7.1 A Single Population Mean using the Normal Distribution
SLO 4
Demonstrate an understanding of, and ability to use, basic ideas of statistical processes, including hypothesis tests and confidence interval estimation.
This is the first confidence interval you build: a point estimate, an error bound, and the five steps that turn one sample mean into a range for the population mean. It also pins down what a confidence level actually promises, which every later interval and test leans on.
SLO 5
Identify appropriate statistical techniques and use technology-based statistical analysis to describe, interpret, and communicate results.
Picking the normal curve here rests on one condition, that sigma is known, and invNorm or a ZInterval supplies the critical value. The last step is always a plain sentence naming the population, the units, and the confidence.
Learning Objectives
By the end of this section, you will be able to:
- explain what a confidence interval estimates and what the confidence level actually promises;
- find the \(z\)-score that matches a stated confidence level, and calculate the error bound for a population mean when the population standard deviation is known;
- construct a confidence interval for a population mean and write the interpretation in the words of the problem;
- predict how changing the confidence level or the sample size changes the width of the interval;
- work backwards from a reported interval to recover the sample mean and the error bound, and calculate the sample size a target error bound requires.
Almost every number a survey reports is a guess. A poll says the average American household watches 151 hours of television a month; a lab says the average newborn elephant weighs 244 pounds. Nobody weighed every newborn elephant. Somebody weighed fifty of them, took the average, and reported it. The honest version of that report is not a single number — it is a range, plus a statement of how confident anyone is that the range caught the truth. That range is a confidence interval, and this section is about building one.
The population mean is one fixed number that never budges. It is the interval that jumps around from sample to sample, sometimes catching \(\mu\) and sometimes missing it.
The whole thing rests on the central limit theorem from the last chapter. Sample means follow an approximately normal distribution centred at the population mean \(\mu\), with a standard deviation of \(\frac{\sigma}{\sqrt{n}}\). Because we know the shape of that curve, we know how far a typical sample mean lands from \(\mu\) — and if we know how far the sample mean usually is from \(\mu\), we can turn that around and say how far \(\mu\) probably is from the sample mean we actually got.
7.1.1 Calculating the Confidence Interval
A point estimate is a single value calculated from sample data that is used as the best guess for an unknown population parameter. When the parameter is a population mean \(\mu\), the point estimate is the sample mean \(\overline{x}\).
Definition 7.1.1 - A point estimate: a whole sample collapses to the one number that stands in for the unknown population mean.
The error bound for a population mean, written \(EBM\), is the distance from the point estimate to either endpoint of the confidence interval. The interval is the point estimate plus or minus the error bound, so \(EBM\) is exactly half the interval's total width.
When a poll reports "52%, plus or minus 3 points", that 3 is the error bound. The news usually calls it the margin of error. Same idea, different name.
The size of the error bound depends on how sure we want to be. That level of certainty is the confidence level, abbreviated CL, and it is chosen by whoever is building the interval — not calculated from the data. Most people pick 90% or higher, because the point of the exercise is to be reasonably certain of the conclusion.
Definition 7.1.2 - The error bound: one length reaches out both ways from the point estimate, so it is exactly half the interval's width.
The confidence level \(CL\) is the percent of confidence intervals that contain the true population parameter when the sampling and interval-building process is repeated many times.
That definition is worded carefully, and the wording matters. It is tempting to say that the confidence level is the probability that your particular interval contains \(\mu\). That is not quite right. Once you have collected your sample and calculated your interval, that interval either contains \(\mu\) or it does not — there is no chance left in it. The 95% describes the method, not the result. If a hundred different researchers each drew their own sample and each built a 95% confidence interval, about ninety-five of those hundred intervals would contain the true \(\mu\) and about five would miss it. You do not get to know which kind you have.
Alongside the confidence level there is a second probability called alpha, written \(\alpha\). It is the probability that the interval does not contain the unknown population parameter — the leftovers.
$$ \alpha + CL = 1 $$So a 95% confidence level means \(\alpha = 0.05\), and a 90% confidence level means \(\alpha = 0.10\).
Every confidence interval in this chapter is built from the same two ingredients: the sample mean \(\overline{x}\) as the best guess for \(\mu\), and the error bound \(EBM\) as the measure of how far off that guess might be. This is the method to reach for whenever the population standard deviation \(\sigma\) is known. Put together, the two ingredients always make the same shape — the point estimate, then the same distance out on each side of it:
$$ (\text{point estimate} - \text{error bound},\ \text{point estimate} + \text{error bound}) = (\overline{x} - EBM,\ \overline{x} + EBM) $$Definition 7.1.3 - The confidence level describes the method: build twenty 90% intervals and about two of them miss the population mean.
Mei Nakashima has data from a sample. Their sample mean is 15, and the error bound for the mean is 3.2. What is the confidence interval estimate for the population mean they should report?
Solution — an interval from a point estimate
Step 1 — Apply the form of the interval.
$$ (\overline{x} - EBM,\ \overline{x} + EBM) = (15 - 3.2,\ 15 + 3.2) $$Step 2 — Do the arithmetic.
$$ (11.8,\ 18.2) $$Answer: \((11.8, 18.2)\). We estimate with the stated confidence that the true population mean is between 11.8 and 18.2.
Now for where the error bound actually comes from. Suppose a sample has mean \(\overline{x} = 10\) and we have built the 90% confidence interval \((5, 15)\), so \(EBM = 5\). To get a 90% confidence interval, we have to capture the central 90% of the probability under the normal curve. If we take the central 90%, we leave out \(\alpha = 10\%\) altogether — 5% in the left tail and 5% in the right tail.
Figure 7.1.1 - The same 95% confidence interval drawn twice: on the sample-mean scale above and on the z-scale below.
The arithmetic behind that picture is the whole method in four lines:
$$ \begin{aligned} \mu &= \overline{X} \pm z_{\frac{\alpha}{2}}\left(\frac{\sigma}{\sqrt{n}}\right) \\ &= 2 \pm 1.96(0.1) \\ &= 2 \pm 0.196 \\ 1.804 &\leq \mu \leq 2.196 \end{aligned} $$To capture the central 90% instead, we go out 1.645 "standard deviations" on either side of the sample mean. The value 1.645 is the \(z\)-score from a standard normal distribution that puts an area of 0.90 in the centre, 0.05 in the far left tail, and 0.05 in the far right tail.
Averaging ten measurements smooths out the flukes; averaging a thousand smooths them out more. The \(\sqrt{n}\) in the denominator is that smoothing, written down.
The word "standard deviation" in that sentence is doing something specific, and getting it wrong is the most common way to wreck this calculation. The standard deviation has to be the one that applies to the thing we are estimating. We are estimating a mean, so we need the standard deviation of sample means, which the central limit theorem tells us is \(\frac{\sigma}{\sqrt{n}}\) — not \(\sigma\) by itself. That quantity has its own name, the standard error of the mean, precisely so nobody confuses it with the population standard deviation \(\sigma\).
In summary, here is what the central limit theorem gives us:
- \(\overline{X}\) is normally distributed, that is, \(\overline{X} \sim N\!\left(\mu_X, \frac{\sigma}{\sqrt{n}}\right)\).
- When the population standard deviation \(\sigma\) is known, we use a normal distribution to calculate the error bound.
To construct a confidence interval estimate for an unknown population mean, we need data from a random sample. The steps are always the same five:
- Calculate the sample mean \(\overline{x}\) from the sample data. In this section the population standard deviation \(\sigma\) is already known.
