Applications: Statistical Inference

A point estimate is a single number, worked out from a sample, that stands for a quantity of the whole population. An estimate is unbiased when its average over all possible samples equals the true value. The sample mean is an unbiased estimate of the population mean. For the variance, the squared deviations are measured from the sample's own mean, which makes them a little too small, so their sum is divided by n − 1 rather than n.

1. The unbiased estimate of the population mean is the sample mean: x = 60525 = 24.2 g.
2. The sum of squares about the mean is S_xx = ∑ x² − (∑ x)²n = 14941 − 605²25 = 14941 − 14641 = 300.
3. (a) The unbiased estimate of the variance divides by n − 1 = 24: s² = 30024 = 12.5. The estimates are 24.2 g for the mean and 12.5 g² for the variance.
4. Dividing by 25 gives 30025 = 12. The deviations are measured from x, the center of this sample itself, so their squares are smaller on average than squares measured from the true mean μ. Dividing by n − 1 instead of n makes up for that.
5. (b) The worker's value is 12 g², which tends to underestimate the variance. The standard error of the mean is estimated as √s²n = √12.525 = √0.5 = 0.707 g. Check: 24 × 12.5 = 25 × 12 = 300.

answer(a) the mean 24.2 g and the variance 12.5 g²; (b) 12 g², too small because the deviations are measured from the sample's own mean; the standard error is about 0.707 g

techniqueUnbiased Sample Variance · Point Estimates

examsH2

A hypothesis test starts from a null hypothesis H₀, here that nothing has changed, and asks how surprising the sample would be if H₀ were true. The p-value is the probability, under H₀, of a result at least as extreme as the one observed. When the question is whether a mean has changed, in either direction, the test is two-tailed and extreme results in both tails count.

1. Let μ be the mean volume after the repair. The question is whether it has changed in either direction, so the test is two-tailed: H₀: μ = 250 and H₁: μ ≠ 250.
2. Under H₀ the mean of 36 cups is normal with mean 250 and standard deviation 6√36 = 1 ml. The test statistic is z = 252.1 − 2501 = 2.1.
3. The p-value counts results at least this far from 250 in either direction: 2 × P(Z > 2.1) = 2 × (1 − 0.98214) = 0.0357.
4. (a) 0.0357 < 0.05, or equally 2.1 > 1.96, so reject H₀. There is evidence at the 5% level that the mean volume has changed from 250 ml since the repair.
5. (b) At the 1% level the critical values are ± 2.576, and 0.0357 > 0.01, so do not reject H₀. At 1% there is not enough evidence that the mean volume has changed. The same data can be significant at one level and not at another, which is why the level is chosen before the data are seen.

answer(a) H₀: μ = 250, H₁: μ ≠ 250; z = 2.1 and the p-value is 0.0357, below 0.05 (2.1 > 1.96): reject H₀, there is evidence that the mean volume has changed; (b) no: 0.0357 > 0.01 (2.1 < 2.576), so at the 1% level there is not enough evidence of a change

techniqueThe Z-Test · P-Values · One-Tailed and Two-Tailed

examsH2

The critical region is the set of results that would lead to rejecting H₀. For a discrete distribution the probability of the region cannot usually be exactly 5%, so the region is the largest tail whose probability does not exceed 5%, and that probability is the actual significance level. A result outside the region means that H₀ is not rejected, which is not the same as showing that it is true.

1. Let X be the number of free throws she scores out of 20. Under H₀, X ∼ B(20, 0.4). The alternative is p > 0.4, so the test is one-tailed and the critical region is at the top: X ≥ k for the smallest k with P(X ≥ k) ≤ 0.05.
2. Add the binomial terms from the top: P(X ≥ 13) = 0.0210 and P(X ≥ 12) = 0.0565. The first is below 0.05 and the second is above it.
3. (a) The critical region is X ≥ 13. The actual significance level is P(X ≥ 13) = 0.0210, about 2.1%: that is the chance of rejecting H₀ when in fact she has not improved.
4. She scored 12, which is not in the critical region. In the same way, P(X ≥ 12) = 0.0565 is more than 0.05.
5. (b) Do not reject H₀. There is not enough evidence at the 5% level that her success rate has risen above 40%. This does not show that she has not improved: 12 out of 20 is 60%, but 20 throws are too few to tell that rate apart from chance.

