Chapter 7: Confidence Intervals

7.1 How Much Does a Customer Spend?

Dana Whitfield is planning next year's budget. Two numbers drive most of it. The first is how much a customer spends on a typical visit. The second is how many customers would recommend Ridgeline to a friend, because word of mouth is how a small chain grows. She asks you for both.

You already know the survey answers. In Chapter 4. Describing Data, you found that the 600 surveyed customers spent an average of $8.48 per visit. In Chapter 1. Statistics, Data, and AI in Business, you saw that 407 of them (67.8%) would recommend Ridgeline. But Dana isn't budgeting for 600 people. She's budgeting for every customer who walks in next year. So her real question is: what do all Ridgeline customers spend on average, and what share of all of them would recommend us?

Nobody knows those population values exactly. In Chapter 6. Sampling and the Central Limit Theorem, you saw that a different sample of 600 customers would have given a slightly different average. This chapter turns that idea into a tool you'll use constantly: a range of plausible values for a population parameter, with a stated level of confidence. By the end, you'll be able to tell Dana something like "the average spend is between $8.27 and $8.68, with 95% confidence," and explain exactly what that sentence means.

7.2 The Point Estimate and Why It Isn't Enough

The simplest answer to Dana's question is the sample mean: $8.48. It's the best single guess you have for the population mean. A single number used this way is called a point estimate.

A point estimate is useful, but on its own it hides something important: how far off it might be. Consider two analysts who both report "average spend is $8.48."

• The first surveyed 600 customers.
• The second surveyed 5.

Both reports look identical. You'd trust the first far more, because you know from Chapter 6. Sampling and the Central Limit Theorem that averages from small samples bounce around much more than averages from large ones. A point estimate says nothing about that. It doesn't tell Dana whether the true average could plausibly be $8.00 or $9.00, and that difference matters. With about 1.56 million transactions a year, a 50-cent error in average spend is a budget error of roughly three-quarters of a million dollars.

What Dana needs is the point estimate plus a statement of its precision. That's what a confidence interval gives her.

7.3 From Standard Error to an Interval

You already have the key ingredient. In Chapter 6. Sampling and the Central Limit Theorem, you met the standard error: the standard deviation of the sampling distribution of a statistic. For a sample mean, it measures how far a typical sample mean lands from the population mean. You also learned that, thanks to the central limit theorem, sample means from reasonably large samples follow a sampling distribution that is close to normal and centered on the population mean.

Put those two facts together with the empirical rule from Chapter 5. Probability and Distributions. About 95% of the values in a normal distribution fall within about 2 standard deviations of its center. Here the "values" are possible sample means, the center is the population mean μ, and the standard deviation is the standard error. So:

In about 95% of all possible samples, the sample mean lands within about 2 standard errors of the population mean.

Now flip that sentence around. If the sample mean is usually within 2 standard errors of μ, then μ is usually within 2 standard errors of the sample mean. So if you start at your sample mean and reach out 2 standard errors in each direction, the range you mark off will contain μ for about 95% of all samples. That range is a confidence interval.

Every confidence interval in this book has the same shape:

estimate ± critical value × standard error

• The estimate is the point estimate: a sample mean or a sample proportion.
• The standard error measures how much that estimate varies from sample to sample.
• The critical value sets how many standard errors to reach out. It comes from a distribution (normal or t) and depends on how confident you want to be. For 95% with a normal distribution, it's 1.96, the precise version of "about 2."

The part after the ± is called the margin of error.

In symbols, for a mean:

x̄ ± t* × s / √n

where x̄ is the sample mean, s is the sample standard deviation, n is the sample size, s / √n is the standard error, and t* is the critical value. The next section explains why the critical value comes from a t distribution rather than the normal distribution.

7.4 Why t Instead of z

The standard error of a mean is σ / √n, where σ is the population standard deviation. There's a catch: you almost never know σ. If you knew everything about the population, you wouldn't need a sample. So you use the sample standard deviation s in its place.

That substitution adds a second source of uncertainty. The sample mean varies from sample to sample, and so does s. In a small sample, s can easily come out too small, by chance. When it does, the standard error looks smaller than it really is, and an interval built with 1.96 standard errors would be too narrow. It would miss μ more often than 5% of the time.

