Your thirty prices averaged CHF 19.93. That single number is a point estimate of the true Geneva average — and it says nothing about how reliable it is. This week we add a margin of error E and form the interval x̄ ± E, built so that 95% of all intervals made this way contain the true mean. Large samples use z; small samples use Student's t; proportions have their own formula; and the last section answers the question a budget asks — how many pizzerias for ± 1 franc? This is the textbook's chapter 7.
The session is three hours in two parts. Part 1 (~1 h) is the lesson: your instructor presents the deck and you follow the same numbered sections here — every slide names its section. Part 2 (~2 h) is ten exercises taken straight from the textbook, worked together; solutions open as your instructor reveals them.
A bank asks you for the average Margherita price in Geneva. "19.93" is a point estimate: one number, and no hint of how far off it might be. The bank wants a range. What ± would you put on it — so that you would be right nineteen times out of twenty?
If we wish to estimate the mean µ of a population for which a census is impractical, a reasonable strategy is to take a sample, compute its mean x̄, and estimate the unknown number µ by the known number x̄. Estimating a population parameter by a single number like this is called point estimation.
A problem with a point estimate is that it gives no indication of how reliable the estimate is. In contrast, in interval estimation we use a formula to compute from the data a number E, called the margin of error of the estimate, and form the interval [x̄ − E, x̄ + E]. We do this in such a way that a certain proportion, say 95%, of all the intervals constructed from sample data by means of this formula contain the unknown parameter µ. Such an interval is called a 95% confidence interval for µ.
The book's own example: 100 men aged 18, mean height 70.6 inches, s = 1.7. Then E = 0.33 and "we are 95% confident that the average height of all 18-year-old men is between 70.27 and 70.93 inches". With only 50 men the same statistics give 70.6 ± 0.47 — a longer interval, because less data means less precision. Everything this week is that one move, four times over: a mean with a large sample, a mean with a small sample, a proportion, and the sample size that buys a chosen E.
Week 5 asked: if µ is 20, where will x̄ land? This week turns it around: x̄ landed at 19.93 — where is µ?
The Central Limit Theorem says that, for large samples (n ≥ 30), the sample mean X̄ is normally distributed with mean µ and standard deviation σ/√n. The Empirical Rule says that we must go about two standard deviations from the mean to capture 95% of the values of X̄ generated by sample after sample. A more precise distance based on the normality of X̄ is 1.960 standard deviations, which is E = 1.960·σ/√n.
Because in sample after sample 95% of the values of X̄ lie in the interval [µ − E, µ + E], if we adjoin to each side of the point estimate x̄ a "wing" of length E, 95% of the intervals formed by the winged dots contain µ. The 95% confidence interval is thus x̄ ± 1.960·σ/√n. For a different level of confidence, say 90% or 99%, the number 1.960 will change, but the idea is the same.
The book's Figure 7.2 is a computer simulation: 40 samples, 40 intervals, and about (0.05)(40) = 2 of them miss µ. Here is that figure, live. The population is the Geneva market of Week 5 — µ = 20, σ = 4.20 — and each click draws fresh samples of thirty and gives each mean its wings.
"95% confident" is a statement about the procedure: of all the intervals it produces, 95% catch µ. Your one interval either does or does not — and you will never know which.
It is standard practice to identify the level of confidence in terms of the area α in the two tails of the distribution of X̄ when the middle part specified by the level of confidence is taken out. For 100(1 − α)% confidence the area in each tail is α/2, and the number that cuts off a right tail of area α/2 is zα/2. The number 1.960 is z0.025, which is zα/2 for α = 1 − 0.95 = 0.05. The most common levels are 90% (α = 0.10), 95% (α = 0.05) and 99% (α = 0.01).
While sometimes the population standard deviation σ is known, typically it is not. If not, for n ≥ 30 it is generally safe to approximate σ by the sample standard deviation s.
