Questions on confidence intervals and tests with the t-distribution

Question 1

State when the t-distribution is used in place of the normal, and how the two differ.

Accepted Answer

It is used when the population is normal but its variance is unknown and estimated by s² from the same sample, particularly when that sample is small. The t-distribution is symmetric about zero like the normal but has heavier tails, so its critical values are further out, and it approaches the normal as the degrees of freedom grow. B1 for the condition of use, B1 for the heavier tails.

Question 2

A random sample of 9 observations from a normal population has Σx = 216 and s = 4.5. Write down the number of degrees of freedom, and find the standard error of the mean and the value of the test statistic for testing H₀: μ = 22.

Accepted Answer

The sample mean is 216/9 = 24, and the degrees of freedom are n − 1 = 8. One degree of freedom is spent estimating s from the same sample, so the divisor is n − 1 = 8, not n = 9.
Standard error = s/√n = 4.5/3 = 1.5.
t = (24 − 22)/1.5 = 1.33.
B1 for the degrees of freedom, M1 for the standard error, A1 for the statistic.

Question 3

State the assumption a one-sample t test makes about the population, and what it requires of the sample.

Accepted Answer

The population is assumed to be normally distributed. B1. The sample must be a random sample from that population, so that the observations are independent of one another. B1. Neither assumption is about the sample looking normal: the condition is on the population the sample came from, and with a small sample the data can look ragged while the assumption holds. Where the population variance is known as well, the normal distribution replaces t and no variance has to be estimated.

Question 4

A random sample of 12 observations from a normal population has mean 18.4 and s = 2.4. Write down the number of degrees of freedom and find a 95% confidence interval for the population mean, using the t value 2.201.

Accepted Answer

The degrees of freedom are n − 1 = 11. B1.
Standard error = 2.4/√12 = 0.6928, so the interval is 18.4 ± 2.201 × 0.6928 = 18.4 ± 1.525, that is (16.88, 19.92). M1 for the form of the interval, A1 for the interval.
Divide s by √12, not by 12: dividing by the sample size gives a standard error of 0.2 and an interval less than one unit wide.

Question 5

A random sample of 16 observations from a normal population has mean 52.5 and s = 6. Test at the 5% level, in one tail, whether the population mean exceeds 50. The critical value is 1.753.

Accepted Answer

H₀: μ = 50; H₁: μ > 50, on 16 − 1 = 15 degrees of freedom. B1 for both hypotheses and the degrees of freedom.
Standard error = 6/√16 = 1.5, so t = (52.5 − 50)/1.5 = 1.667.
Since 1.667 < 1.753, do not reject H₀. There is insufficient evidence at the 5% level that the mean exceeds 50.
M1 for the standard error, A1 for the statistic, M1 for the comparison with the critical value, A1 for the conclusion in context. Write 'do not reject H₀' rather than 'accept H₀', because the test never establishes that μ is 50.

Question 6

A random sample of 16 observations from a normal population has mean 52.5 and s = 6. Find a 95% confidence interval for the population mean, using the t value 2.131, and state whether the interval supports the claim that the mean is 50.

Accepted Answer

Standard error = 6/√16 = 1.5, so the interval is 52.5 ± 2.131 × 1.5 = 52.5 ± 3.20, that is (49.30, 55.70).
The value 50 lies inside the interval, so the data are consistent with a mean of 50.
M1 for the form of the interval, A1 for the half-width, A1 for the interval, B1 for the verdict on 50.
Using 1.96 in place of 2.131 gives (49.56, 55.44), which is too narrow. The extra width is the price of estimating the variance from the sample, and it is why the t value is always the larger of the two.

Question 7

Eight subjects are measured before and after a treatment. The differences, after minus before, are 2, 5, −1, 3, 4, 0, 2, 1. Test at the 5% level, in one tail, whether the treatment raises the measurement. The critical value is 1.895.

Accepted Answer

Work with the differences alone, never with the raw before and after readings. Subtracting within pairs removes the variation between subjects, which is usually far larger than the effect being measured.
H₀: μ_d = 0; H₁: μ_d > 0, on 8 − 1 = 7 degrees of freedom.
The differences total 16, so the mean difference is 16/8 = 2. Their squared deviations from 2 total 28, so s² = 28/7 = 4 and s = 2. Divide by 7, not by 8; the divisor n − 1 is what makes s² unbiased.
Standard error = 2/√8 = 0.707, so t = 2/0.707 = 2.83.
Since 2.83 > 1.895, reject H₀. There is evidence at the 5% level that the treatment raises the measurement.
B1 for the hypotheses and the degrees of freedom, M1 A1 for the mean difference and the sample standard deviation, M1 for the standard error, A1 for the statistic, A1 for the conclusion in context.
Treating these data as two independent samples of 8 would inflate the standard error and hide the effect.