The fix, worked out by William Gosset at the Guinness brewery in 1908, is to use a slightly wider multiplier taken from a different distribution: the t distribution. It looks like the standard normal distribution (bell-shaped, centered on 0, symmetric), but it has heavier tails, which means more of its area is far from the center. Heavier tails mean larger critical values, which mean wider intervals. That extra width pays for not knowing σ.

How much heavier the tails are depends on how much information you have about spread, which is measured by the degrees of freedom. For a single mean, the degrees of freedom are n − 1.

Why n − 1 and not n? To calculate s, you first calculate the sample mean, and then you measure how far each value is from it. Those distances always add up to zero. So once you know n − 1 of them, the last one is fixed: it has to be whatever makes the total zero. Only n − 1 of the distances are free to vary, so only n − 1 pieces of information go into estimating spread. With three numbers whose mean is 5, if two of them are 4 and 7, the third must be 4. You had three numbers but only two free choices.

The table below shows the 95% critical value for several degrees of freedom. You can reproduce every number in it with =T.INV.2T(0.05, df), and the last row with =NORM.S.INV(0.975).

Sample size n	Degrees of freedom	95% critical value t*
5	4	2.776
10	9	2.262
25	24	2.064
100	99	1.984
600	599	1.964
(σ known)		z* = 1.960

With 5 observations, you reach out 2.78 standard errors instead of 1.96. That's a big penalty for having so little information. By 100 observations, the difference is small, and at 600 it's almost gone: 1.964 versus 1.960. With large samples, t and z give practically the same interval. Statistical software uses t for means anyway, because it's correct at every sample size, and this book does the same.

7.5 Prepare the Data in Excel

Open the Ridgeline workbook. You'll work with the Survey Table, which has one row per survey response. Two columns matter in this chapter:

• spend: dollars spent on the visit, a continuous numeric variable.
• would_recommend: Yes or No, a categorical variable with two values.

Before building any interval, check the data. Confirm that spend has no blanks and no text: =COUNT(Survey[spend]) should return 600, the same as =ROWS(Survey[spend]). Confirm that would_recommend holds only "Yes" and "No": =COUNTIF(Survey[would_recommend],"Yes")+COUNTIF(Survey[would_recommend],"No") should also return 600. A quick histogram of spend (Insert > Insert Statistic Chart > Histogram) shows a roughly symmetric, single-peaked shape, which matches the skewness of −0.03 you found earlier. That shape matters later, when you check the interval's assumptions.

Now make a working sheet. Right-click the Survey sheet tab, choose Insert… > Worksheet, and rename the new sheet CI (double-click its tab and type). You'll build both intervals on this sheet.

7.6 How the Interval Works: Mean Spend

This is the core of the chapter. You'll build a 95% confidence interval for the mean spend of all Ridgeline customers, one cell at a time, so you can see every piece.

7.6.1 The worked example, cell by cell

On the CI sheet, type the labels in column A and the formulas in column B:

Cell	Label (column A)	Formula (column B)	Result
B1	Sample mean	=AVERAGE(Survey[spend])	8.475
B2	Sample SD	=STDEV.S(Survey[spend])	2.584
B3	n	=COUNT(Survey[spend])	600
B4	Standard error	=B2/SQRT(B3)	0.1055
B5	Confidence level	0.95 (typed in)	0.95
B6	Critical value t*	=T.INV.2T(1-B5,B3-1)	1.9639
B7	Margin of error	=B6*B4	0.207
B8	Lower limit	=B1-B7	8.268
B9	Upper limit	=B1+B7	8.682

Read the cells as a story:

1. B1 to B3 summarize the sample: the average customer in the survey spent $8.475, individual spends typically varied by about $2.58 from that average, and there are 600 responses.
2. B4 is the standard error. Individual customers vary by $2.58, but an average of 600 customers varies much less: by about $0.11 from sample to sample. Dividing by √600 (about 24.5) is what shrinks it.
3. B6 is the critical value for 95% confidence with 599 degrees of freedom. 1-B5 turns the confidence level into the 0.05 that T.INV.2T expects. Using a cell for the level, instead of typing 0.05 into the formula, lets you change the level later and watch everything update.
4. B7 is the margin of error: 1.964 standard errors, or about 21 cents.
5. B8 and B9 are the interval: $8.27 to $8.68.

