Questions on hypothesis testing: correlation and the normal

Question 1

Observations are drawn independently from N(μ, σ²). State the distribution of the mean of n such observations.

Accepted Answer

The sample mean is normal with the same centre and a tighter spread, so X̄ ~ N(μ, σ²/n). B1 for the normal distribution with mean μ, B1 for the variance. In a mean, high and low observations partly offset each other, and the variance shrinks by a factor of n, so the standard deviation of the mean is σ/√n rather than σ/n.

Question 2

The masses, in grams, of items from a production line are normally distributed with mean μ and standard deviation 3. A test of H₀: μ = 50 against H₁: μ < 50 uses a random sample of 36 items, whose mean mass is 48.8 g. Find the value of the test statistic.

Accepted Answer

The standard error of the sample mean is 3/√36 = 0.5, so z = (48.8 − 50)/0.5 = −2.4. M1 for the standard error, M1 for the z formula, A1 for −2.4. The denominator is the standard error, not the population standard deviation. Dividing by 3 instead of 0.5 gives −0.4, which tests a single observation rather than the mean of 36.

Question 3

A random sample of 9 observations is taken from a population distributed as N(50, 6²). Write down the distribution of the sample mean X̄, and state its standard deviation.

Accepted Answer

Averaging leaves the centre where it was and divides the variance by n, so X̄ ~ N(50, 6²/9), which is N(50, 4). B1 for the distribution with both parameters. Its standard deviation is √4 = 2, which is 6/√9 and not 6/9. B1 for 2.

Question 4

Explain what is meant by the p-value of a hypothesis test, state the rule for comparing it with the significance level, and state what a p-value does not measure.

Accepted Answer

The p-value is the probability, calculated on the assumption that H₀ is true, of a result at least as extreme as the one observed. B1 for the definition with that assumption stated. If the p-value is smaller than the significance level, reject H₀; otherwise there is insufficient evidence against it. B1 for the comparison. It is not the probability that H₀ is true: H₀ is a statement about the population, and nothing in the calculation attaches a probability to it. B1. The conditioning runs from the hypothesis to the data, and reading it backwards is the same reversal as reading P(A|B) as P(B|A).

Question 5

The masses, in grams, of bags of flour are normally distributed with mean μ and standard deviation 4. A supplier claims that μ = 100. A random sample of 25 bags has mean mass 98.6 g. Test, at the 5% significance level, whether the mean mass is less than the supplier claims. State your hypotheses clearly.

Accepted Answer

H₀: μ = 100 and H₁: μ < 100. Under H₀, X̄ ~ N(100, 4²/25), so the standard error is 0.8 and z = (98.6 − 100)/0.8 = −1.75. Then P(Z ≤ −1.75) = 0.0401. Since 0.0401 < 0.05, reject H₀. There is significant evidence at the 5% level that the mean mass of the bags is below 100 g. The five marks are B1 hypotheses, M1 distribution of X̄, M1 probability, A1 0.0401, A1 comparison and conclusion in context. All three parts of the conclusion are needed: the comparison, the decision, and a sentence about flour rather than about z.

Question 6

Bags filled by a machine have masses, in grams, distributed as N(μ, 4²). A test of H₀: μ = 30 against H₁: μ > 30 at the 5% significance level uses the mean of a random sample of 16 bags. Find the critical region for the test, and state the conclusion when the sample mean is 31.2 g.

Accepted Answer

Under H₀, X̄ ~ N(30, 16/16), so the standard error is 1. A one-tailed 5% test rejects when z > 1.6449, that is when X̄ > 30 + 1.6449 × 1 = 31.6449 g. The critical region is X̄ > 31.6449. The observed mean 31.2 is well below that boundary, so it is not in the critical region. Do not reject H₀. There is insufficient evidence at the 5% level that the mean mass has risen. M1 for the standard error, B1 for z = 1.6449, M1 for the boundary, A1 for 31.6449, A1 for the conclusion. As a cross-check, z = 1.2 gives a tail probability of 0.115, comfortably above 0.05, and the two methods agree.
Carry the boundary at full accuracy. Rounding it to 31.6 hands the critical region every mean between 31.6 and 31.6449, and a sample mean of 31.62 would then be declared significant when its tail probability is 0.0526, above 0.05. A rounded boundary is what makes the critical-region and p-value methods appear to disagree. Round the boundary only for reporting, and only when the observed mean is nowhere near it.