- Find the \(z\)-score that corresponds to the confidence level.
- Calculate the error bound \(EBM\).
- Construct the confidence interval.
- Write a sentence that interprets the estimate in the context of the problem, in the words of the problem.
The rest of this section walks each step in more detail, then runs the whole process on real data.
Suppose we have collected data from a sample. We know the sample mean but we do not know the mean for the entire population. The sample mean is seven, and the error bound for the mean is 2.5. Find the confidence interval, and state the interpretation if the confidence level is 95%.
Solution — reading an interval off a point estimate
Step 1 — Write down what we have.
$$ \overline{x} = 7 \quad \text{and} \quad EBM = 2.5 $$Step 2 — Apply the form of the interval. The interval is the point estimate plus or minus the error bound.
$$ (\overline{x} - EBM,\ \overline{x} + EBM) = (7 - 2.5,\ 7 + 2.5) = (4.5,\ 9.5) $$Step 3 — Say it in words. If the confidence level is 95%, we say, "We estimate with 95% confidence that the true value of the population mean is between 4.5 and 9.5."
Answer: The confidence interval is \((4.5, 9.5)\).
7.1.2 Finding the z-score for the Stated Confidence Level
The critical value \(z_{\frac{\alpha}{2}}\) is the \(z\)-score that has an area of \(\frac{\alpha}{2}\) to its right under the standard normal curve. Equivalently, the area to its left is \(1 - \frac{\alpha}{2}\).
invNorm wants the area below the value, not the tail above it. For a 95% interval that is 0.975, not 0.025. The last two arguments stay 0 and 1 because this is the standard normal curve.
Work an example of the notation before using it. When \(CL = 0.95\), we get \(\alpha = 0.05\) and \(\frac{\alpha}{2} = 0.025\), so we write \(z_{\frac{\alpha}{2}} = z_{0.025}\). The area to the right of \(z_{0.025}\) is 0.025, which means the area to the left is \(1 - 0.025 = 0.975\). Looking that up gives \(z_{\frac{\alpha}{2}} = z_{0.025} = 1.96\), using a calculator, a computer, or a standard normal probability table.
On a TI-83/83+/84/84+: invNorm(0.975, 0, 1) = 1.96.
The three confidence levels you will meet most often have critical values worth recognising on sight: \(z = 1.645\) for 90%, \(z = 1.96\) for 95%, and \(z = 2.326\) for 98%. Every one of them comes from the same two moves — split \(\alpha\) in half, then ask for the \(z\)-score with that much area above it.
Try it in rāSHio
To get a critical value without a calculator, open rāSHio and choose Distributions → Normal, set the mean to 0 and the standard deviation to 1, and ask for the value with an area of 0.975 below it — the panel returns the 1.96 this section looks up, and swapping in 0.95 or 0.99 gives the 1.645 and 2.326 the other two confidence levels need. The walkthrough beside this note runs the same panel in the opposite direction, entering a cutoff and reading the area; the inverse read is the same two fields, filled the other way round.
Figure 7.1.2 — The rāSHio Normal distribution panel: Distributions → Normal. The walkthrough sets a mean and a standard deviation and reads an area off the curve. It uses its own demonstration numbers, not this section's.
When the population standard deviation \(\sigma\) is known, we use the standard normal distribution \(Z \sim N(0, 1)\) to find the error bound. What we are looking for is the value of \(z\) that puts an area equal to the confidence level (written as a decimal) in the middle of the standard normal curve.
The confidence level \(CL\) is the area in the middle. Since \(CL = 1 - \alpha\), the leftover area \(\alpha\) is split equally between the two tails, so each tail contains an area of \(\frac{\alpha}{2}\).
Definition 7.1.4 - The critical value: as the confidence level climbs, the two boundaries slide outward to leave alpha/2 in each tail.
Find the critical value \(z_{\frac{\alpha}{2}}\) for a 99% confidence level.
Solution — the 99% critical value
Step 1 — Find \(\alpha\). The confidence level is \(CL = 0.99\), so
$$ \alpha = 1 - CL = 1 - 0.99 = 0.01. $$Step 2 — Split it between the two tails.
$$ \frac{\alpha}{2} = 0.005 $$Step 3 — Ask for the \(z\)-score with 0.005 of the area above it. That is the same as the \(z\)-score with \(1 - 0.005 = 0.995\) of the area below it, so run invNorm(0.995, 0, 1).
Answer: \(z_{\frac{\alpha}{2}} = 2.576\). It is larger than the 1.96 used for a 95% interval, which is what we should expect — being more confident means reaching further out into the tails.
7.1.3 Writing the Interpretation
An interval by itself does not say anything. The sentence that goes with it has to do three things: state the confidence level, name the population parameter being estimated, and give both endpoints with their units. The template is
"We estimate with ____% confidence that the true population mean (include the context of the problem) is between ____ and ____ (include appropriate units)."
Fill in all four blanks every time. An answer that reports \((67.18, 68.82)\) and stops has skipped the part that carries the meaning, because a bare pair of numbers does not say what they measure, how sure anybody is, or what units they are in. The two worked examples that follow run the full five-step process end to end, and each one closes with that sentence written out.
Suppose average pizza delivery times are normally distributed with an unknown population mean and a population standard deviation of six minutes. A random sample of 28 pizza delivery restaurants is taken and has a sample mean delivery time of 36 minutes. Find a 90% confidence interval estimate for the population mean delivery time.
Solution — pizza delivery at 90%
Step 1 — Collect what we know. \(\overline{x} = 36\), \(\sigma = 6\), \(n = 28\), \(CL = 0.90\).
Step 2 — Find the critical value. For 90% confidence, \(\alpha = 0.10\) and \(\frac{\alpha}{2} = 0.05\), so \(z_{0.05} = 1.645\).
Step 3 — Calculate the error bound.
$$ EBM = z_{\frac{\alpha}{2}}\left(\frac{\sigma}{\sqrt{n}}\right) = 1.645\left(\frac{6}{\sqrt{28}}\right) = 1.645(1.1339) = 1.8653 $$Step 4 — Build the interval.
$$ (36 - 1.8653,\ 36 + 1.8653) = (34.13,\ 37.87) $$Answer: \((34.13, 37.87)\). We estimate with 90% confidence that the true mean pizza delivery time is between 34.13 and 37.87 minutes.
Adaeze Nwosu teaches statistics, and scores on her exams are normally distributed with an unknown population mean and a population standard deviation of three points. She takes a random sample of 36 scores, which gives a sample mean score of 68.
Find a 90% confidence interval for the true (population) mean of statistics exam scores.
Solution — statistics exam scores at 90%
Step 1 — Collect what we know. We are given \(\sigma = 3\), \(n = 36\), and \(\overline{x} = 68\).
Step 2 — Find the \(z\)-score for a 90% confidence level. For a 90% confidence interval, picture an area of 0.90 centred under the normal curve. The remaining area for the two tails is 0.10, so each tail holds 0.05. The \(z\)-score that cuts off an area of 0.05 in the left tail is 1.645.
Figure 7.1.3 - The central 90% of a normal curve: on the z-scale the area 0.90 runs from -1.645 to 1.645, leaving 0.05 in each tail.