7.6.2 Check it with CONFIDENCE.T

Excel has a function that computes the margin of error in one step. In B10, type the label "Margin (CONFIDENCE.T)" in A10 and this formula in B10:

=CONFIDENCE.T(1-B5,B2,B3)

It takes the significance (0.05), the sample standard deviation and the sample size, and returns 0.2072, the same margin as B7. This is the book's "two tools, one answer" habit at small scale: you computed the margin from its parts, and Excel's built-in function agrees.

7.6.3 Changing the confidence level

Change B5 to 0.90, then to 0.99, and watch the interval change. Here is what you'll see:

Confidence level	Critical value t*	Margin of error	Interval
90%	1.647	$0.174	$8.30 to $8.65
95%	1.964	$0.207	$8.27 to $8.68
99%	2.584	$0.273	$8.20 to $8.75

Higher confidence means a wider interval. That's the basic trade-off. To be more sure your net catches the true mean, you need a bigger net. A 100% confidence interval would have to run from the smallest possible spend to the largest, which is useless. Nobody can be completely sure and precise from a sample.

95% is the business and scientific convention. It's a sensible balance, but it isn't a law of nature. A manager deciding on something cheap and reversible might be happy with 90%. A decision that is expensive or hard to undo might call for 99%.

7.6.4 What "95% confident" actually means

This is the most misunderstood idea in introductory statistics, so take it slowly.

The population mean μ is a fixed number. It isn't random. You don't know it, but it doesn't change from sample to sample. What changes is your interval. Each possible sample of 600 customers gives a different mean, a different standard deviation, and so a different interval. Some of those intervals contain μ, and some miss it.

The 95% describes that whole collection of possible intervals. If you repeated the survey over and over, drawing a new random sample each time and building a 95% interval each time, about 95% of those intervals would contain μ. The 95% is the method's long-run hit rate.

Your one interval, $8.27 to $8.68, either contains μ or it doesn't. You can't know which. But you used a method that works 95% of the time, and that's why you can say you're "95% confident" in it. It's like trusting a forecaster who is right 95% of the time: you never know whether today is one of the misses, but you have good reason to trust the forecast.

7.6.5 See the 95% for yourself

You can watch the long-run hit rate happen with a simulation like the ones in Chapter 6. Sampling and the Central Limit Theorem. Treat the 600 survey spends as if they were the whole population, so the "true" mean is known: μ = 8.475. Then draw many small samples from it, build an interval from each, and count how many contain 8.475.

1. Insert a new sheet named Coverage.

2. In A1 type Lower, in B1 Upper, in C1 Hit?.

3. In A2, enter this formula, which draws 25 random spends, computes their mean and margin, and returns the lower and upper limits side by side:

   =LET(x,INDEX(Survey[spend],RANDARRAY(25,1,1,600,TRUE)),m,AVERAGE(x),e,CONFIDENCE.T(0.05,STDEV.S(x),25),HSTACK(m-e,m+e))

4. In C2, enter =AND(A2<=8.475,B2>=8.475). It shows TRUE if the interval contains the true mean.

5. Select A2:C2 and drag the fill handle down to row 101, giving 100 intervals.

6. In E1, count the hits: =COUNTIF(C2:C101,TRUE).

Press F9 to draw 100 new samples. The count changes every time, but it usually lands somewhere in the low-to-high 90s, centered near 95. A few intervals miss, through no fault of the method. That's the 5%.

7.6.6 Assumptions behind the t interval

A t interval for a mean gives honest 95% coverage when these conditions hold:

1. The sample is random, or close to it. The interval accounts for random sampling error only. It does nothing about bias. If Ridgeline had surveyed only its happiest customers, the interval would be precise and wrong.
2. Observations are independent. One customer's spend doesn't affect another's. That's reasonable for a survey of different customers on different days.
3. The sampling distribution of the mean is close to normal. This holds if the data themselves are roughly normal, or if the sample is large enough for the central limit theorem to do its work. A common rule of thumb is that n of 30 or more is enough unless the data are strongly skewed or have extreme outliers. With 600 roughly symmetric spends, this condition is easily met.

