Questions on comparing two normal means

Question 1

Two independent samples of size 25 come from populations with variances 50 and 30. Find the standard error of the difference of the sample means.

Accepted Answer

Each mean has variance σ²/n, so 50/25 = 2 and 30/25 = 1.2. Adding gives 3.2, so the standard error is √3.2 = 1.789 to three decimal places. M1 for adding the two variances of the means, A1 for the standard error. The variances are added, never subtracted, even though the means are subtracted.

Question 2

Sample 1: n = 36, mean 21.4, population variance 9. Sample 2: n = 49, mean 19.6, population variance 16. Find the standard error of the difference and the test statistic for equal means.

Accepted Answer

Standard error = √(9/36 + 16/49) = √(0.25 + 0.3265) = √0.5765 = 0.759. The difference of means is 1.8, so z = 1.8/0.759 = 2.37. M1 A1 for the standard error, A1 for the statistic.

Question 3

Independent random samples of sizes n_x and n_y are taken from N(μ_x, σ_x²) and N(μ_y, σ_y²). Write down the distribution of X̄ − Ȳ.

Accepted Answer

X̄ − Ȳ is normal, with mean μ_x − μ_y and variance σ_x²/n_x + σ_y²/n_y. B1 for the mean, B1 for the variance. Each sample mean carries variance σ²/n, and those two variances add because the samples are independent.

Question 4

Two independent samples, of sizes 40 and 60, come from populations with variances 20 and 45. Find the standard error of the difference of the sample means, and state its value if both sample sizes were doubled.

Accepted Answer

The two variances of the means are 20/40 = 0.5 and 45/60 = 0.75, so the standard error is √1.25 = 1.118. M1 for adding the two, A1 for 1.118.
Doubling both sample sizes halves each term, so the variance under the root halves and the standard error divides by √2: it becomes 1.118/√2 = 0.791. B1.
Halving the standard error would need four times the readings in each sample, since n sits under a square root.

Question 5

Sample 1: n = 36, mean 21.4, population variance 9. Sample 2: n = 49, mean 19.6, population variance 16, giving a standard error of 0.759 and a test statistic of 2.37. Test at the 5% level whether the population means differ.

Accepted Answer

H₀: μ₁ = μ₂; H₁: μ₁ ≠ μ₂, where μ₁ and μ₂ are the population means. Two-tailed at 5%, so the critical values are ±1.96. The statistic 2.37 exceeds 1.96 and lies in the critical region, so reject H₀. There is evidence at the 5% level that the two population means differ, with the first appearing larger. B1 for the hypotheses in terms of population means, B1 for the critical values, M1 for the comparison, A1 for the conclusion in context. Write the hypotheses about μ₁ and μ₂: the sample means are known, so no hypothesis is being made about them.

Question 6

Two samples give means 21.4 and 19.6, with a standard error of 0.759 for their difference. Find a 95% confidence interval for the difference of the population means, and check it agrees with a two-tailed test at the 5% level.

Accepted Answer

The point estimate is 1.8 and the standard error 0.759, so the interval is 1.8 ± 1.96 × 0.759 = 1.8 ± 1.488, that is (0.31, 3.29). M1 for the form estimate plus or minus z times the standard error, A1 for the half-width, A1 for the interval, B1 for the comparison with the test. Zero lies outside, which is exactly the verdict the two-tailed test at 5% reached. The interval says more: the difference could plausibly be anywhere from a third of a unit to over three.

Question 7

Explain what changes if the population variances are unknown but both samples are large.

Accepted Answer

Replace each σ² by the sample variance s² from that sample. The Central Limit Theorem keeps the sample means approximately normal whatever the population shape, and with large samples the estimated variances are close enough to the real ones for the normal critical values to remain usable. The conclusion becomes approximate rather than exact, and this should be stated. B1 for replacing each variance by its sample estimate, B1 for the Central Limit Theorem, B1 for saying the result is approximate.