answer(a) the critical region is X ≥ 13, and the actual significance level is 0.0210; (b) 12 is not in the critical region, since P(X ≥ 12) = 0.0565, so do not reject H₀: there is not enough evidence that her success rate has risen above 40%

techniqueCritical Regions · Testing a Proportion with the Binomial · Null and Alternative Hypotheses

examsH2

A Type I error is rejecting H₀ when it is true; its probability is the significance level. A Type II error is failing to reject H₀ when it is false. Its probability depends on the true value, so it is found in two stages: first the critical region is fixed using H₀, then the probability of a sample mean outside that region is worked out using the true mean.

1. Under H₀ the mean of 25 loaves is normal with mean 400 g and standard deviation 10√25 = 2 g. The alternative is μ < 400, so the test is one-tailed with the critical region in the lower tail.
2. The lowest 5% of the standard normal distribution lies below z = −1.645, so the critical value is 400 − 1.645 × 2 = 396.71 g.
3. (a) The critical region is x < 396.71 g. A Type I error is rejecting H₀ when it is true, and its probability is the significance level, 0.05.
4. With the true mean 395 g, the sample mean is normal with mean 395 and standard deviation 2. The test fails to reject H₀ when x ≥ 396.71, which is z = 396.71 − 3952 = 0.855 standard deviations above 395.
5. (b) P(x ≥ 396.71) = 1 − 0.8037 = 0.196. This is a Type II error, failing to reject H₀ when it is false: about one inspection in five would pass loaves that are on average 5 g light. Weighing more loaves would make this probability smaller.

answer(a) the critical region is x < 396.71 g, and the probability of a Type I error is 0.05; (b) 0.196, the probability of a Type II error

techniqueType One and Type Two Errors · Critical Regions · One-Tailed and Two-Tailed

examsH2

A confidence interval is a range of plausible values for a population mean, built around the sample mean. A 95% interval reaches 1.96 standard errors either side, because 95% of a normal distribution lies within 1.96 standard deviations of its mean. The 95% describes the method: of many intervals made this way from different samples, about 95% contain the true mean.

1. The sample is large, so by the central limit theorem the sample mean is approximately normal, and s = 8 can be used in place of σ. The standard error is 8√64 = 1 minute.
2. A 95% interval reaches 1.96 standard errors either side of the sample mean: 34.5 ± 1.96 × 1 = 34.5 ± 1.96.
3. (a) The 95% confidence interval is (32.54, 36.46) minutes.
4. 32 minutes lies below the interval, so a mean of 32 minutes is not a plausible value at this level: the timetable allows too little time. A two-tailed test of μ = 32 at the 5% level would reject it.
5. (b) The interval is about the mean journey time. It does not say that 95% of journeys take between 32.54 and 36.46 minutes: with a standard deviation of 8 minutes, single journeys vary far more than that. Nor is there a 95% chance that this one interval contains μ; the 95% describes how often intervals made this way contain the true mean.

answer(a) (32.54, 36.46) minutes; (b) 32 is below the interval, so the timetable allows too little time; the interval is for the mean, not for single journeys, and the 95% describes how often intervals made this way contain the true mean

techniqueConfidence Intervals · Point Estimates

examsSAT

If two classifications are independent, the share of each row that falls in a column is the same for every row, so the expected frequency of a cell is its row total times its column total divided by the grand total. The statistic χ² = ∑ (O − E)²E measures how far the observed table is from those expected frequencies. For a table of r rows and c columns it is compared with the χ² distribution on (r − 1)(c − 1) degrees of freedom.