With small samples (say, under 30), look at a histogram first. If it's strongly skewed or has an extreme outlier, the t interval may not be trustworthy, and you should say so.

7.7 A Confidence Interval for a Proportion

Dana's second question is about a Yes/No variable: what share of all Ridgeline customers would recommend it? The parameter here is a population proportion, usually written p. The point estimate is the sample proportion, p̂ ("p-hat"): 407 Yes answers out of 600, or 0.678.

The interval has the same shape as before: estimate ± critical value × standard error. Two pieces change.

• The standard error of a proportion is √(p̂(1 − p̂) / n). You met this formula in Chapter 6. Sampling and the Central Limit Theorem, where the standard error of the recommend proportion for n = 600 came out to 0.019. Notice that it doesn't need a separate standard deviation: for a Yes/No variable, the spread is determined by the proportion itself.
• The critical value comes from the normal distribution, not the t distribution. There's no separately estimated standard deviation here, so the reason for using t doesn't apply. For 95%, the critical value is z* = NORM.S.INV(0.975) = 1.960. The 0.975 leaves 2.5% in the upper tail, and by symmetry another 2.5% in the lower tail.

7.7.1 The worked example, cell by cell

On the CI sheet, starting in D1 (labels) and E1 (formulas):

Cell	Label (column D)	Formula (column E)	Result
E1	Yes count	=COUNTIF(Survey[would_recommend],"Yes")	407
E2	n	=COUNTA(Survey[would_recommend])	600
E3	Sample proportion p̂	=E1/E2	0.678
E4	Standard error	=SQRT(E3*(1-E3)/E2)	0.0191
E5	Critical value z*	=NORM.S.INV(0.975)	1.960
E6	Margin of error	=E5*E4	0.0374
E7	Lower limit	=E3-E6	0.641
E8	Upper limit	=E3+E6	0.716

The 95% confidence interval for the share of all customers who would recommend Ridgeline runs from 0.641 to 0.716, or about 64% to 72%. That's the inferential statement you saw previewed in Chapter 1. Statistics, Data, and AI in Business. Now you know where it came from.

To check the margin, use CONFIDENCE.NORM. It's built for means with a known σ, but for a proportion the "σ" is √(p̂(1 − p̂)), so you can feed it that:

=CONFIDENCE.NORM(0.05,SQRT(E3*(1-E3)),E2)

It returns 0.0374, matching E6.

7.7.2 Conditions for the proportion interval

This formula (often called the Wald or normal-approximation interval) relies on the sampling distribution of p̂ being close to normal. That's true when:

1. The sample is random and the observations are independent, as for the mean.
2. There are enough of each answer. A common rule is at least 10 "successes" and at least 10 "failures" in the sample. Here there are 407 Yes and 193 No, far more than enough.

When either count is small, for example if you're estimating the share of customers who complain and only 4 of 80 did, the simple formula can give poor intervals, sometimes even below 0. Software offers better methods for those cases, which you'll meet when you run this in R.

7.8 Sample Size and Precision

7.8.1 Why 600 responses beat 25

Suppose the survey had stopped after the first 25 responses (R0001 through R0025). You can build that interval yourself. In a spare area of the CI sheet, use TAKE to grab the first 25 spends:

• Mean: =AVERAGE(TAKE(Survey[spend],25)) → 9.506
• SD: =STDEV.S(TAKE(Survey[spend],25)) → 3.022
• Margin: =CONFIDENCE.T(0.05,STDEV.S(TAKE(Survey[spend],25)),25) → 1.247

The 95% interval from 25 responses runs from $8.26 to $10.75. From all 600, it runs from $8.27 to $8.68.

Three things stand out.