Step 3 — Write the formula. The confidence interval for an unknown population mean, when the population standard deviation is known, is
$$ \overline{X} - z_{\frac{\alpha}{2}}\left(\frac{\sigma}{\sqrt{n}}\right) \leq \mu \leq \overline{X} + z_{\frac{\alpha}{2}}\left(\frac{\sigma}{\sqrt{n}}\right). $$Step 4 — Substitute and calculate.
$$ \begin{aligned} 68 - 1.645\left(\frac{3}{\sqrt{36}}\right) &\leq \mu \leq 68 + 1.645\left(\frac{3}{\sqrt{36}}\right) \\ 68 - 0.8225 &\leq \mu \leq 68 + 0.8225 \\ 67.1775 &\leq \mu \leq 68.8225 \end{aligned} $$On a TI-83/83+/84/84+: press STAT, arrow over to TESTS, arrow down to 7:ZInterval, and press ENTER. Arrow to Stats and press ENTER. Enter 3 for \(\sigma\), 68 for \(\overline{x}\), 36 for \(n\), and .90 for C-level. Arrow down to Calculate and press ENTER. The confidence interval comes back as \((67.178, 68.822)\).
Step 5 — Interpret it. We estimate with 90% confidence that the true population mean exam score for all statistics students is between 67.18 and 68.82.
Answer: \((67.18, 68.82)\). Ninety percent of all confidence intervals constructed this way contain the true mean statistics exam score. If we built 100 of them, we would expect about 90 to contain the true population mean.
Try it in rāSHio
The whole interval comes out in one step. In rāSHio choose Stats → Z Stats, switch it to the confidence-interval option rather than the hypothesis test, and enter this example's summary values — 68 for the sample mean, 3 for the population standard deviation, 36 for the sample size, and 0.90 for the confidence level. It reports the same \((67.18, 68.82)\) the four lines of algebra above produce, which makes it a good way to check work rather than replace it.
Figure 7.1.4 — The rāSHio Z Stats panel: Stats → Z Stats. One panel covers both jobs a known \(\sigma\) allows — a confidence interval and a hypothesis test. The walkthrough runs the hypothesis-test side, with its own demonstration numbers; the note above uses the confidence-interval option and this example's numbers.
Table 7.1.1 shows a different random sampling of 20 cell phone models. Use this data to calculate a 93% confidence interval for the true mean SAR for cell phones certified for use in the United States. As before, assume that the population standard deviation is \(\sigma = 0.337\).
| 1.48 | 1.53 |
| 0.8 | 0.68 |
| 1.15 | 1.4 |
| 1.36 | 1.24 |
| 0.77 | 0.57 |
| 0.462 | 0.2 |
| 1.36 | 0.51 |
| 1.39 | 0.3 |
| 1.3 | 0.73 |
| 0.7 | 0.869 |
Solution — twenty phones at 93%
Step 1 — Find the point estimate. The 20 values total 18.801, so
$$ \overline{x} = \frac{18.801}{20} = 0.9401. $$Step 2 — Find the critical value. With \(CL = 0.93\), we get \(\alpha = 0.07\) and \(\frac{\alpha}{2} = 0.035\). The area to the left is \(1 - 0.035 = 0.965\), so invNorm(0.965, 0, 1) gives
Step 3 — Calculate the error bound.
$$ EBM = 1.812\left(\frac{0.337}{\sqrt{20}}\right) = 1.812(0.0754) = 0.1366 $$Step 4 — Build the interval.
$$ (0.9401 - 0.1366,\ 0.9401 + 0.1366) = (0.8035,\ 1.0766) $$Answer: \((0.8035, 1.0766)\) watts per kilogram. We estimate with 93% confidence that the true mean SAR for certified cell phones is in this range.
Try it in rāSHio
This example hands you \(\overline{x} = 1.024\) without showing the addition. To get it yourself, open rāSHio, paste the thirty SAR values from Table 7.1.2 in through File → Delimited List... — it takes the spaces and line breaks as they stand — then choose Stats → Summary Statistics and read the mean straight off the panel.
Figure 7.1.5 — The rāSHio Summary Statistics panel: Stats → Summary Statistics. It reports the mean, median, standard deviation and spread of a column at once. The walkthrough loads its own demonstration data through File → Load Sample Data rather than the Delimited List... paste the note describes, and its numbers are not Table 7.1.1's — what it shows you is the Summary Statistics panel itself.
Notice the difference between the intervals in Example 7.1.3 and Try It Now 7.1.4. They are different for several reasons: they were calculated from different samples, the samples were different sizes, and the intervals were built for different confidence levels. Different does not mean contradictory — both intervals are honest reports from the data that produced them. Sorting out which of those three changes does what to the width of an interval is the subject of the next part of this section.
The Specific Absorption Rate (SAR) for a cell phone measures the amount of radio frequency (RF) energy absorbed by the user's body when using the handset. Every cell phone emits RF energy, and different phone models have different SAR measures. To receive certification from the Federal Communications Commission (FCC) for sale in the United States, the SAR level for a cell phone must be no more than 1.6 watts per kilogram. Table 7.1.2 shows the highest SAR level for a random selection of cell phone models as measured by the FCC.
| 1.11 | 1.36 | 0.74 |
| 1.48 | 1.34 | 0.5 |
| 1.43 | 1.18 | 0.4 |
| 1.3 | 1.3 | 0.867 |
| 1.09 | 1.26 | 0.68 |
| 0.455 | 1.29 | 0.51 |
| 1.41 | 0.36 | 1.13 |
| 0.82 | 0.52 | 0.3 |
| 0.78 | 1.6 | 1.48 |
| 1.25 | 1.39 | 1.38 |
Find a 98% confidence interval for the true (population) mean of the Specific Absorption Rates for cell phones. Assume that the population standard deviation is \(\sigma = 0.337\).
Solution — cell phone radiation at 98%
Step 1 — Find the point estimate. Averaging the 30 values in Table 7.1.2 gives
$$ \overline{x} = 1.024. $$Step 2 — Find the critical value. We are building a 98% confidence interval, so \(CL = 0.98\) and
$$ \alpha = 1 - CL = 1 - 0.98 = 0.02, \qquad \frac{\alpha}{2} = 0.01. $$We need \(z_{0.01}\), the value with an area of 0.01 under the normal density curve to its right and 0.99 to its left. A calculator, a computer, or a standard normal probability table gives \(z_{0.01} = 2.326\).
Figure 7.1.6 - Locating z(0.01) for a 98% confidence interval: 0.01 sits in the shaded right tail and 0.99 lies to the left.
Step 3 — Calculate the error bound.
$$ EBM = (z_{0.01})\frac{\sigma}{\sqrt{n}} = (2.326)\frac{0.337}{\sqrt{30}} = 0.1431 $$Step 4 — Build the interval by finding \(\overline{x} \pm EBM\).
$$ \overline{x} - EBM = 1.024 - 0.1431 = 0.8809 $$ $$ \overline{x} + EBM = 1.024 + 0.1431 = 1.1671 $$On a TI-83/83+/84/84+: press STAT, arrow over to TESTS, arrow down to 7:ZInterval, press ENTER, arrow to Stats and press ENTER. Enter \(\sigma\): 0.337, \(\overline{x}\): 1.024, \(n\): 30, C-level: 0.98. Arrow down to Calculate and press ENTER. The confidence interval comes back as \((0.881, 1.167)\).
Step 5 — Interpret it. We estimate with 98% confidence that the true SAR mean for the population of cell phones in the United States is between 0.8809 and 1.1671 watts per kilogram.
Answer: \((0.8809, 1.1671)\) watts per kilogram.
