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Confidence intervals and tests with the t-distribution questions
Estimating the variance from the same small sample adds uncertainty, and the t-distribution accounts for it. It has heavier tails than the normal, so intervals built from it are wider, and this lesson covers the three standard situations that call for it.
6 original questions · 26 marks · the confidence intervals and tests with the t-distribution notes · Further Statistics 2
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State when the t-distribution is used in place of the normal, and how the two differ.
Worked 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.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.
Worked answer
The sample mean is 216/9 = 24, and the degrees of freedom are n − 1 = 8. One is lost because s is estimated from the same sample, and quoting 9 here is the commonest slip in the topic.
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.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.
Worked answer
H₀: μ = 50; H₁: μ > 50, on 16 − 1 = 15 degrees of freedom. B1 for both hypotheses and the degrees of freedom, which is a mark before any arithmetic.
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. 'Accept H₀' does not earn that last mark, because the test never establishes that μ is 50.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.
Worked 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.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.
Worked 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, and that is the whole reason for pairing.
H₀: μd = 0; H₁: μd > 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.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.
Worked answer
H₀: μ₁ = μ₂; H₁: μ₁ ≠ μ₂, on n₁ + n₂ − 2 = 6 + 9 − 2 = 13 degrees of freedom. Two are lost, one for each mean estimated, and 14 is the usual wrong answer here.
Pool the variances, weighting each by its own degrees of freedom rather than by its sample size: sp² = [5(2.4²) + 8(3.1²)]/13 = (28.8 + 76.88)/13 = 105.68/13 = 8.129, so sp = 2.851. A plain average of 5.76 and 9.61 would give 7.685 and lose both marks for this stage.
Standard error of the difference = sp√(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.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise confidence intervals and tests with the t-distribution one question at a time
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