1. H₀: the preferred kind of film is independent of age group. H₁: it is not independent. The column totals are 64 + 26 = 90 for action, 28 + 32 = 60 for comedy and 28 + 22 = 50 for drama.
2. For customers under 30 the expected frequencies are 120 × 90200 = 54, 120 × 60200 = 36 and 120 × 50200 = 30; for customers aged 30 or over they are 36, 24 and 20. Every one is at least 5, so the test can be used.
3. (a) χ² = ∑ (O − E)²E = 10²54 + 8²36 + 2²30 + 10²36 + 8²24 + 2²20 = 9.407.
4. The table has 2 rows and 3 columns, so there are (2 − 1)(3 − 1) = 2 degrees of freedom. The critical value of χ²(2) at the 5% level is 5.991.
5. (b) 9.407 > 5.991, so reject H₀. There is evidence at the 5% level that the kind of film customers prefer depends on their age group: customers under 30 choose action more often than independence would give, and older customers choose comedy more often. Check: in each row the differences O − E add up to 0, since 10 − 8 − 2 = 0.

answer(a) the expected frequencies are 54, 36 and 30 under 30, and 36, 24 and 20 at 30 or over; χ² = 9.407; (b) 9.407 > 5.991 on 2 degrees of freedom, so reject H₀: the kind of film preferred depends on age group

techniqueThe Chi-Squared Test for Independence · Expected Frequencies

A goodness-of-fit test compares observed frequencies with the frequencies a stated distribution predicts. The statistic χ² = ∑ (O − E)²E is compared with the χ² distribution whose degrees of freedom are the number of classes minus the number of constraints; here the only constraint is that the expected frequencies add up to the same total as the observed ones.

1. H₀: the die is fair, so each score has probability 16. H₁: the die is not fair. The expected frequency of each score is 120 × 16 = 20.
2. The differences O − E are −6, 2, −2, 6, −3 and 3, which add up to 0 as they must. Their squares are 36, 4, 4, 36, 9 and 9, a total of 98.
3. (a) Every expected frequency is 20, so χ² = 9820 = 4.9.
4. There are 6 classes and one constraint, that the frequencies add up to 120, so there are 6 − 1 = 5 degrees of freedom. The critical value of χ²(5) at the 5% level is 11.070.
5. (b) 4.9 < 11.070, so do not reject H₀. There is not enough evidence at the 5% level that the die is unfair: differences as large as these often happen with a fair die in 120 throws. Had the statistic been above 11.070, the club would have had evidence that the die is biased.

answer(a) each expected frequency is 20, and χ² = 4.9; (b) 4.9 < 11.070 on 5 degrees of freedom, so do not reject H₀: there is not enough evidence that the die is unfair

techniqueThe Chi-Squared Goodness of Fit Test · Expected Frequencies · Null and Alternative Hypotheses

examsH2

When the population standard deviation is unknown it is estimated by s from the sample. For a small sample from a normal population, x − μs / √n then follows the t distribution with n − 1 degrees of freedom, which has heavier tails than the normal distribution, so its critical values are further out.

1. Let μ be the mean life of the batteries. H₀: μ = 20 and H₁: μ < 20, a one-tailed test. The variance is unknown and the sample is small, so the test uses the t distribution with 9 − 1 = 8 degrees of freedom.
2. The nine lives add up to 172.8, so x = 172.89 = 19.2 hours. Their deviations from 19.2 are −1.6, −1.2, −0.8, −0.6, −0.2, 0.2, 1.0, 1.2 and 2.0 hours, and their squares add up to 11.52.
3. (a) s² = 11.528 = 1.44, so s = 1.2 hours, and the test statistic is 19.2 − 201.2 / √9 = −0.80.4 = −2.0.
4. (b) For a lower-tailed test at 5% on 8 degrees of freedom the critical value is −1.860. Since −2.0 < −1.860, reject H₀: there is evidence at the 5% level that the batteries last less than 20 hours on average.
5. At the 1% level the critical value is −2.896, and −2.0 > −2.896, so H₀ is not rejected: at 1% there is not enough evidence against the manufacturer's claim. Check: the deviations add up to 0, as deviations from a mean must.