7.1.4 Changing the Confidence Level or Sample Size
Two dials control the width of a confidence interval, and each one pulls in a predictable direction. Raising the confidence level widens the interval, because being surer means reaching further out. Raising the sample size narrows it, because more data means a tighter sampling distribution. The next two examples change one dial at a time, holding everything else fixed.
Refer back to the pizza-delivery exercise. The population standard deviation is six minutes and the sample mean delivery time is 36 minutes. Use a sample size of 20. Find a 95% confidence interval estimate for the true mean pizza delivery time.
Solution — pizza delivery at 95%
Step 1 — Find the critical value. For \(CL = 0.95\), \(\frac{\alpha}{2} = 0.025\) and \(z_{0.025} = 1.96\).
Step 2 — Calculate the error bound.
$$ EBM = 1.96\left(\frac{6}{\sqrt{20}}\right) = 1.96(1.3416) = 2.6296 $$Step 3 — Build the interval.
$$ (36 - 2.6296,\ 36 + 2.6296) = (33.37,\ 38.63) $$Answer: \((33.37, 38.63)\) minutes. It is wider than the 90% interval in Try It Now 7.1.3 for two reasons at once — a higher confidence level and a smaller sample.
Suppose we change the problem in Example 7.1.2 by using a 95% confidence level instead of 90%. Find a 95% confidence interval for the true (population) mean statistics exam score.
Solution — the same exam data at 95%
Step 1 — Collect what we know. Nothing about the data changed: \(\overline{x} = 68\), \(\sigma = 3\), \(n = 36\). Only the confidence level moved, to \(CL = 0.95\).
Step 2 — Find the new critical value.
$$ \alpha = 1 - CL = 1 - 0.95 = 0.05, \qquad \frac{\alpha}{2} = 0.025, \qquad z_{\frac{\alpha}{2}} = z_{0.025} $$The area to the right of \(z_{0.025}\) is 0.025 and the area to the left is \(1 - 0.025 = 0.975\), so
$$ z_{\frac{\alpha}{2}} = z_{0.025} = 1.96 $$using invNorm(0.975, 0, 1).
Figure 7.1.7 - The central 95% of a normal curve, bounded by z = -1.96 and 1.96, with 0.025 left in each tail.
Step 3 — Calculate the error bound.
$$ EBM = \left(z_{\frac{\alpha}{2}}\right)\left(\frac{\sigma}{\sqrt{n}}\right) = (1.96)\left(\frac{3}{\sqrt{36}}\right) = 0.98 $$Step 4 — Build the interval.
$$ \overline{x} - EBM = 68 - 0.98 = 67.02 $$ $$ \overline{x} + EBM = 68 + 0.98 = 68.98 $$Step 5 — Interpret it and compare. We estimate with 95% confidence that the true population mean for all statistics exam scores is between 67.02 and 68.98. Ninety-five percent of all confidence intervals built this way contain the true value.
The 90% interval was \((67.18, 68.82)\); the 95% interval is \((67.02, 68.98)\). The 95% interval is wider, and the error bound grew from 0.8225 to 0.98. That is what has to happen: capturing an area of 0.95 takes more room than capturing an area of 0.90, so to be more confident that the interval really does contain \(\mu\), the interval has to be wider.
Answer: \((67.02, 68.98)\), which is wider than the 90% interval.
Summary: effect of changing the confidence level.
- Increasing the confidence level increases the error bound, making the confidence interval wider.
- Decreasing the confidence level decreases the error bound, making the confidence interval narrower.
Refer back to the pizza-delivery exercise. The mean delivery time is 36 minutes and the population standard deviation is six minutes. Assume the sample size is changed to 50 restaurants with the same sample mean. Find a 90% confidence interval estimate for the population mean delivery time.
Solution — fifty restaurants at 90%
Step 1 — Find the critical value. For \(CL = 0.90\), \(z_{0.05} = 1.645\).
Step 2 — Calculate the error bound.
$$ EBM = 1.645\left(\frac{6}{\sqrt{50}}\right) = 1.645(0.8485) = 1.3958 $$Step 3 — Build the interval.
$$ (36 - 1.3958,\ 36 + 1.3958) = (34.60,\ 37.40) $$Answer: \((34.60, 37.40)\) minutes. Compare this to the \((34.13, 37.87)\) we got from 28 restaurants at the same confidence level: more data, narrower interval.
Suppose we change the problem in Example 7.1.2 again, this time leaving everything the same except the sample size. Use the original 90% confidence level. What happens to the error bound and the confidence interval if we increase the sample size to \(n = 100\) instead of 36? What happens if we decrease it to \(n = 25\)?
We know \(\overline{x} = 68\), \(\sigma = 3\), \(CL = 0.90\), and \(z_{\frac{\alpha}{2}} = z_{0.05} = 1.645\).
Solution — two new sample sizes
Step 1 — Increase the sample size to \(n = 100\).
$$ EBM = \left(z_{\frac{\alpha}{2}}\right)\left(\frac{\sigma}{\sqrt{n}}\right) = (1.645)\left(\frac{3}{\sqrt{100}}\right) = 0.4935 $$The error bound dropped from 0.8225 to 0.4935, so the interval got narrower.
Step 2 — Decrease the sample size to \(n = 25\).
$$ EBM = (1.645)\left(\frac{3}{\sqrt{25}}\right) = 0.987 $$The error bound rose from 0.8225 to 0.987, so the interval got wider.
Answer: With \(n = 100\), \(EBM = 0.4935\) and the interval narrows. With \(n = 25\), \(EBM = 0.987\) and the interval widens. Note that nearly tripling the sample size from 36 to 100 only cut the error bound by about 40% — the \(\sqrt{n}\) makes precision expensive.
Summary: effect of changing the sample size.
- Increasing the sample size causes the error bound to decrease, making the confidence interval narrower.
- Decreasing the sample size causes the error bound to increase, making the confidence interval wider.
7.1.5 Working Backwards to Find the Error Bound or Sample Mean
So far we have gone forwards: find the sample mean, calculate the error bound, build the interval. Published research often runs the other way. A study reports its confidence interval and nothing else, and we want to recover the two ingredients that went into it. Both are easy to get back, because an interval is symmetric around its centre — the sample mean sits exactly in the middle, and the error bound is exactly half the width.
Finding the error bound. Either subtract the sample mean from the upper value of the interval, or subtract the lower value from the upper value and divide the difference by two.
Finding the sample mean. Either subtract the error bound from the upper value of the interval, or average the upper and lower endpoints.
Each quantity has two routes, so pick whichever one matches the information you already have. If the study printed the sample mean, one subtraction gets you the error bound. If it printed nothing but the interval, the halving and the averaging get you both.
Suppose we know that a confidence interval is \((42.12, 47.88)\). Find the error bound and the sample mean.
Solution — working backwards from an interval
Step 1 — Find the error bound by halving the width of the interval.
$$ EBM = \frac{47.88 - 42.12}{2} = \frac{5.76}{2} = 2.88 $$Step 2 — Find the sample mean by averaging the endpoints.
$$ \overline{x} = \frac{42.12 + 47.88}{2} = \frac{90}{2} = 45 $$Answer: \(EBM = 2.88\) and \(\overline{x} = 45\).
Suppose we know that a confidence interval is \((67.18, 68.82)\) and we want to find the error bound. We may know that the sample mean is 68, or our source may have given only the confidence interval without telling us the sample mean.
Solution — recovering the error bound and the mean
Step 1 — Calculate the error bound, both ways.