Question 8

Six students sit a test before and after a revision course. Their scores are: before 62, 58, 65, 60, 57, 63; after 65, 59, 69, 62, 57, 68. Assuming the differences are normally distributed, find a 95% confidence interval for the mean improvement, using the t value 2.571, and state what the interval says about the course.

Accepted Answer

The same student is measured twice, so this is paired data: work with the differences, after minus before, which are 3, 1, 4, 2, 0, 5, on 6 − 1 = 5 degrees of freedom.
The differences total 15, so the mean improvement is 15/6 = 2.5. The squared deviations from 2.5 total 17.5, so s² = 17.5/5 = 3.5 and s = 1.871.
Standard error = √(3.5/6) = 0.764, so the interval is 2.5 ± 2.571 × 0.764 = 2.5 ± 1.964, that is (0.54, 4.46).
Zero lies outside the interval, and the whole of it is positive, so at the level the interval matches, the two-tailed 5% test, there is evidence that the course improves scores. The interval is wide because six pairs is a small sample, so the size of the improvement is poorly pinned down.
B1 for the differences and the degrees of freedom, M1 for the mean and standard deviation of the differences, A1 for 2.5 and 1.871, M1 for the form of the interval, A1 for the interval and the verdict on the course.

Question 9

A machine is set to cut rods to a mean length of 25 cm. A random sample of 10 rods, from a normal population of lengths, has mean 25.34 cm and s = 0.42 cm. Test at the 5% level, in two tails, whether the mean length has changed. The critical value is 2.262.

Accepted Answer

H₀: μ = 25; H₁: μ ≠ 25, on 10 − 1 = 9 degrees of freedom, with 2.5% in each tail. B1 for the hypotheses and the degrees of freedom.
Standard error = 0.42/√10 = 0.1328. M1.
t = (25.34 − 25)/0.1328 = 2.56. A1.
Since 2.56 > 2.262, reject H₀. There is evidence at the 5% level that the mean length cut by the machine is no longer 25 cm, the sample pointing to a longer mean. A1 for the conclusion in context.
The alternative names no direction, so the 5% is split between the tails and 2.262 is the value to beat. A one-tailed test here would compare with 1.833 and reach the same verdict on weaker grounds, so settle the tail from the wording before looking anything up.

Question 10

State the assumptions required by each of the three t procedures: the one-sample t test, the paired t test, and the two-sample t test using a pooled estimate of variance. State which assumption is the one most worth checking before it is relied on, and how it could be checked.

Accepted Answer

One-sample. The population is normal and the sample is random, so the observations are independent. B1. Paired. The pairs are independent of one another and the population of differences is normal. Nothing at all is assumed about the two original populations, which may be skewed and may have different variances, because the test never looks at them separately. B1. Two-sample pooled. Both populations are normal, the two samples are independent, and the two population variances are equal, that common value being what the pooled estimate estimates. B1 for normality and independence, B1 for the equal variances. The equal-variance condition is the one worth checking, because it is the only one the arithmetic depends on numerically and it is the only one of the four with a test of its own: an F test on the ratio of the two sample variances, on n₁ − 1 and n₂ − 1 degrees of freedom. B1. Where it fails, the pooled procedure is the wrong one, and two sample standard deviations that differ by half as much again are worth putting through that F test before any pooling goes ahead.

Question 11

A random sample of 20 observations from a normal population gives Σx = 96 and Σx² = 480. Calculate unbiased estimates of the population mean and variance, and find the value of the t statistic for testing H₀: μ = 5.

Accepted Answer

The sample mean is 96/20 = 4.8, unbiased for μ as it stands. Build S_xx from the two totals: S_xx = Σx² − (Σx)²/n = 480 − 96²/20 = 480 − 460.8 = 19.2. M1 for the formula. The unbiased estimate divides by n − 1 = 19, so s² = 19.2/19 = 1.011 and s = 1.005. A1. Standard error = s/√20 = 1.005/4.472 = 0.2248. M1. t = (4.8 − 5)/0.2248 = −0.890, on 19 degrees of freedom. A1. The sign is part of the answer: the sample mean sits below the value under test, so any one-tailed conclusion must be about μ being smaller, not larger. Dividing S_xx by 20 rather than 19 gives s² = 0.96 and a statistic of −0.913.

Question 12

Two independent random samples are taken from normal populations. The first has n = 6, mean 18.2 and s = 2.4; the second has n = 9, mean 15.8 and s = 3.1. Assuming the two populations have equal variances, test at the 5% level, in two tails, whether the population means differ. The critical value is 2.160.