Question 8

A farmer trials a new feed. Herd A, of 50 animals on the new feed, has mean mass gain 512 g per week, from a population with variance 64; herd B, of 40 animals on the old feed, has mean gain 508 g, with population variance 81. The two herds are independent. Test, at the 5% level, whether the new feed increases the mean gain, using the critical value 1.6449.

Accepted Answer

The claim is an increase, so the test is one-tailed: H₀: μ_A = μ_B; H₁: μ_A > μ_B. B1 for one-tailed hypotheses about the population means. The standard error adds the two variances of the means: √(64/50 + 81/40) = √3.305 = 1.818. M1 for the sum under the root, A1 for 1.818. The statistic is z = (512 − 508)/1.818 = 2.200. M1. Since 2.200 > 1.6449, reject H₀: there is evidence at the 5% level that the new feed increases the mean weekly gain. A1 for the conclusion drawn in context. The independence of the herds is what licenses adding the variances.

Question 9

A supermarket compares queueing times at two branches. At branch P a random sample of 100 customers waits a mean of 4.6 minutes with s = 1.8 minutes; at branch Q an independent random sample of 150 customers waits a mean of 4.2 minutes with s = 2.1 minutes. The population variances are unknown. Test at the 5% level whether the mean waiting time at P is longer than at Q, using the critical value 1.6449.

Accepted Answer

The claim names a direction, so the test is one-tailed: H₀: μ_P = μ_Q; H₁: μ_P > μ_Q, where the μ are the population mean waiting times. B1.
The population variances are unknown, so each is replaced by its sample estimate. Both samples are large, so the Central Limit Theorem keeps the two sample means approximately normal and the normal critical values remain usable; the conclusion is approximate rather than exact.
Standard error = √(1.8²/100 + 2.1²/150) = √(0.0324 + 0.0294) = √0.0618 = 0.2486. M1 for the sum under the root, A1 for 0.2486.
z = (4.6 − 4.2)/0.2486 = 1.609. A1.
Since 1.609 < 1.6449, do not reject H₀. There is insufficient evidence at the 5% level that customers wait longer at branch P than at branch Q. A1 for the conclusion in context.
The statistic sits just below its critical value, so report the verdict as a near miss rather than as a finding of no difference: the test has not established that the two branches are alike.

Question 10

Two independent random samples are taken from normal populations. The first has n = 64, mean 82.5 and population variance 36; the second has n = 81, mean 78.9 and population variance 45. Find a 99% confidence interval for the difference between the population means, and state whether it supports a claim that the two means differ.

Accepted Answer

Standard error = √(36/64 + 45/81) = √(0.5625 + 0.5556) = √1.1181 = 1.0574. M1 for the standard error.
The point estimate is 82.5 − 78.9 = 3.6, and the 99% multiplier is 2.5758, so the half-width is 2.5758 × 1.0574 = 2.724. A1.
The interval is 3.6 ± 2.724, that is (0.876, 6.324). A1.
Zero lies outside the interval, so the data support the claim that the population means differ, at the level this interval matches, the two-tailed 1% test. B1 for the verdict.
Zero clears the lower limit by less than a unit, so the support is not emphatic; the interval also says that the true difference could be anything from under 1 to over 6.

Question 11

Explain why a 95% confidence interval for the difference of two population means and a two-tailed test at the 5% level always reach the same verdict about whether the means are equal, and state which confidence interval pairs with a one-tailed test at the 5% level.

Accepted Answer

Both are built from the same point estimate and the same standard error, and both use ±1.96. The interval holds exactly the hypothesised differences that the test would not reject, so zero lies inside it precisely when the statistic falls short of 1.96. B1 for the shared multiplier and standard error, B1 for the interval being the set of values not rejected.
A one-tailed test at 5% puts the whole 5% in one tail, which matches a 90% two-sided interval, whose tails carry 5% each, or equivalently a one-sided 95% bound. B1.
Reading the lower limit of a 95% two-sided interval as a one-tailed 5% test quietly performs a 2.5% test instead.