If we know the sample mean is 68:
$$ EBM = 68.82 - 68 = 0.82 $$If we do not know the sample mean:
$$ EBM = \frac{68.82 - 67.18}{2} = \frac{1.64}{2} = 0.82 $$Step 2 — Calculate the sample mean, both ways.
If we know the error bound is 0.82:
$$ \overline{x} = 68.82 - 0.82 = 68 $$If we do not know the error bound:
$$ \overline{x} = \frac{67.18 + 68.82}{2} = 68 $$Answer: \(EBM = 0.82\) and \(\overline{x} = 68\). Both routes to each answer agree, which is a useful check that the interval was read correctly.
7.1.6 Calculating the Sample Size
To estimate a population mean to within an error bound \(EBM\) at a given confidence level, when the population standard deviation \(\sigma\) is known, the required sample size is
$$ n = \frac{z^2\sigma^2}{EBM^2}, $$where \(z\) is \(z_{\frac{\alpha}{2}}\) for the desired confidence level. Always round the result up to the next whole number.
The \(EBM\) is squared in the denominator, so halving the error bound multiplies the required sample size by four. Doubling your precision costs four times the data.
The rounding rule is not a convention, it is arithmetic. Rounding down would leave the sample slightly too small, which makes the error bound slightly larger than the one the study promised. Rounding up costs one extra observation and keeps the promise.
Researchers planning a study usually work in the opposite direction from everything above. They decide in advance how precise they need to be and how confident they want to be, and then they ask how much data that will take. Since the error bound formula already links precision, confidence, and sample size, solving it for \(n\) answers the question.
The error bound formula for a population mean when the population standard deviation is known is
$$ EBM = \left(z_{\frac{\alpha}{2}}\right)\left(\frac{\sigma}{\sqrt{n}}\right). $$Solving that for \(n\) gives the sample size formula.
Definition 7.1.5 - Precision is bought at a square-rate: quadrupling the sample size only halves the error bound.
The population standard deviation for the height of high school basketball players is three inches. If we want to be 95% confident that the sample mean height is within one inch of the true population mean height, how many randomly selected students must be surveyed?
Solution — the sample size for one inch
Step 1 — Collect what we know. \(\sigma = 3\), \(EBM = 1\), and for 95% confidence \(z = 1.96\).
Step 2 — Substitute into the formula.
$$ n = \frac{z^2\sigma^2}{EBM^2} = \frac{(1.96)^2(3)^2}{1^2} = (3.8416)(9) = 34.57 $$Step 3 — Round up.
$$ n = 35 $$Answer: 35 students must be surveyed.
The population standard deviation for the age of Foothill College students is 15 years. If we want to be 95% confident that the sample mean age is within two years of the true population mean age of Foothill College students, how many randomly selected Foothill College students must be surveyed?
Solution — how many students to survey
Step 1 — Collect what we know. From the problem, \(\sigma = 15\) and \(EBM = 2\). The confidence level is 95%, so \(z = z_{0.025} = 1.96\).
Step 2 — Substitute into the sample size formula.
$$ n = \frac{z^2\sigma^2}{EBM^2} = \frac{(1.96)^2(15)^2}{2^2} = \frac{(3.8416)(225)}{4} = 216.09 $$Step 3 — Round up. Use \(n = 217\). Always round up to the next higher integer so the sample is large enough to deliver the promised error bound.
Answer: 217 Foothill College students should be surveyed to be 95% confident that the sample mean age is within two years of the true population mean age.
Problem Set 7.1
Use the following information to answer the next five exercises: The standard deviation of the weights of elephants is known to be approximately 15 pounds. We wish to construct a 95% confidence interval for the mean weight of newborn elephant calves. Fifty newborn elephants are weighed. The sample mean is 244 pounds. The sample standard deviation is 11 pounds.
Problem 1. Identify the following:
a) \(\overline{x} =\) ____
b) \(\sigma =\) ____
c) \(n =\) ____
Problem 1 Solution
Step 1 — Read the point estimate off the problem. The sample mean is the average of the 50 weights that were actually measured.
$$ \overline{x} = 244 $$Step 2 — Read the population standard deviation. The problem states that the standard deviation of elephant weights is known to be approximately 15 pounds. Because it describes the whole population and not just this sample, it is \(\sigma\), not \(s\).
$$ \sigma = 15 $$Step 3 — Read the sample size. Fifty newborn elephants were weighed.
$$ n = 50 $$The sample standard deviation of 11 pounds is given as a distractor. When \(\sigma\) is known, this section's method never uses \(s\).
Answer: a) \(\overline{x} = 244\) b) \(\sigma = 15\) c) \(n = 50\)
Problem 2. In words, define the random variables \(X\) and \(\overline{X}\).
Problem 2 Solution
Step 1 — Define \(X\), the individual variable. A capital \(X\) with no bar always refers to one member of the population, measured once.
\(X\) = the weight, in pounds, of one randomly selected newborn elephant calf.
Step 2 — Define \(\overline{X}\), the sample mean. A bar over the letter means we averaged a whole sample, so \(\overline{X}\) is a random variable in its own right — it changes every time a new sample of 50 is drawn.
\(\overline{X}\) = the mean weight, in pounds, of a sample of 50 randomly selected newborn elephant calves.
Step 3 — Say why the distinction matters. The confidence interval estimates the population mean \(\mu\), and the tool that gets us there is the distribution of \(\overline{X}\), not of \(X\). Individual calves vary by a lot more than sample means of 50 calves do.
Answer: \(X\) is the weight of one newborn elephant calf; \(\overline{X}\) is the mean weight of a sample of 50 newborn elephant calves.
Problem 3. Which distribution should you use for this problem?
Problem 3 Solution
Step 1 — Check which distribution the conditions point to. The population standard deviation \(\sigma = 15\) is known, and the sample size \(n = 50\) is greater than 30, so the central limit theorem applies and \(\overline{X}\) is approximately normal.
Step 2 — Write down its two parameters. The centre is \(\mu\) and the spread is the standard error \(\frac{\sigma}{\sqrt{n}}\).
$$ \frac{\sigma}{\sqrt{n}} = \frac{15}{\sqrt{50}} = 2.1213 $$Step 3 — State the distribution.
$$ \overline{X} \sim N\!\left(244, \frac{15}{\sqrt{50}}\right) = N(244, 2.1213) $$Answer: Use the normal distribution \(\overline{X} \sim N\!\left(244, \frac{15}{\sqrt{50}}\right)\), because \(\sigma\) is known and \(n > 30\).
Problem 4. Construct a 95% confidence interval for the population mean weight of newborn elephants. State the confidence interval, sketch the graph, and calculate the error bound.
Problem 4 Solution
Step 1 — Find the critical value for 95% confidence. With \(CL = 0.95\), we get \(\alpha = 0.05\) and \(\frac{\alpha}{2} = 0.025\), so \(z_{0.025} = 1.96\).
Step 2 — Calculate the error bound.
$$ EBM = z_{\frac{\alpha}{2}}\left(\frac{\sigma}{\sqrt{n}}\right) = 1.96\left(\frac{15}{\sqrt{50}}\right) = 1.96(2.1213) = 4.16 $$Step 3 — Build the interval.
$$ (\overline{x} - EBM,\ \overline{x} + EBM) = (244 - 4.16,\ 244 + 4.16) = (239.84,\ 248.16) $$Step 4 — Sketch the graph. Draw a normal curve centred at 244. Mark 239.84 and 248.16 on the horizontal axis and shade the region between them; that shaded area is 0.95. Each unshaded tail holds an area of 0.025.