• The small-sample interval is about six times wider. It spans about $2.50, which is too vague to budget with.
• Its point estimate is noticeably off. The first 25 customers happened to spend more, averaging $9.51. That's exactly the sampling variability you studied in Chapter 6. Sampling and the Central Limit Theorem.
• Its interval still contains the larger sample's estimate. The wide interval was honest about its own uncertainty. It didn't claim more precision than 25 responses could give.

The margin of error shrinks with √n, not with n. To cut the margin in half, you need four times as many observations. Going from 600 to 2,400 responses would shrink the spend margin from about $0.21 to about $0.10. Each extra bit of precision costs more than the last, which is why surveys rarely go beyond a few thousand responses.

7.8.2 Planning a sample size

Often you know the precision you need before you collect any data. Suppose Dana says: "Next year, I want to know average spend to within 25 cents, with 95% confidence." How many customers should you survey?

Start from the margin of error, E = z* × σ / √n, and solve it for n:

n = (z* × σ / E)²

Then round up, because rounding down would give a margin slightly bigger than the target. Two practical points:

• Use z*, not t*. The t critical value depends on n, which is the thing you're solving for. With samples in the hundreds, the difference between t* and z* is negligible.
• You need a guess for σ. Use the standard deviation from a previous survey or a small pilot. This year's survey gives 2.584.

In Excel:

=ROUNDUP((NORM.S.INV(0.975)*STDEV.S(Survey[spend])/0.25)^2,0)

returns 411. A survey of 411 customers should give a margin of error of about $0.25. This year's 600 gave a margin of $0.21, which is consistent: more responses than needed, so a slightly smaller margin.

For a proportion, the formula is the same with √(p(1 − p)) in place of σ:

n = (z* / E)² × p(1 − p)

Suppose Marcus wants next year's recommend share to within ±3 percentage points (E = 0.03). If you have no idea what p will be, use p = 0.5, which gives the largest possible p(1 − p) and so the safest (largest) sample size:

=ROUNDUP((NORM.S.INV(0.975)/0.03)^2*0.5*(1-0.5),0) → 1,068

If you trust this year's 0.678 as a guide, use it instead (here, from cell E3):

=ROUNDUP((NORM.S.INV(0.975)/0.03)^2*E3*(1-E3),0) → 932

The 1,068 is why national polls so often survey "about 1,000 people": it's the sample size that guarantees ±3 points at 95% whatever the true proportion is.

7.9 Intervals by Region

Marcus wants to know whether average spend differs by region. Confidence intervals give you a useful first look. The survey doesn't record region, so you'll look it up from the Stores Table.

Add a region column to the Survey Table.

1. Go to the Survey sheet and click the first empty cell to the right of the Table's header row (the cell right of would_recommend).
2. Type region and press Enter. Excel extends the Table to include the new column.
3. In the first data cell of the new column, type =XLOOKUP([@store_id],Stores[store_id],Stores[region]) and press Enter. The Table fills the formula down all 600 rows.

Build the region summary. On the CI sheet, starting in G1, type the headers Region, n, Mean, SD, Margin, Lower, Upper. In G2:G4, type Metro, Mountain and Suburban. Then in row 2:

Column	Formula in row 2
n (H)	=COUNTIFS(Survey[region],G2)
Mean (I)	=AVERAGEIFS(Survey[spend],Survey[region],G2)
SD (J)	=STDEV.S(FILTER(Survey[spend],Survey[region]=G2))
Margin (K)	=CONFIDENCE.T(0.05,J2,H2)
Lower (L)	=I2-K2
Upper (M)	=I2+K2

Copy row 2 down to rows 3 and 4. FILTER returns only the spends for that row's region, and STDEV.S measures their spread. Excel has no STDEV.SIFS function, so FILTER fills that gap.

Region	n	Mean spend	95% CI
Metro	217	$8.33	$7.97 to $8.69
Mountain	190	$8.40	$8.03 to $8.76
Suburban	193	$8.72	$8.36 to $9.08

Each region's interval is wider than the overall one (about ±$0.36 versus ±$0.21), because each is based on about a third of the data.