90%: α = 0.10, so zα/2 = z0.05. The area to the left of z0.05 is 0.95. The closest entries in the normal table are 0.9495 and 0.9505, for z = 1.64 and 1.65; since 0.95 is exactly halfway, we use the average 1.645.
99%: α = 0.01, so zα/2 = z0.005; area to the left 0.9950, halfway between 0.9949 (2.57) and 0.9951 (2.58): 2.575.
The faster way (Example 2): the last line of the t table (Figure 12.3), headed ∞ [z], gives zc directly for every column: z0.05 = 1.645, z0.005 = 2.576. When a column exists, use it; it is both faster and more accurate than reading Figure 12.2 in reverse.
A sample of size 49 has sample mean 35 and sample standard deviation 14. Construct a 98% confidence interval for the population mean. Interpret its meaning.
98% means α = 0.02, so zα/2 = z0.01 = 2.326 from the t table's last line. Thus x̄ ± zα/2·s/√n = 35 ± 2.326 × 14/√49 = 35 ± 4.652 ≈ 35 ± 4.7.
We are 98% confident that the population mean µ lies in the interval [30.3, 39.7], in the sense that in repeated sampling 98% of all intervals constructed from the sample data in this manner will contain µ.
Example 4 is the same move on 120 students with GPA mean 2.71, s = 0.51, at 90%: 2.71 ± 1.645 × 0.51/√120 = 2.71 ± 0.0766 → (2.63, 2.79). Keep it in mind: section 6.3 redoes it with twelve students instead of 120.
Type a sample mean and standard deviation, pick n and a confidence level. The interval is drawn on the number line and the arithmetic is written out the way the book writes it. It opens on the Geneva data.
Two dials, two directions. More confidence → bigger z → wider interval. You pay for certainty with vagueness. Bigger n → smaller s/√n → narrower interval — but by √n, so four times the pizzerias only halve the width (Week 5's most expensive fact, again).
The confidence interval formulas in the previous section are based on the Central Limit Theorem. When the population mean µ is estimated with a small sample (n < 30), the Central Limit Theorem does not apply. In order to proceed we assume that the numerical population from which the sample is taken has a normal distribution to begin with. If this condition is satisfied then when the population standard deviation σ is known the old formula x̄ ± zα/2(σ/√n) can still be used.
If the population standard deviation is unknown and the sample size n is small then when we substitute the sample standard deviation s for σ the normal approximation is no longer valid. The solution is to use a different distribution, called Student's t-distribution with n − 1 degrees of freedom. It is very much like the standard normal distribution in that it is centered at 0 and has the same qualitative bell shape, but it has heavier tails. As the sample size n increases, Student's t-distribution ever more closely resembles the standard normal distribution; once the sample size is 30 or more it is typically acceptable to use the standard normal distribution instead, as we will always do in this text.
Why heavier tails? With a small sample, s is itself a poor estimate of σ — sometimes much too small — so dividing by s/√n produces more extreme values than dividing by σ/√n would. The t curve admits that: it pushes probability out into the tails, and the number that cuts off 2.5% moves out from 1.960 to 2.365 (df = 7) or 4.303 (df = 2). The price of a small sample is a wider interval.
Just as zc stands for the value that cuts off a right tail of area c under the standard normal curve, tc cuts off a right tail of area c under the t curve with the given degrees of freedom. The book's Figure 12.3 lists tc for df = 1 to 50 and for the columns c = 0.100, 0.050, 0.025, 0.010, 0.005, 0.0025, 0.001, 0.0005. Pick a row and a confidence level and the cell you need lights up.
A sample of size 15 drawn from a normally distributed population has sample mean 35 and sample standard deviation 14. Construct a 95% confidence interval for the population mean, and interpret its meaning.
Since the population is normally distributed, the sample is small, and σ is unknown, the formula that applies is x̄ ± tα/2·s/√n. Confidence level 95% means α/2 = 0.025. Since n = 15 there are n − 1 = 14 degrees of freedom, and by Figure 12.3 t0.025 = 2.145. Thus 35 ± 2.145 × 14/√15 = 35 ± 7.8.