Accepted Answer

H₀: μ₁ = μ₂; H₁: μ₁ ≠ μ₂, on n₁ + n₂ − 2 = 6 + 9 − 2 = 13 degrees of freedom. Two degrees of freedom are spent, one for each mean estimated, so the total is n₁ + n₂ − 2 = 13, not 14.
Pool the variances, weighting each by its own degrees of freedom rather than by its sample size: s_p² = [5(2.4²) + 8(3.1²)]/13 = (28.8 + 76.88)/13 = 105.68/13 = 8.129, so s_p = 2.851. A plain average of 5.76 and 9.61 would give 7.685, which ignores the different sample sizes.
Standard error of the difference = s_p√(1/n₁ + 1/n₂) = 2.851 × √(1/6 + 1/9) = 2.851 × 0.527 = 1.503.
t = (18.2 − 15.8)/1.503 = 2.4/1.503 = 1.60.
Since 1.60 < 2.160, do not reject H₀. There is insufficient evidence at the 5% level that the population means differ.
B1 for the hypotheses and the degrees of freedom, M1 A1 for the pooled variance, M1 for the standard error, A1 for the statistic, A1 for the conclusion in context.
The equal-variance assumption is doing a lot of work, and with sample standard deviations of 2.4 and 3.1 it would be worth testing with an F test before relying on it.

Question 13

Two independent random samples come from normal populations with a common variance. Variety P: n = 10, mean yield 74.2, s = 3.0. Variety Q: n = 8, mean yield 70.7, s = 3.5. Find a 95% confidence interval for the difference between the population mean yields, using the t value 2.120, and state whether it supports a claim that the varieties differ.

Accepted Answer

The degrees of freedom are n₁ + n₂ − 2 = 10 + 8 − 2 = 16.
Pool the variances, weighting each by its own degrees of freedom: s_p² = [9(3.0²) + 7(3.5²)]/16 = (81 + 85.75)/16 = 166.75/16 = 10.42, so s_p = 3.228.
Standard error of the difference = s_p√(1/n₁ + 1/n₂) = 3.228 × √(1/10 + 1/8) = 1.531.
The difference in sample means is 74.2 − 70.7 = 3.5, so the interval is 3.5 ± 2.120 × 1.531 = 3.5 ± 3.246, that is (0.25, 6.75).
Zero lies outside the interval, so at the matching level, the two-tailed 5% test, the data support the claim that the mean yields differ. The lower limit sits close to zero, though, so the support is not emphatic, and a slightly different sample could pull zero inside.
B1 for the degrees of freedom, M1 A1 for the pooled variance, M1 for the standard error of the difference, A1 for the interval, B1 for the verdict on the claim.

Question 14

A manufacturer claims its batteries last more than 3 hours. A random sample of 25 batteries, from a population of normally distributed lifetimes, has mean 3.42 hours and s = 0.75. Find the one-sided 95% lower confidence bound for the mean lifetime, using the one-tailed t value 1.711, state what it says about the claim, and explain why the 95% two-sided interval would be the wrong interval to pair with a one-tailed 5% test here.

Accepted Answer

Standard error = 0.75/√25 = 0.15, on 25 − 1 = 24 degrees of freedom. The one-sided bound keeps all 5% in the lower tail: μ > 3.42 − 1.711 × 0.15 = 3.42 − 0.257 = 3.16 with 95% confidence. Since 3.16 > 3, the one-tailed 5% test rejects μ = 3 in favour of μ > 3, so the data support the claim. The 95% two-sided interval puts only 2.5% in each tail, so reading its lower limit as a one-tailed test quietly performs a 2.5% test, not the 5% one intended. The one-tailed 5% test pairs with this one-sided bound, or equivalently with the 90% two-sided interval, whose tails carry 5% each. M1 for the standard error, M1 for the bound as mean minus t times the standard error, A1 for 3.16, B1 for the verdict on the claim, B1 for the tail-matching explanation.

Question 15

A 95% confidence interval for the mean of a normal population, built from a random sample of 16 observations using the t value 2.131, is (48.2, 53.8). Find the sample mean and the value of s, and hence find the 99% confidence interval for the population mean from the same sample, using the t value 2.947.

Accepted Answer

A confidence interval is symmetric about the sample mean, so the mean is the midpoint: (48.2 + 53.8)/2 = 51. B1.
The half-width is (53.8 − 48.2)/2 = 2.8, and that half-width is t × standard error, so the standard error is 2.8/2.131 = 1.314. M1 for the half-width read as t times the standard error, A1 for 1.314.
The standard error is s/√16 = s/4, so s = 4 × 1.314 = 5.26. A1.
The sample has not changed, so the standard error stays at 1.314 and only the multiplier moves: 51 ± 2.947 × 1.314 = 51 ± 3.872, that is (47.13, 54.87). M1 for reusing the same standard error, A1 for the interval.
Check the direction before writing it down. The 99% interval must contain the 95% one, and (47.13, 54.87) does. An interval that came out narrower would say the 99% multiplier was smaller than the 95% one, which no t distribution allows.