Chart it. Select G1:G4 and I1:I4 (hold Ctrl to select the second range), then choose Insert > Insert Column or Bar Chart > Clustered Column. Click the chart, then Chart Design > Add Chart Element > Error Bars > More Error Bar Options. Under Error Amount, choose Custom, click Specify Value, and set both the positive and negative error values to =CI!$K$2:$K$4. Each bar now shows its 95% interval.

What does the chart say? Suburban customers spent the most in the sample, $8.72 versus about $8.33 to $8.40 elsewhere. But all three intervals overlap substantially. Metro's interval reaches up to $8.69, and Suburban's reaches down to $8.36. The data are consistent with all three regions having the same average spend, or with real differences of a few tens of cents. This sample can't tell those apart.

7.10 Describe the Data to AI

Now do the same analysis with R for Excel, so you can compare it with your hand-built version. Open the app in JOLT. In 1. Data, click Add a table… and choose Survey. The app lists the columns and their types. If you added the region column, it appears too; that's fine.

In 2. Ask your AI, type your data description. The app already tells the AI each column's name, type and range. Your job is to say what the columns mean and what you want:

Each row of Survey is one customer's response to Ridgeline Coffee's 2025 survey
(600 responses, one per customer visit). spend is the dollars the customer spent
on that visit. would_recommend is Yes or No: whether the customer would recommend
Ridgeline to a friend.
Question: What is the average spend per visit for ALL Ridgeline customers, and what
share of ALL customers would recommend us? I want a range of plausible values for
each, not just the sample numbers.
I'm in an intro business statistics course and have learned descriptive statistics,
the normal distribution, standard error, and confidence intervals for a mean (using t)
and for a proportion.
Please compute a 95% confidence interval for the mean of spend and a 95% confidence
interval for the proportion who answered Yes to would_recommend, send the key results
to Excel with excel_out(), and explain how to read them.

Click Copy prompt, paste it into your AI assistant, and send it.

7.11 Read the Script

Before you run anything, read what the AI sent. You're checking three things: which data it uses, which method it runs, and where the results go. A script for this usually looks like the one below.

# 95% confidence interval for the average spend (uses the t distribution)
spend_ci <- t.test(Survey$spend, conf.level = 0.95)
excel_out(spend_ci, "95% CI for mean spend")

# Count the Yes answers and the total number of answers
yes <- sum(Survey$would_recommend == "Yes")
n <- nrow(Survey)

# 95% confidence interval for the share who would recommend
rec_ci <- prop.test(yes, n, conf.level = 0.95)
excel_out(rec_ci, "95% CI for share who would recommend")

In plain language:

• The first block runs t.test on the spend column of the Survey data frame. t.test is R's function for t procedures, and it reports a confidence interval for the mean along with other results. conf.level = 0.95 asks for 95%. Then excel_out sends the whole result to an output card titled "95% CI for mean spend."
• The second block counts how many rows have would_recommend equal to "Yes" (407) and how many rows there are in total (600). Check that the script compares with "Yes" spelled exactly as it is in the data. A script that tested for "yes" would count zero and still run.
• The third block runs prop.test on 407 out of 600, which reports a confidence interval for a proportion, and sends it to a second card.

In the AI's version, check that it uses spend, counts "Yes," asks for 95% and ends with excel_out. If it used binom.test instead of prop.test, that's fine too; you'll see below why the choice matters a little.

7.12 Run It in R for Excel

1. Paste the AI's whole reply into 3. Run R code. The app keeps just the code.
2. Click Run.
3. Two output cards appear under the results. On the first, click New sheet. Do the same for the second.

If you get an error instead, click Copy a prompt to fix it, paste that prompt back into your AI, and run the corrected code. The fix prompt already contains the code, the error message and your data.

7.13 Interpret and Verify

7.13.1 Reading the mean output

The first sheet has a two-column block headed "One Sample t-test." The rows you need are:

• mean of x: 8.475, the point estimate.
• Lower 95% and Upper 95%: 8.268 and 8.682, the confidence interval.

These match cells B8 and B9 on your CI sheet to every decimal shown. Two tools, one answer.