Step 5 — Interpret it. We estimate with 95% confidence that the true mean weight of all newborn elephant calves is between 239.84 and 248.16 pounds.
Answer: \((239.84, 248.16)\) pounds, with \(EBM = 4.16\) pounds.
Problem 5. What will happen to the confidence interval obtained, if 500 newborn elephants are weighed instead of 50? Why?
Problem 5 Solution
Step 1 — Find what changes in the formula. Only \(n\) moves, from 50 to 500. It sits under a square root in the denominator of the standard error, so a larger \(n\) makes the standard error smaller.
$$ \frac{15}{\sqrt{500}} = 0.6708 \quad \text{versus} \quad \frac{15}{\sqrt{50}} = 2.1213 $$Step 2 — Recalculate the error bound at the same 95% confidence level.
$$ EBM = 1.96(0.6708) = 1.31 $$Step 3 — Compare the intervals. The new interval is \((244 - 1.31,\ 244 + 1.31) = (242.69,\ 245.31)\), which is much narrower than \((239.84, 248.16)\).
Answer: The confidence interval gets narrower, dropping from a width of about 8.3 pounds to about 2.6 pounds. More data means less variability in the sample mean, so a tighter range still captures \(\mu\) with the same 95% confidence.
Problem 6. Among various ethnic groups, the standard deviation of heights is known to be approximately three inches. We wish to construct a 95% confidence interval for the mean height of male Swedes. Forty-eight male Swedes are surveyed. The sample mean is 71 inches. The sample standard deviation is 2.8 inches.
a) Identify the following: i. \(\overline{x} =\) ____ ii. \(\sigma =\) ____ iii. \(n =\) ____
b) In words, define the random variables \(X\) and \(\overline{X}\).
c) Which distribution should you use for this problem? Explain your choice.
d) Construct a 95% confidence interval for the population mean height of male Swedes. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound.
e) What will happen to the level of confidence obtained if 1,000 male Swedes are surveyed instead of 48? Why?
Problem 6 Solution
Step 1 — Read off the given values (part a).
i. \(\overline{x} = 71\) inches ii. \(\sigma = 3\) inches iii. \(n = 48\)
The sample standard deviation of 2.8 inches is not used, because \(\sigma\) is known.
Step 2 — Define the variables (part b). \(X\) is the height, in inches, of one randomly selected male Swede. \(\overline{X}\) is the mean height of a sample of 48 randomly selected male Swedes.
Step 3 — Choose the distribution (part c). Use the normal distribution. We know the standard deviation for the population, and the sample size 48 is greater than 30, so the central limit theorem gives
$$ \overline{X} \sim N\!\left(71, \frac{3}{\sqrt{48}}\right) = N(71, 0.4330). $$Step 4 — Build the 95% confidence interval (part d). For 95% confidence, \(z_{0.025} = 1.96\).
$$ EBM = 1.96\left(\frac{3}{\sqrt{48}}\right) = 1.96(0.4330) = 0.849 $$ $$ (71 - 0.849,\ 71 + 0.849) = (70.151,\ 71.849) $$For the sketch, draw a normal curve centred at 71, mark 70.151 and 71.849, and shade between them to represent an area of 0.95.
Step 5 — Answer the sample-size question (part e). Surveying 1,000 male Swedes instead of 48 would shrink the standard error from 0.4330 to \(\frac{3}{\sqrt{1000}} = 0.0949\). Holding the confidence level at 95%, the interval would become far narrower. Turned around: if we insisted on keeping the same error bound of 0.849, the extra data would let us report a much higher confidence level for that same width.
Answer: a) 71, 3, 48. b) \(X\) is one male Swede's height; \(\overline{X}\) is the mean height of 48 of them. c) Normal, because \(\sigma\) is known and \(n > 30\). d) \((70.151, 71.849)\) inches with \(EBM = 0.849\). e) The interval narrows sharply, because increasing \(n\) decreases variability in the sample mean.
Problem 7. Rosa Delgado randomly picked announcements for 84 upcoming engineering conferences from her stack of IEEE Spectrum magazines. The mean length of the conferences was 3.94 days, with a standard deviation of 1.28 days. Assume the underlying population is normal.
a) In words, define the random variables \(X\) and \(\overline{X}\).
b) Which distribution should you use for this problem? Explain your choice.
c) Construct a 95% confidence interval for the population mean length of engineering conferences. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound.
Problem 7 Solution
Step 1 — Define the variables (part a). \(X\) is the length, in days, of one randomly selected engineering conference. \(\overline{X}\) is the mean length of a sample of 84 randomly selected engineering conferences.
Step 2 — Choose the distribution (part b). The underlying population is assumed normal, and the standard deviation of 1.28 days is being treated as the population standard deviation, so use the normal distribution.
$$ \overline{X} \sim N\!\left(3.94, \frac{1.28}{\sqrt{84}}\right) = N(3.94, 0.1397) $$Step 3 — Find the critical value for 95% confidence (part c). \(\alpha = 0.05\), \(\frac{\alpha}{2} = 0.025\), so \(z_{0.025} = 1.96\).
Step 4 — Calculate the error bound.
$$ EBM = 1.96\left(\frac{1.28}{\sqrt{84}}\right) = 1.96(0.1397) = 0.2737 $$Step 5 — Build the interval and sketch it.
$$ (3.94 - 0.2737,\ 3.94 + 0.2737) = (3.67,\ 4.21) $$Draw a normal curve centred at 3.94, mark 3.67 and 4.21, and shade between them for an area of 0.95.
Answer: a) \(X\) is one conference's length in days; \(\overline{X}\) is the mean length of 84 conferences. b) Normal, \(N\!\left(3.94, \frac{1.28}{\sqrt{84}}\right)\). c) \((3.67, 4.21)\) days with \(EBM = 0.2737\) days.
Problem 8. Suppose that an accounting firm does a study to determine the time needed to complete one person's tax forms. It randomly surveys 100 people. The sample mean is 23.6 hours. There is a known standard deviation of 7.0 hours. The population distribution is assumed to be normal.
a) Identify the following: i. \(\overline{x} =\) ____ ii. \(\sigma =\) ____ iii. \(n =\) ____
b) In words, define the random variables \(X\) and \(\overline{X}\).
c) Which distribution should you use for this problem? Explain your choice.
d) Construct a 90% confidence interval for the population mean time to complete the tax forms. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound.
e) If the firm wished to increase its level of confidence and keep the error bound the same by taking another survey, what changes should it make?
f) If the firm did another survey, kept the error bound the same, and only surveyed 49 people, what would happen to the level of confidence? Why?
g) Suppose that the firm decided that it needed to be at least 96% confident of the population mean length of time to within one hour. How would the number of people the firm surveys change? Why?
Problem 8 Solution
Step 1 — Read off the given values (part a).
i. \(\overline{x} = 23.6\) hours ii. \(\sigma = 7\) hours iii. \(n = 100\)
Step 2 — Define the variables (part b). \(X\) is the time needed to complete one person's tax forms. \(\overline{X}\) is the mean time to complete tax forms for a sample of 100 people.
Step 3 — Choose the distribution (part c). The population standard deviation is known and the population is assumed normal, so
$$ \overline{X} \sim N\!\left(23.6, \frac{7}{\sqrt{100}}\right) = N(23.6, 0.7). $$Step 4 — Build the 90% confidence interval (part d). For 90% confidence, \(\alpha = 0.10\) and \(\frac{\alpha}{2} = 0.05\), so \(z_{0.05} = 1.645\).
$$ EBM = 1.645(0.7) = 1.1515 $$ $$ (23.6 - 1.1515,\ 23.6 + 1.1515) = (22.45,\ 24.75) $$Sketch a normal curve centred at 23.6, mark 22.45 and 24.75, and shade between them for an area of 0.90.
