Worked solution: The first test only looks for positive correlation and puts the whole 5% in the upper tail, giving a lower critical value. The second allows for correlation in either direction, so it needs stronger evidence in each direction. With a borderline value such as 0.42, one test can be significant and the other not.
Question 3Hard9 marks
A machine fills bottles with water. The volume, in ml, of water in a bottle is normally distributed with mean \(\mu\) and standard deviation 4 ml. The machine is set so that \(\mu = 500\). After the machine is serviced, a random sample of 25 bottles has a mean volume of 498.3 ml.
(a) Test, at the 5% level of significance, whether there is evidence that the mean volume has decreased. State your hypotheses clearly.[5]
(b) A student writes: “The test proves that the machine now under-fills, and its new mean volume is 498.3 ml.” Give two criticisms of this statement.[2]
(c) Explain whether the conclusion would be different if the test had been two-tailed at the 1% level of significance.[2]
Show the answer and mark scheme
(a)Answer: \(H_0: \mu = 500\), \(H_1: \mu \lt 500\); \(\bar{X} \sim \mathrm{N}(500, 0.8^2)\), \(P(\bar{X} \le 498.3) = 0.0168 \lt 0.05\): reject \(H_0\). There is evidence that the mean volume of water in the bottles has decreased.
B1 for \(H_0: \mu = 500\), \(H_1: \mu \lt 500\)
M1 for \(\bar{X} \sim \mathrm{N}\left(500, \frac{4^2}{25}\right)\)
M1 for \(P(\bar{X} \le 498.3)\) or \(z = \frac{498.3 - 500}{0.8} = -2.125\)
A1 for 0.0168 (or \(z = -2.125 \lt -1.6449\), or critical value 498.68)
A1 for a conclusion in context: reject \(H_0\); there is evidence that the mean volume has decreased
Worked solution: \(H_0: \mu = 500\), \(H_1: \mu \lt 500\). Under \(H_0\), \(\bar{X} \sim \mathrm{N}\left(500, \frac{16}{25}\right) = \mathrm{N}(500, 0.8^2)\). \(P(\bar{X} \le 498.3) = P(Z \le -2.125) = 0.0168 \lt 0.05\). Reject \(H_0\): there is evidence at the 5% level that the mean volume of water in the bottles has decreased.
(b)Answer: A test gives evidence, not proof (there is a chance of rejecting a true \(H_0\)); 498.3 ml is only the sample mean, an estimate of the new population mean.
B1 for a significant result is evidence, not proof: there is up to a 5% chance of rejecting \(H_0\) when it is true
B1 for 498.3 is the mean of this sample only; the population mean is unknown (498.3 is an estimate)
Worked solution: A hypothesis test cannot prove anything: with a 5% significance level there is a chance of getting such a sample even if \(\mu = 500\). Also 498.3 ml is the mean of this sample, which only estimates the new population mean; another sample would give a different value.
(c)Answer: Yes: two-tailed at 1% needs a probability below 0.005 in the lower tail (or \(z \lt -2.5758\)); \(0.0168 \gt 0.005\), so \(H_0\) would not be rejected.
M1 for comparing 0.0168 with 0.005 (or \(2 \times 0.0168 = 0.0336\) with 0.01, or \(-2.125\) with \(-2.5758\))
A1 for not significant: insufficient evidence that the mean volume has changed, so the conclusion would be different
Worked solution: For a two-tailed test at 1%, each tail has 0.5%. \(P(\bar{X} \le 498.3) = 0.0168 \gt 0.005\) (equivalently \(z = -2.125 \gt -2.576\)), so \(H_0\) would not be rejected: there would be insufficient evidence that the mean volume has changed.