One may be 95% confident that the true value of µ is contained in the interval (27.2, 42.8).
A random sample of 12 students from a large university yields mean GPA 2.71 with sample standard deviation 0.51. Construct a 90% confidence interval for the mean GPA of all students at the university. Assume the population of GPAs is normal.
α/2 = 0.05, df = 11, t0.05 = 1.796. Thus 2.71 ± 1.796 × 0.51/√12 = 2.71 ± 0.26, so (2.45, 2.97).
Compare Example 4: same summary statistics, 120 students, (2.63, 2.79). The small-sample interval is longer, partly because there is less information in twelve observations than in 120 — and partly because t0.05 = 1.796 is larger than z0.05 = 1.645.
Your partner priced the Margherita at eight pizzerias along the lake — too few for z. Assume lakeside prices are roughly normal.
Since from Section 6.3 we know the mean, standard deviation, and sampling distribution of the sample proportion p̂, the ideas of the previous two sections can be applied to produce a confidence interval for a population proportion.
The book's key takeaway is worth saying twice: the condition that a sample be large is not that n ≥ 30, but that the bell centred at p̂ fits inside [0, 1]. A sample of 25 can pass; a sample of 450 can fail (section 6.7, Exercise 7 of the book: 448 readers out of 450).
To estimate the proportion of students at a large college who are female, a random sample of 120 students is selected. There are 69 female students in the sample. Construct a 90% confidence interval for the proportion of all students at the college who are female.
p̂ = 69/120 = 0.575. Confidence 90% means α/2 = 0.05, and from the last line of Figure 12.3 z0.05 = 1.645. Thus 0.575 ± 1.645 × √(0.575 × 0.425 / 120) = 0.575 ± 0.074.
One may be 90% confident that the true proportion of all students at the college who are female is contained in the interval (0.501, 0.649).
Move the counts and the confidence level. The bar shows the book's check — does p̂ ± 3σp̂ stay inside [0, 1]? — and the interval below it.
Thirty pizzerias give a share "between 19% and 54%". That is honest, and almost useless — which is exactly why section 6.5 exists.
Sampling is typically done with a set of clear objectives in mind. For example, an economist might wish to estimate the mean yearly income of workers in a particular industry at 90% confidence and to within $500. Since sampling costs time, effort, and money, it would be useful to be able to estimate the smallest size sample that is likely to meet these criteria.
The confidence interval formulas for estimating µ have the form x̄ ± E with E = zα/2·σ/√n. To say that we wish to estimate the mean to within a certain number of units means that we want the margin of error E to be no larger than that number. Thus we obtain the minimum sample size needed by solving the displayed equation for n. To apply the formula we must have prior knowledge of the population in order to have an estimate of σ.
Example 8. Find the minimum sample size necessary to construct a 99% confidence interval for µ with a margin of error E = 0.2. Assume σ = 1.3. → z0.005 = 2.576, so n = (2.576)²(1.3)²/(0.2)² = 280.36, which we round up to 281, since it is impossible to take a fractional observation.
Example 9. An economist wishes to estimate, with a 95% confidence interval, the yearly income of welders with at least five years experience to within $1,000. He estimates that the range of incomes is no more than $24,000, so using the Empirical Rule he estimates σ to be about one-sixth as much, or about $4,000. → n = (1.960)²(4000)²/(1000)² = 61.47, rounded up to 62.
Example 9 hides a useful trick for the pizzeria: when nobody has measured σ, range ÷ 6 is a serviceable guess, because almost all of a normal population lies within three standard deviations either side of its mean.
± 1 franc at 95% is 68 pizzerias. ± 0.50 is 272. The margin is in the denominator squared: halve it and the bill quadruples.