The block also contains rows you can ignore for now: "Hypothesized mean" (0), "t" (about 80), "df" (599), "P-value" (a tiny number) and "Alternative." t.test always tests something, and with no target given it tests whether the mean spend could be $0. Nobody thinks it could, so that test tells you nothing. You'll learn to read those rows in Chapter 8. Hypothesis Testing, where the hypothesized value is chosen to answer a real question.

7.13.2 Reading the proportion output, and why it differs slightly

The second sheet's block is headed something like "1-sample proportions test with continuity correction." The estimate row, p, shows 0.678. But look at the limits:

Method	Lower	Upper
Your Excel formula (Wald)	0.641	0.716
R prop.test (Wilson, with continuity correction)	0.639	0.715
R binom.test (exact, Clopper–Pearson)	0.639	0.716

The R interval differs from yours in the third decimal. Neither tool made a mistake. They used different formulas for the same idea.

• Your Excel formula is the simplest one: p̂ ± z* × √(p̂(1 − p̂)/n). It treats the sampling distribution of p̂ as exactly normal and centered on p̂.
• prop.test uses the Wilson method, with a small continuity correction that allows for the fact that counts are whole numbers. It's more accurate than the simple formula, especially when n is small or p̂ is near 0 or 1, and its interval isn't exactly centered on p̂.
• binom.test computes an exact interval directly from the binomial distribution in Chapter 5. Probability and Distributions.

With 600 responses and a proportion near the middle, all three agree to about two decimals: 64% to 72%. In a business report, that's the same answer. With a sample of 40 and a proportion of 0.95, they would disagree noticeably, and the simple formula would be the one to distrust.

This is a general lesson about AI-written analysis. When an AI's R output and your Excel check differ slightly, the usual reason is a different method or a different default setting, not an error. When they differ a lot, something is wrong: a different column, a filter, a typo in "Yes." Either way, find out why before you report.

If you want R to match your Excel formula exactly, for instance so a reviewer can trace every number, ask for that method by name.

Check the AI's explanation against the table above. If it says one of the two answers is "wrong," push back: both are legitimate methods, and they agree closely here.

7.13.3 The conclusion, in business language

Numbers on a sheet aren't the end. Dana needs sentences she can act on:

Average spend. Surveyed customers spent $8.48 per visit on average. We are 95% confident that the average for all Ridgeline customers is between $8.27 and $8.68. For budgeting, a planning figure of about $8.50 per visit is well supported. A conservative budget could use $8.27.

Recommendation. 67.8% of surveyed customers said they would recommend Ridgeline. We are 95% confident that between about 64% and 72% of all customers would. Roughly two in three customers are likely promoters, and about one in three is not.

By region. Suburban customers spent the most in the survey ($8.72 versus about $8.35 elsewhere), but the regions' intervals overlap a lot, so the survey doesn't show a clear difference in spend by region.

Each statement gives the estimate, the interval and what it means for a decision, and each talks about all customers, not just the 600 surveyed. The region sentence says what the data can't show, which is just as useful to a manager as what it can. Notice the rounding, too: two decimals for dollars and whole percentages for proportions.

7.14 Summary

A point estimate, such as the sample mean of $8.48, is the best single guess for a population parameter, but it says nothing about how far off it might be. A confidence interval adds that information. Every interval in this book has the form estimate ± critical value × standard error, and the part after the ± is the margin of error. For a mean, the critical value comes from the t distribution with n − 1 degrees of freedom, because the standard error uses the sample standard deviation in place of the unknown σ. For a proportion, the critical value comes from the standard normal distribution. In Excel, T.INV.2T, NORM.S.INV, CONFIDENCE.T and CONFIDENCE.NORM let you build and check both kinds of interval cell by cell.

The confidence level describes the method, not any single interval: a 95% method captures the true parameter in 95% of samples. Higher confidence gives wider intervals, and larger samples give narrower ones, with the margin shrinking with √n. You can plan the sample size needed for a target margin before collecting data. Ridgeline's 600 responses estimate average spend at $8.27 to $8.68 and the share who would recommend at about 64% to 72%. When R and Excel give slightly different intervals, the usual reason is a different method, such as prop.test's Wilson interval. Find out why before you report, and then say what the interval means in a sentence a manager can use.