Question 7

A random sample of 20 pairs of observations gives a product moment correlation coefficient of r = 0.62. The critical value from the table for a sample of size 20 at the 0.01 level is 0.5155. Test, at the 1% significance level, whether there is positive correlation between the two variables.

Accepted Answer

H₀: ρ = 0 and H₁: ρ > 0, where ρ is the population correlation coefficient. The test is one-tailed, so the 0.01 column of the table is the right one and the critical value is 0.5155. Since 0.62 > 0.5155, reject H₀. There is significant evidence at the 1% level of positive correlation between the two variables in the population. B1 for the hypotheses in terms of ρ, B1 for the critical value, M1 for the comparison, A1 for the conclusion in context.

Question 8

Explain why increasing the sample size makes a fixed difference between the sample mean and the hypothesised mean more likely to be judged significant.

Accepted Answer

The standard error σ/√n shrinks as n grows, so the same difference is measured against a tighter yardstick and the z value rises. B1 for the standard error shrinking, B1 for the effect on z or on the tail probability.

Question 9

The volume, in millilitres, of drink dispensed by a machine is normally distributed with mean μ and standard deviation 2.5. The machine is set to dispense 500 ml. A random sample of 20 cups has mean volume 501.6 ml. Test, at the 5% significance level, whether the mean volume dispensed has changed. State your hypotheses clearly.

Accepted Answer

H₀: μ = 500 and H₁: μ ≠ 500, where μ is the mean volume dispensed in ml. B1 for the hypotheses. Under H₀, X̄ ~ N(500, 2.5²/20), so the standard error is 2.5/√20 = 0.5590. M1 for the distribution of X̄ with its standard error. Then z = (501.6 − 500)/0.5590 = 2.862. M1 for the test statistic. The alternative names no direction, so the test is two-tailed and the p-value is 2 × P(Z > 2.862) = 2 × 0.00210 = 0.00421. A1 for 0.00421. Since 0.00421 < 0.05, reject H₀. There is significant evidence at the 5% level that the mean volume the machine dispenses is no longer 500 ml. A1 for the comparison and a conclusion written about the machine. Doubling the tail is what makes the test two-tailed: setting the single tail of 0.00210 against 0.05 reaches the same decision here by a route that fails whenever the result sits near the boundary.

Question 10

At each of 25 second-hand car sales the age of the car and its resale value are recorded, giving a product moment correlation coefficient of r = −0.42. The critical value from the table for a sample of size 25 at the 0.05 level is 0.3365. Test, at the 5% significance level, whether there is negative correlation between age and resale value.

Accepted Answer

H₀: ρ = 0 and H₁: ρ < 0, where ρ is the correlation coefficient between age and resale value in the population of such cars. B1 for the hypotheses written in terms of ρ. The alternative points downwards, so the test is one-tailed in the lower tail and the critical value is −0.3365. B1 for the critical value with its sign. Since −0.42 < −0.3365, the observed coefficient lies in the critical region, so reject H₀. M1 for the comparison. There is significant evidence at the 5% level of negative correlation between the age of a car and its resale value: older cars sell for less. A1 for the conclusion in context. The table is printed for the positive tail, so a one-tailed test in the negative direction uses the same figure carrying a minus sign; setting −0.42 against +0.3365 compares the coefficient with the wrong end of the distribution.

Question 11

The times taken, in seconds, by a machine to complete a task are distributed as N(μ, 1.5²). A test of H₀: μ = 12 against H₁: μ < 12 at the 1% significance level uses the mean of a random sample of 25 times. Find the critical region for the test.

Accepted Answer

Under H₀, X̄ ~ N(12, 1.5²/25), so the standard error is 1.5/√25 = 0.3. M1 for the standard error. A one-tailed 1% test in the lower tail rejects when z < −2.3263. B1 for that z value. The boundary is 12 − 2.3263 × 0.3 = 12 − 0.6979 = 11.302. M1 for forming the boundary. The critical region is X̄ < 11.302 seconds. A1 for 11.302 with the inequality pointing the right way. Carry the z value to four decimal places: 12 − 2.33 × 0.3 = 11.301 and 12 − 2.3 × 0.3 = 11.31, so a sample mean between 11.302 and 11.31 is classified differently by the three versions of one boundary.