There is a dilemma here: the formula for estimating how large a sample to take contains the number p̂, which we know only after we have taken the sample. There are two ways out. Typically the researcher will have some idea as to the value of the population proportion p: if last month 37% of all voters thought that state taxes are too high, we would use 0.37 for p̂ in the formula. The second approach is simply to replace p̂ in the formula by 0.5. This is because if p̂ is large then 1 − p̂ is small, and vice versa, which limits their product to a maximum value of 0.25, which occurs when p̂ = 0.5. This is called the most conservative estimate, since it gives the largest possible estimate of n.
Example 10. Find the necessary minimum sample size to construct a 98% confidence interval for p with a margin of error E = 0.05, (a) assuming that no prior knowledge about p is available; (b) assuming that prior studies suggest that p is about 0.1. → z0.01 = 2.326. (a) p̂ = 0.5: n = (2.326)²(0.5)(0.5)/(0.05)² = 541.03 → 542. (b) p̂ = 0.1: n = (2.326)²(0.1)(0.9)/(0.05)² = 194.77 → 195.
Example 11. A dermatologist wishes to estimate the proportion of young adults who apply sunscreen regularly before going out in the sun in the summer. Find the minimum sample size required to estimate the proportion to within three percentage points, at 90% confidence. → z0.05 = 1.645, p̂ = 0.5, E = 0.03: n = (1.645)²(0.25)/(0.03)² = 751.67 → 752.
The book interprets every interval the same way: "We are 98% confident that µ lies in the interval, in the sense that in repeated sampling 98% of all intervals constructed from the sample data in this manner will contain µ." The confidence belongs to the method, not to the one interval in front of you. Your Geneva interval is (18.43, 21.43). Which of these statements about it are correct?
µ is a fixed number; it does not have a 95% chance of being anywhere. The interval is the random thing — it moves from sample to sample, and 95% of the time it lands on µ.
The book's key takeaway for choosing a formula: is the population standard deviation σ known or unknown, and is the sample large or small? — and for a small sample, is the population normal? Walk the tree.
In Week 5 we asked how surprising 19.93 would be if the magazine's "CHF 21" were true, and found about one sample in twelve — unusual, not damning. Look at this week's interval: (18.43, 21.43) contains 21. The same verdict, read off the interval: the claim is not contradicted at the 95% level. A value outside the interval would be. Week 7 makes that rule precise — the hypothesis test, with α, the rejection region and the p-value — and the interval you built today is the shadow of the test you will run then.
This week's exercises are not invented: each one is reproduced exactly as it appears at the end of a section of the textbook, labelled with its section and number, so you can find it in the book. Your instructor puts each exercise on the slide; you work it here, by hand, with a calculator, the t table and the normal table. Check marks your number immediately, Hint nudges without giving it away, and the Solution button opens when the room does that exercise together.
About 100 minutes for all ten. Take zα/2 and tα/2 from Figure 12.3 (three decimals), keep σ/√n unrounded, and round the margin E the way the book's answer does — the tolerances allow for it. Parts that ask you to comment or verify are answered in the worked solution. Your answers are saved and reach your instructor's dashboard. Nothing here is graded.
One attempt per question, every answer explained. Your score syncs to your instructor's dashboard as a self-check but does not count toward your grade. Have a calculator and the t table to hand.
In Week 7 the interval becomes a verdict. This week: "µ is somewhere in (18.43, 21.43)." Next week: "a competitor claims µ = 21 — does the data contradict that?" The machinery is the same σ/√n and the same z and t; what changes is the question. Before that, the Week 6 homework: thirteen more exercises from the book, by hand.
Your business plan will quote the Geneva average. Quote it with its interval, choose the sample size that gives the margin you actually need — and say whether the extra visits are worth their Margheritas. This instalment is three sentences a lender could read.
📎 Everything saves automatically and reaches your instructor's dashboard. This is the sixth instalment of the individual project.
Applied Statistics · Week 6 · Estimation · Jan Erik Meidell · HEG Genève
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