Question 12

Sample 1: n = 36, mean 21.4, population variance 9. Sample 2: n = 49, mean 19.6, population variance 16, giving a standard error of 0.7593 and a test statistic of 2.371. Test at the 1% level, one-tailed, with critical value 2.3263. Then find the smallest difference of sample means that would have been significant at that level, and comment on the pair of results.

Accepted Answer

H₀: μ₁ = μ₂; H₁: μ₁ > μ₂. The statistic is 2.371, which exceeds 2.3263, so reject H₀ at the 1% level as well.
The test rejects when the difference d satisfies d/0.7593 > 2.3263, that is d > 2.3263 × 0.7593 = 1.766.
The observed 1.8 clears that by only 0.034. A difference of 1.76 would have been significant at 5% but not at 1%, so the two levels would have disagreed. B1 for the one-tailed hypotheses, M1 for the comparison with 2.3263, A1 for the conclusion in context, M1 A1 for the smallest significant difference, B1 for the comment on how marginal it is. Report a statistic this close to its critical value as marginal rather than as a firm verdict, and quote the level with every conclusion.

Question 13

Explain why the variances of the two sample means are added, and never subtracted, when finding the standard error of the difference between two independent sample means, and state the part independence plays.

Accepted Answer

A difference of two varying quantities varies more than either alone: the two sample means miss their targets separately, and the errors compound rather than cancel, so the spread of the difference has to exceed the spread of a single mean. Subtracting the variances would make it smaller, which is the wrong way round, and could even go negative. B1 for the direction argument. So the variances combine as σ_x²/n_x + σ_y²/n_y under the square root, whatever the sign between the means. B1 for the combination written out. The rule rests on the two samples being independent: correlated samples would share part of their error, and the simple sum would misstate the spread, which is why the independence of the samples is stated as part of the test. B1 for independence named as the licence.

Question 14

Two independent random samples, each of size n, are to be taken from normal populations with variances 12 and 20. Find the smallest n for which a 95% confidence interval for the difference between the population means has total width at most 2.

Accepted Answer

With equal sample sizes the standard error is √(12/n + 20/n) = √(32/n). M1 for the standard error in terms of n, A1 for √(32/n).
The interval is the difference of the sample means plus or minus 1.96 × √(32/n), so its total width is twice that. A total width of at most 2 means a half-width of at most 1: 1.96√(32/n) ≤ 1. M1 for halving the width before writing the inequality.
So √(32/n) ≤ 1/1.96, that is 32/n ≤ 0.2603, and n ≥ 32/0.2603 = 122.93. A1 for 122.9.
The inequality runs upwards and n must be a whole number, so round up to n = 123 in each sample. dM1 for the rounding decision, A1 for 123.
Check both sides: n = 123 gives a half-width of 0.9997 and n = 122 gives 1.0038, so 122 misses the requirement and 123 meets it.

Question 15

A researcher claims that brand A batteries last, on average, more than 5 hours longer than brand B batteries. A random sample of 45 brand A batteries has mean lifetime 38.6 hours, from a population with variance 9; an independent random sample of 40 brand B batteries has mean lifetime 31.9 hours, from a population with variance 12. Test the researcher's claim at the 5% level, using the critical value 1.6449.

Accepted Answer

The claim is about a difference of 5, not about equality, so the hypothesised difference enters the hypotheses: H₀: μ_A − μ_B = 5; H₁: μ_A − μ_B > 5, where the μ are the population mean lifetimes. B1.
Standard error = √(9/45 + 12/40) = √(0.2 + 0.3) = √0.5 = 0.7071. M1 for the sum under the root, A1 for 0.7071.
The statistic subtracts the hypothesised difference from the observed one: the sample means differ by 38.6 − 31.9 = 6.7, so z = (6.7 − 5)/0.7071 = 1.7/0.7071 = 2.404. M1 for removing the 5, A1 for 2.404.
Since 2.404 > 1.6449, reject H₀. There is evidence at the 5% level that brand A batteries last more than 5 hours longer on average than brand B. A1 for the conclusion in context.
Testing 6.7 against zero instead gives a statistic near 9.5 and answers a different question: it would establish only that the two brands differ.