Step 5 — Raise the confidence, keep the error bound (part e). A higher confidence level needs a larger \(z\), which would push the error bound up. The only way to hold \(EBM\) fixed while \(z\) grows is to shrink the standard error, and the only lever for that is \(n\). The firm must survey more people.
Step 6 — Same error bound, only 49 people (part f). With \(n = 49\), the standard error becomes \(\frac{7}{\sqrt{49}} = 1\). Holding \(EBM = 1.1515\) means \(z = \frac{1.1515}{1} = 1.15\), which corresponds to a confidence level of roughly 75%. The level of confidence drops, because a smaller sample carries more variability and the same-width interval now catches \(\mu\) less often.
Step 7 — Sample size for 96% confidence within one hour (part g). For \(CL = 0.96\), \(\alpha = 0.04\) and \(\frac{\alpha}{2} = 0.02\), so \(z_{0.02} = 2.054\).
$$ n = \frac{z^2\sigma^2}{EBM^2} = \frac{(2.054)^2(7)^2}{1^2} = \frac{(4.2189)(49)}{1} = 206.7 $$Rounding up gives \(n = 207\). The firm must survey more people than the 100 it started with, because it is asking for both a higher confidence level and a tighter error bound than the original survey delivered.
Answer: a) 23.6, 7, 100. b) \(X\) is one person's completion time; \(\overline{X}\) is the mean for 100 people. c) \(N\!\left(23.6, \frac{7}{\sqrt{100}}\right)\). d) \((22.45, 24.75)\) hours with \(EBM = 1.1515\). e) Increase the sample size. f) The confidence level falls to about 75%. g) Survey 207 people.
Problem 9. A sample of 16 small bags of the same brand of candies was selected. Assume that the population distribution of bag weights is normal. The weight of each bag was then recorded. The mean weight was two ounces with a standard deviation of 0.12 ounces. The population standard deviation is known to be 0.1 ounce.
a) Identify the following: i. \(\overline{x} =\) ____ ii. \(\sigma =\) ____ iii. \(s_x =\) ____
b) In words, define the random variable \(X\).
c) In words, define the random variable \(\overline{X}\).
d) Which distribution should you use for this problem? Explain your choice.
e) Construct a 90% confidence interval for the population mean weight of the candies. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound.
f) Construct a 98% confidence interval for the population mean weight of the candies. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound.
g) In complete sentences, explain why the confidence interval in part f is larger than the confidence interval in part e.
h) In complete sentences, give an interpretation of what the interval in part f means.
Problem 9 Solution
Step 1 — Read off the given values (part a).
i. \(\overline{x} = 2\) ounces ii. \(\sigma = 0.1\) ounce iii. \(s_x = 0.12\) ounces
Step 2 — Define \(X\) (part b). \(X\) is the weight, in ounces, of one randomly selected small bag of this brand of candies.
Step 3 — Define \(\overline{X}\) (part c). \(\overline{X}\) is the mean weight of a sample of 16 randomly selected small bags.
Step 4 — Choose the distribution (part d). The sample size is only 16, but the problem states that the population of bag weights is normal, and \(\sigma\) is known. Both conditions together make the normal distribution the right choice even at \(n < 30\).
$$ \overline{X} \sim N\!\left(2, \frac{0.1}{\sqrt{16}}\right) = N(2, 0.025) $$Step 5 — Build the 90% interval (part e). \(z_{0.05} = 1.645\).
$$ EBM = 1.645(0.025) = 0.0411 $$ $$ (2 - 0.0411,\ 2 + 0.0411) = (1.9589,\ 2.0411) $$Step 6 — Build the 98% interval (part f). \(\alpha = 0.02\), \(\frac{\alpha}{2} = 0.01\), so \(z_{0.01} = 2.326\).
$$ EBM = 2.326(0.025) = 0.0582 $$ $$ (2 - 0.0582,\ 2 + 0.0582) = (1.9418,\ 2.0582) $$For both sketches, draw a normal curve centred at 2 and shade between the endpoints; the 98% shading is visibly wider.
Step 7 — Explain the difference (part g). The 98% interval is larger because a higher confidence level requires a larger critical value — 2.326 instead of 1.645. The standard error did not change at all, so the entire increase in width comes from the \(z\). Being surer means reaching further out into the tails, and reaching further out means a wider interval.
Step 8 — Interpret the 98% interval (part h). We estimate with 98% confidence that the true mean weight of all small bags of this brand of candies is between 1.9418 and 2.0582 ounces.
Answer: a) 2, 0.1, 0.12. b) \(X\) is one bag's weight. c) \(\overline{X}\) is the mean weight of 16 bags. d) \(N(2, 0.025)\), valid because the population is normal and \(\sigma\) is known. e) \((1.9589, 2.0411)\), \(EBM = 0.0411\). f) \((1.9418, 2.0582)\), \(EBM = 0.0582\). g) The larger \(z\) for 98% widens the interval. h) See Step 8.
Problem 10. Marcus Redcloud, a camp director, is interested in the mean number of letters each child sends during their camp session. He knows the population standard deviation is 2.5. His survey of 20 campers has a sample mean of 7.9 with a sample standard deviation of 2.8.
a) Identify the following: i. \(\overline{x} =\) ____ ii. \(\sigma =\) ____ iii. \(n =\) ____
b) Define the random variables \(X\) and \(\overline{X}\) in words.
c) Which distribution should you use for this problem? Explain your choice.
d) Construct a 90% confidence interval for the population mean number of letters campers send home. i. State the confidence interval. ii. Sketch the graph. iii. Calculate the error bound.
e) What will happen to the error bound and confidence interval if 500 campers are surveyed? Why?
Problem 10 Solution
Step 1 — Read off the given values (part a).
i. \(\overline{x} = 7.9\) letters ii. \(\sigma = 2.5\) letters iii. \(n = 20\)
The sample standard deviation of 2.8 is not needed, because \(\sigma\) is known.
Step 2 — Define the variables (part b). \(X\) is the number of letters one camper sends home during the session. \(\overline{X}\) is the mean number of letters sent home by a sample of 20 campers.
Step 3 — Choose the distribution (part c). The population standard deviation is known, so use the normal distribution with the standard error in place of \(\sigma\).
$$ \overline{X} \sim N\!\left(7.9, \frac{2.5}{\sqrt{20}}\right) = N(7.9, 0.5590) $$Step 4 — Build the 90% confidence interval (part d). \(z_{0.05} = 1.645\).
$$ EBM = 1.645\left(\frac{2.5}{\sqrt{20}}\right) = 1.645(0.5590) = 0.92 $$ $$ (7.9 - 0.92,\ 7.9 + 0.92) = (6.98,\ 8.82) $$Sketch a normal curve centred at 7.9, mark 6.98 and 8.82, and shade between them for an area of 0.90.
Step 5 — Survey 500 campers instead (part e). The standard error would drop from 0.5590 to \(\frac{2.5}{\sqrt{500}} = 0.1118\), so the error bound would drop to \(1.645(0.1118) = 0.18\) and the interval would narrow to about \((7.72, 8.08)\).