Question 12

A student measures two variables on a random sample of 30 subjects and obtains a product moment correlation coefficient of r = 0.42. The table of critical values for a sample of size 30 gives 0.3061 at the 0.05 level, 0.3610 at the 0.025 level and 0.4226 at the 0.01 level. Test, at the 5% significance level, whether there is any correlation between the two variables, and state whether the conclusion would change at the 2% level.

Accepted Answer

H₀: ρ = 0 and H₁: ρ ≠ 0. The alternative hypothesis names no direction, so the test is two-tailed and the 5% is split between the two ends. The table is indexed by one-tail probability, so the row must be read at 0.025, giving critical values ±0.3610. Since 0.42 > 0.3610, reject H₀. There is significant evidence at the 5% level of correlation between the two variables. At the 2% level the relevant column is 0.01, where the critical value is 0.4226, and 0.42 < 0.4226, so H₀ would not be rejected and the conclusion changes. B1 hypotheses, B1 identifying the test as two-tailed, B1 reading the 0.025 column, A1 first conclusion, A1 second conclusion. Reaching for the 0.05 column on a two-tailed 5% test doubles the true significance level of the test.

Question 13

The lengths, in centimetres, of components are distributed as N(μ, 2²). A two-tailed test of H₀: μ = 50 at the 5% significance level uses the mean of a random sample of 16 components. Find the critical region for the test, and state, with a reason, whether a sample mean of 51.1 cm leads to rejection of H₀.

Accepted Answer

Under H₀, X̄ ~ N(50, 4/16), so the standard error is 0.5. A two-tailed 5% test puts 2.5% in each tail, so the critical z values are ±1.96 and the boundaries are 50 ± 1.96 × 0.5, that is 49.02 and 50.98. The critical region is X̄ < 49.02 or X̄ > 50.98. Since 51.1 > 50.98 it lies inside the critical region, so reject H₀. There is significant evidence at the 5% level that the mean length is not 50 cm. M1 standard error, B1 z = ±1.96, M1 forming a boundary, A1 both boundaries, A1 conclusion. Using 1.6449 here would give a critical region starting at 50.82, and 51.1 would still be rejected, so the right decision would arrive from the wrong critical value.

Question 14

The masses, in grams, of items from a production line are distributed as N(μ, 5²). A test of H₀: μ = 80 against H₁: μ > 80 at the 5% significance level uses the mean of a random sample of n items. Find the smallest value of n for which a sample mean of 81.5 g would lead to rejection of H₀.

Accepted Answer

Rejection needs the test statistic to reach the one-tailed 5% value, so (81.5 − 80)/(5/√n) > 1.6449. M1 for the inequality, B1 for z = 1.6449. Multiplying out, 1.5√n/5 > 1.6449, so √n > 5 × 1.6449/1.5 = 5.483. M1 for rearranging to √n, A1 for 5.483. Squaring gives n > 30.06, so the smallest whole number is n = 31. A1 for 30.06, A1 for n = 31. Check either side of that boundary: with n = 30 the standard error is 0.9129 and z = 1.643, just short of 1.6449, so H₀ stands; with n = 31 the standard error is 0.8980 and z = 1.670, which clears it. Rounding 30.06 down to 30 lands on the one sample size that fails.

Question 15

At each of 18 orchards the number of hours of sunshine in a season and the mass of fruit picked are recorded. The product moment correlation coefficient is r = 0.44, and statistical software reports a one-tailed p-value of 0.034. Test, at the 5% significance level, whether there is positive correlation between sunshine and yield, state whether the conclusion would change at the 1% level, and explain why a significant result does not establish that sunshine causes a higher yield.

Accepted Answer

H₀: ρ = 0 and H₁: ρ > 0, where ρ is the population correlation coefficient between hours of sunshine and mass of fruit picked. B1 for the hypotheses in terms of ρ. The alternative names a direction, so the one-tailed p-value is the one to use: 0.034 < 0.05. B1 for the comparison. Reject H₀: there is significant evidence at the 5% level of positive correlation between sunshine hours and the mass of fruit picked at these orchards. A1 for the conclusion in context. At the 1% level the comparison reads 0.034 > 0.01, so H₀ would not be rejected. B1 for the second comparison. The conclusion therefore changes: at the 1% level there is insufficient evidence of positive correlation. B1. Even at 5% the test says only that a coefficient this far from zero would be unusual if the population coefficient were zero. The orchards were not assigned their sunshine, so temperature, rainfall and soil vary along with it, and any of those could carry the effect. B1 for causation not following from the test.