answer(a) x = 19.2 hours, s² = 1.44, and the test statistic is −2.0; (b) −2.0 < −1.860, so at 5% reject H₀: there is evidence that the batteries last less than 20 hours on average; at 1% the critical value is −2.896 and H₀ is not rejected

techniqueThe One-Sample T-Test · One-Tailed and Two-Tailed · Null and Alternative Hypotheses

examsH2

When two normal populations share one variance, the two sample estimates are combined into a pooled estimate, each weighted by its degrees of freedom. The difference in sample means, divided by its standard error √s_p² (1n₁ + 1n₂), follows the t distribution with n₁ + n₂ − 2 degrees of freedom when H₀ is true.

1. Let μ_A and μ_B be the mean heights. H₀: μ_A = μ_B and H₁: μ_A ≠ μ_B, a two-tailed test. The sample means are x_A = 1446 = 24 cm and x_B = 1266 = 21 cm.
2. The squared deviations add up to 4 + 4 + 0 + 1 + 1 + 4 = 14 for A and 4 + 4 + 0 + 0 + 4 + 4 = 16 for B, so s_A² = 145 = 2.8 and s_B² = 165 = 3.2.
3. The pooled variance weights each estimate by its degrees of freedom: s_p² = 5 × 2.8 + 5 × 3.210 = 3010 = 3.
4. (a) The standard error of the difference is √3 × (16 + 16) = √1 = 1 cm, so the test statistic is 24 − 211 = 3.0, on 6 + 6 − 2 = 10 degrees of freedom.
5. (b) At 5%, two-tailed, the critical value of t(10) is 2.228, and 3.0 > 2.228: reject H₀. There is evidence that the fertilizers give different mean heights, with the seedlings given A taller. At 1% the critical value is 3.169, and 3.0 < 3.169, so at that level the difference is not significant.

answer(a) the pooled variance is 3 and the test statistic is 3.0, on 10 degrees of freedom; (b) 3.0 > 2.228, so at 5% reject H₀: the fertilizers give different mean heights; at 1% the critical value is 3.169, so there the difference is not significant

techniqueThe Two-Sample T-Test

For a sample of n from a normal population with variance σ², the quantity (n − 1)s²σ² follows the χ² distribution with n − 1 degrees of freedom, which gives a test of one variance. To compare two variances, the ratio of the two estimates follows the F distribution, with the degrees of freedom of the top estimate first and those of the bottom estimate second.

1. For machine A, H₀: σ_A² = 0.04 and H₁: σ_A² > 0.04. The test statistic is (n − 1)s_A²0.04 = 10 × 0.080.04 = 20.0, on the χ²(10) distribution.
2. (a) The upper 5% critical value of χ²(10) is 18.307. Since 20.0 > 18.307, reject H₀: there is evidence at the 5% level that machine A's variance is greater than the 0.04 mm² the specification allows.
3. For the comparison, H₀: σ_A² = σ_B² and H₁: σ_A² > σ_B². The test statistic is the ratio of the estimates, the larger on top: F = 0.080.032 = 2.5.
4. The degrees of freedom are 11 − 1 = 10 for the top and 13 − 1 = 12 for the bottom. The upper 5% critical value of F(10, 12) is 2.753.
5. (b) 2.5 < 2.753, so do not reject H₀. There is not enough evidence at the 5% level that machine A's lengths vary more than machine B's. The two results are consistent: A is shown to be outside the specification, but samples of 11 and 13 bolts are too small to show that A is worse than B.

answer(a) the test statistic is 20.0, above the critical value 18.307 of χ²(10), so reject H₀: machine A's variance is greater than 0.04 mm²; (b) F = 2.5, below the critical value 2.753 of F(10, 12), so do not reject H₀: there is not enough evidence that A's lengths vary more than B's

techniqueTesting a Variance with Chi-Squared · Comparing Two Variances with the F-Test