Answer: a) 7.9, 2.5, 20. b) \(X\) is one camper's letter count; \(\overline{X}\) is the mean for 20 campers. c) \(N\!\left(7.9, \frac{2.5}{\sqrt{20}}\right)\). d) \((6.98, 8.82)\) with \(EBM = 0.92\). e) Both the error bound and the interval shrink, because a larger sample lowers the standard error.
Problem 11. What is meant by the term "90% confident" when constructing a confidence interval for a mean?
a) If we took repeated samples, approximately 90% of the samples would produce the same confidence interval.
b) If we took repeated samples, approximately 90% of the confidence intervals calculated from those samples would contain the sample mean.
c) If we took repeated samples, approximately 90% of the confidence intervals calculated from those samples would contain the true value of the population mean.
d) If we took repeated samples, the sample mean would equal the population mean in approximately 90% of the samples.
Problem 11 Solution
Step 1 — Recall what the confidence level describes. The confidence level is a property of the method, not of any one interval. It reports the percent of intervals built this way that would contain the true population parameter if the whole sampling process were repeated many times.
Step 2 — Test each option against that.
Option a is wrong: repeated samples produce different sample means, so they produce different intervals, not the same one.
Option b is wrong in a subtle but important way: every interval contains its own sample mean by construction, since the interval is built as \(\overline{x} \pm EBM\). That statement would be true 100% of the time, which makes it useless.
Option c matches the definition exactly.
Option d is wrong: a sample mean almost never lands exactly on the population mean, whatever the confidence level.
Answer: c. If we took repeated samples, approximately 90% of the confidence intervals calculated from those samples would contain the true value of the population mean.
Problem 12. The Federal Election Commission collects information about campaign contributions and disbursements for candidates and political committees each election cycle. During a certain campaign season, there were 1,619 candidates for the House of Representatives across the United States who received contributions from individuals. Table 7.1.3 shows the total receipts from individuals for a random selection of 40 House candidates rounded to the nearest $100. The standard deviation for this data to the nearest hundred is \(\sigma = \$909,200\).
| $3,600 | $1,243,900 | $10,900 | $385,200 | $581,500 |
| $7,400 | $2,900 | $400 | $3,714,500 | $632,500 |
| $391,000 | $467,400 | $56,800 | $5,800 | $405,200 |
| $733,200 | $8,000 | $468,700 | $75,200 | $41,000 |
| $13,300 | $9,500 | $953,800 | $1,113,500 | $1,109,300 |
| $353,900 | $986,100 | $88,600 | $378,200 | $13,200 |
| $3,800 | $745,100 | $5,800 | $3,072,100 | $1,626,700 |
| $512,900 | $2,309,200 | $6,600 | $202,400 | $15,800 |
a) Find the point estimate for the population mean.
b) Using 95% confidence, calculate the error bound.
c) Create a 95% confidence interval for the mean total individual contributions.
d) Interpret the confidence interval in the context of the problem.
Problem 12 Solution
Step 1 — Find the point estimate (part a). Add the 40 contribution totals in Table 7.1.3 and divide by 40. The values sum to $22,754,900, so
$$ \overline{x} = \frac{22{,}754{,}900}{40} = \$568,873. $$Step 2 — Find the critical value for 95% confidence (part b). \(\alpha = 1 - 0.95 = 0.05\) and \(\frac{\alpha}{2} = 0.025\), so \(z_{0.025} = 1.96\).
Step 3 — Calculate the error bound.
$$ EBM = z_{0.025}\frac{\sigma}{\sqrt{n}} = 1.96\frac{909{,}200}{\sqrt{40}} = 1.96(143{,}757) = \$281,764 $$Step 4 — Build the interval (part c).
$$ \overline{x} - EBM = 568{,}873 - 281{,}764 = \$287,109 $$ $$ \overline{x} + EBM = 568{,}873 + 281{,}764 = \$850,637 $$A calculator running ZInterval with \(\sigma = 909{,}200\), \(\overline{x} = 568{,}873\), \(n = 40\), and C-level 0.95 returns \((\$287,114, \$850,632)\). The small difference is rounding in the hand calculation, not a different answer.
Step 5 — Interpret it (part d). We estimate with 95% confidence that the mean amount of contributions received from individuals by all House candidates is between $287,109 and $850,637. The interval is enormously wide relative to its centre, which is the data telling us something real: contributions ranged from $400 to over $3.7 million, so a sample of 40 pins the mean down only loosely.
Answer: a) $568,873. b) $281,764. c) \((\$287,109, \$850,637)\). d) See Step 5.
Problem 13. The American Community Survey (ACS), part of the United States Census Bureau, conducts a yearly census similar to the one taken every ten years, but with a smaller percentage of participants. The recent survey estimates with 90% confidence that the mean household income in the U.S. falls between $69,720 and $69,922. Find the point estimate for mean U.S. household income and the error bound for mean U.S. household income.
Problem 13 Solution
Step 1 — Recognise which direction we are working. We are given the finished interval and asked for the two ingredients, so this is a working-backwards problem.
Step 2 — Find the point estimate by averaging the endpoints.
$$ \overline{x} = \frac{69{,}720 + 69{,}922}{2} = \frac{139{,}642}{2} = \$69,821 $$Step 3 — Find the error bound by halving the width.
$$ EBM = \frac{69{,}922 - 69{,}720}{2} = \frac{202}{2} = \$101 $$Step 4 — Check the two against each other. The point estimate plus the error bound should return the upper endpoint: \(69{,}821 + 101 = 69{,}922\). It does.
Answer: The point estimate for mean U.S. household income is $69,821, and the error bound is $101.
Problem 14. The average height of young adult males has a normal distribution with standard deviation of 2.5 inches. You want to estimate the mean height of students at your college or university to within one inch with 93% confidence. How many male students must you measure?
Problem 14 Solution
Step 1 — Collect what we know. The population standard deviation is \(\sigma = 2.5\) inches, the target error bound is \(EBM = 1\) inch, and the confidence level is \(CL = 0.93\).
Step 2 — Find the critical value. \(\alpha = 1 - 0.93 = 0.07\) and \(\frac{\alpha}{2} = 0.035\), so the area to the left is \(1 - 0.035 = 0.965\) and invNorm(0.965, 0, 1) gives
Step 3 — Substitute into the sample size formula.
$$ n = \frac{z^2\sigma^2}{EBM^2} = \frac{(1.812)^2(2.5)^2}{1^2} = \frac{(3.2833)(6.25)}{1} = 20.52 $$Step 4 — Round up. A sample of 20 would leave the error bound slightly above one inch, so round up to the next whole student.
$$ n = 21 $$Answer: You must measure 21 male students.
Key Terms
confidence interval (CI) — an interval estimate for an unknown population parameter, built as the point estimate plus or minus the error bound.
point estimate — a single value calculated from sample data used as the best guess for a population parameter; for a population mean it is \(\overline{x}\).
error bound for a population mean (EBM) — the distance from the point estimate to either endpoint of the interval; half the interval's width.
confidence level (CL) — the percent of intervals built by the same method that contain the true parameter when the sampling is repeated.
alpha (\(\alpha\)) — the probability that the interval misses the parameter, equal to \(1 - CL\), split evenly between the two tails.
critical value (\(z_{\frac{\alpha}{2}}\)) — the \(z\)-score with an area of \(\frac{\alpha}{2}\) to its right under the standard normal curve.
standard error of the mean — the standard deviation of the sampling distribution of \(\overline{X}\), equal to \(\frac{\sigma}{\sqrt{n}}\).